Question 15 Marks
Evaluate the following integrals:$\int\frac{2\text{x}+5}{\text{x}^2-\text{x}-2}\text{ dx}$
AnswerLet $\text{I}=\int\frac{2\text{x}+5}{\text{x}^2-\text{x}-2}\text{ dx}$
Let $2\text{x}+5=\lambda\frac{\text{d}}{\text{dx}}\big(\text{x}^2-\text{x}-2\big)+\mu$
$=\lambda(2\text{x}-1)+\mu$
$2\text{x}+5=(2\lambda)\text{x}-\lambda+\mu$
Comparing the coefficients of like power of x,
$2\lambda=2\Rightarrow\lambda=1$
$-\lambda+\mu=5\Rightarrow-1+\mu=5$
$\mu=6$
So, $\text{I}=\int\frac{(2\text{x}-1)+6}{\text{x}^2-\text{x}-2}\text{ dx}$
$\text{I}=\int\frac{(2\text{x}-1)}{\text{x}^2-\text{x}-2}\text{ dx}+6\int\frac{1}{\text{x}^2-2\text{x}\big(\frac{1}{2}\big)+\big(\frac{1}{2}\big)^2-\big(\frac{1}{2}\big)^2-2}\text{ dx}$
$\text{I}=\int\frac{2\text{x}-1}{\text{x}^2-\text{x}-2}\text{ dx}+6\int\frac{1}{\big(\text{x}-\frac{1}{2}\big)^2-\frac{9}{4}}\text{ dx}$
$\text{I}=\int\frac{2\text{x}-1}{\text{x}^2-\text{x}-2}\text{ dx}+6\int\frac{1}{\big(\text{x}-\frac{1}{2}\big)^2-\big(\frac{3}{2}\big)^2}\text{ dx}$
$\text{I}=\log\big|\text{x}^2-\text{x}-2\big|+\frac{6}{2\big(\frac{3}{2}\big)}\log\Bigg|\frac{\text{x}-\frac{1}{2}-\frac{3}{2}}{\text{x}-\frac{1}{2}+\frac{3}{2}}\Bigg|+\text{C}$ $\Big[\text{Since }\int\frac{1}{\text{x}^2+\text{a}^2}\text{ dx}=\frac{1}{2\text{a}}\log\Big|\frac{\text{x}-\text{a}}{\text{x}+\text{a}}\Big|+\text{C}\Big]$
$\text{I}=\log\big|\text{x}^2-\text{x}-2\big|+2\log\Big|\frac{\text{x}-2}{\text{x}+1}\Big|+\text{C}$
View full question & answer→Question 25 Marks
Evaluate the following integrals:
$\int(\text{x}-3)\sqrt{\text{x}^2+3\text{x}-18}\text{dx}$
AnswerConsider the integral $\text{I}=\int(\text{x}-3)\sqrt{\text{x}^2+3\text{x}-18}\text{dx}$Let us express $\text{x}-3=\lambda\frac{\text{d}}{\text{dx}}(\text{x}^2+3\text{x}-18)+\mu$
$\Rightarrow\text{x}-3=\lambda(2\text{x}+3)+\mu$
$\Rightarrow\text{x}-3=2\lambda\text{x}+3\lambda+\mu$
Comparing the co-efficients, we have,
$2\lambda=1\text{ and }3\lambda+\mu=-3$
$\Rightarrow\lambda=\frac{1}{2}\text{ and }3\times\frac{1}{2}+\mu=-3$
$\Rightarrow\lambda=\frac{1}{2}\text{ and }\mu-3-\frac{3}{2}$
$\Rightarrow\lambda=\frac{1}{2}\text{ and }\mu=-\frac{9}{2}$
Then
$\text{x}-3=\lambda(2\text{x}+3)+\mu$
Now the integral $\text{I}=\int(\text{x}-3)\sqrt{\text{x}^2+3\text{x}-18}\text{dx}$
$=\int\Big(\frac{1}{2}(2\text{x}+3)-\frac{9}{2}\Big)\sqrt{\text{x}^2+3\text{x}-18}\text{dx}$
$\text{I}=\frac{1}{2}\int(2\text{x}+3)\sqrt{\text{x}^2+3\text{x}-18}\text{dx}\\-\frac{9}{2}\int\sqrt{\text{x}^2+3\text{x}-18}\text{dx}$
$\Rightarrow\text{I}=\text{I}_1+\text{I}_2$
where, $\text{I}_1=\frac{1}{2}\int(2\text{x}+3)\sqrt{\text{x}^2+3\text{x}-18}\text{dx}$
and $\text{I}_2=-\frac{9}{2}\int\sqrt{\text{x}^2+3\text{x}-18}\text{dx}$
Let us consider the integral, $I_1:$
$\text{I}_1=\frac{1}{2}\int(2\text{x}+3)\sqrt{\text{x}^2+3\text{x}-18}\text{dx}$
Substituting, $\text{x}^2+3\text{x}-18=\text{t}$
$\Rightarrow(2\text{x}+3)\text{dx = dt}$
Thus,
$\text{I}_1=\frac{1}{2}\int\sqrt{\text{t}}\text{dt}$
$=\frac{1}{2}\times\frac{\text{t}^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\text{C}$
$=\frac{1}{2}\times\frac{\text{t}^{\frac{3}{2}}}{\frac{3}{2}}+\text{C}$
$=\frac{1}{2}\times\frac{2}{3}\times\text{t}^{\frac{3}{2}}+\text{C}$
$=\frac{1}{3}\times\text{t}^{\frac{3}{2}}+\text{C}$
$=\frac{1}{3}\times(\text{x}^2+3\text{x}-18)^{\frac{3}{2}}+\text{C}$
Now consider the integral
$\text{I}_2=-\frac{9}{2}\int\sqrt{\text{x}^2+3\text{x}-18}\text{dx}$
$=-\frac{9}{2}\int\sqrt{\text{x}^2+2\times\frac{3}{2}\text{x}+\Big(\frac{3}{2}\Big)^2-\Big(\frac{3}{2}\Big)^2-18}\text{dx}$
$=-\frac{9}{2}\int\sqrt{\Big(\text{x}+\frac{3}{2}\Big)^2-\frac{9}{4}-18}\text{dx}$
$=-\frac{9}{2}\int\sqrt{\Big(\text{x}+\frac{3}{2}\Big)^2-\Big(\frac{9}{4}+18\Big)}\text{dx}$
$=-\frac{9}{2}\int\sqrt{\Big(\text{x}+\frac{3}{2}\Big)^2-\Big(\frac{9+72}{4}\Big)}\text{dx}$
$=-\frac{9}{2}\int\sqrt{\Big(\text{x}+\frac{3}{2}\Big)^2-\Big(\frac{81}{4}\Big)}\text{dx}$
$=-\frac{9}{2}\int\sqrt{\Big(\text{x}+\frac{3}{2}\Big)^2-\Big(\frac{9}{2}\Big)^2}\text{dx}$
We know that,
$\int\sqrt{\text{x}^2-\text{a}^2}\text{dx}=\frac{1}{2}\text{x}\sqrt{\text{x}^2-\text{a}^2}-\frac{1}{2}\text{a}^2\log\Big|\text{x}+\sqrt{\text{x}^2-\text{a}^2}\Big|+\text{C}$
$\therefore\ \text{I}_2=-\frac{9}{2}\begin{Bmatrix}\frac{1}{2}\Big(\text{x}+\frac{3}{2}\Big)\sqrt{\Big(\text{x}+\frac{3}{2}\Big)^2-\Big(\frac{9}{2}\Big)^2}\\-\frac{1}{2}\Big(\frac{9}{2}\Big)^2\log\Big|\Big(\text{x}+\frac{3}{2}\Big)+\sqrt{\Big(\text{x}+\frac{3}{2}\Big)^2-\Big(\frac{9}{2}\Big)^2}\Big|\end{Bmatrix}+\text{C}$
$=-\frac{9}{4}\begin{Bmatrix}\Big(\frac{2\text{x}+3}{2}\Big)\sqrt{\text{x}^2+3\text{x}-18}-\Big(\frac{729}{4}\Big)\\\log\Big|\Big(\text{x}+\frac{3}{2}\Big)+\sqrt{\text{x}^2+3\text{x}-18}\Big|\end{Bmatrix}+\text{C}$
$=-\frac{9}{8}(2\text{x}+3)\sqrt{\text{x}^2+3\text{x}-18}+\frac{729}{16}\\\log\Big|\Big(\text{x}+\frac{3}{2}\Big)+\sqrt{\text{x}^2+3\text{x}-18}\Big|+\text{C}$
Thus,
$\text{I}=-\frac{1}{3}(\text{x}^2+3\text{x}-18)^{\frac{3}{2}}-\frac{9}{8}(2\text{x}+3)\sqrt{\text{x}^2+3\text{x}-18}\\+\frac{729}{16}\log\Big|\Big(\text{x}+\frac{3}{2}\Big)+\sqrt{\text{x}^2+3\text{x}-18}\Big|+\text{C}$
View full question & answer→Question 35 Marks
Evaluvate the following intregals:
$\int\frac{1}{3+4\cot\text{x}}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{1}{3+4\cot\text{x}}\ \text{dx}$
$=\int\frac{1}{3+\frac{4\cos\text{x}}{\sin\text{x}}}\ \text{dx}$
$=\int\frac{\sin\text{x}}{3\sin\text{x}+4\cos\text{x}}\ \text{dx}$
Let $\sin\text{x}=\text{A}(3\sin\text{x}+4\cos\text{x})+\text{B}(3\cos\text{x}-4\sin\text{x})\dots(1)$
$\Rightarrow\sin\text{x}=(3\text{A}-4\text{B})\sin\text{x}+(4\text{A}+3\text{B})\cos\text{x}$
By compairing the coefficient of both sides we get,
$3\text{A}-4\text{B}=1\dots(2)$
$4\text{A}+3\text{B}=0\dots(3)$
Multiplying eq (2) by (3) and equation (3) by 4, then by adding them we get
$9\text{A}-12\text{B}+16\text{A}+12\text{B}=3+0$
$\Rightarrow25\text{A}=3$
$\text{A}=\frac{3}{25}$
Putting value of A in eq (3) we get,
$4\times\frac{3}{25}+3\text{B}=0$
$\Rightarrow3\text{B}=-\frac{12}{25}$
$\Rightarrow\text{B}=-\frac{4}{25}$
Thus, by substituting the value of A and B in eq (1) we get
$\text{I}=\int\bigg[\frac{\frac{3}{25}(3\sin\text{x}+4\cos\text{x})-\frac{4}{25}(3\cos\text{x}-4\sin\text{x})}{3\sin\text{x}+4\cos\text{x}}\bigg]\text{dx}$
$=\int\text{dx}-\frac{4}{25}\int\Big(\frac{3\cos\text{x}-4\sin\text{x}}{3\sin\text{x}+4\cos\text{x}}\Big)\ \text{dx}$
Putting $3\sin\text{x}+4\cos\text{x}=\text{t}$
$\Rightarrow(3\cos\text{x}-4\sin\text{x})\text{ dx}=\text{dt}$
$\therefore\text{I}=\frac{3}{25}\int\text{dx}-\frac{4}{25}\int\frac{\text{dt}}{\text{t}}$
$=\frac{3}{25}\text{x}-\frac{4}{25}\ln|\text{t}|+\text{C}$
$=\frac{3\text{x}}{25}-\frac{4}{25}\ln|3\sin\text{x}+4\cos\text{x}|+\text{C}$
View full question & answer→Question 45 Marks
Evaluate the following integrals:
$\int\text{x}\sqrt{\text{x}^2+\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\text{x}\sqrt{\text{x}^2+\text{x}}\text{dx}$
Let $\text{x}=\lambda\frac{\text{d}}{\text{dx}}(\text{x}^2+\text{x})+\mu$
$=\lambda(2\text{x}+1)+\mu$
Equating similar terms, we get,
$2\lambda=1\Rightarrow\lambda=\frac{1}{2}$
$\lambda+\mu=0\Rightarrow\mu=-\frac{1}{2}$
So,
$\text{I}=\frac{1}{2}\int\Big(\frac{1}{2}(2\text{x}+1)-\frac{1}{2}\Big)\sqrt{\text{x}^2+\text{x}}\text{dx}$
$=\frac{1}{2}\int(2\text{x}+1)\sqrt{\text{x}^2+\text{x}}-\frac{1}{2}\int\sqrt{\text{x}^2+\text{x}}\text{dx}$
Let $\text{x}^2+\text{x}=\text{t}$
$\Rightarrow(2\text{x}+1)\text{dx = dt}$
So,
$\text{I}=\frac{1}{2}\int\sqrt{\text{t}}\text{dt}-\frac{1}{2}\int\sqrt{\Big(\text{x}+\frac{1}{2}\Big)^2-\Big(\frac{1}{2}\Big)^2}$
$\text{I}=\frac{1}{2}.\frac{\text{t}^{\frac{3}{2}}}{\frac{3}{2}}-\frac{1}{2}\begin{Bmatrix}\frac{\big(\text{x}+\frac{1}{2}\big)}{2}\sqrt{\text{x}^2+\text{x}}\\-\frac{1}{8}\log\Big|\Big(\text{x}+\frac{1}{2}\Big)+\sqrt{\text{x}^2+\text{x}}\Big|\end{Bmatrix}+\text{C}$
Hence,
$\text{I}=\frac{1}{3}(\text{x}^2+\text{x})^{\frac{3}{2}}-\frac{1}{8}{\Big(\text{x}+\frac{1}{2}\Big)}{2}\sqrt{\text{x}^2+\text{x}}\\+\frac{1}{16}\log\Big|\Big(\text{x}+\frac{1}{2}\Big)+\sqrt{\text{x}^2+\text{x}}\Big|+\text{C}$
View full question & answer→Question 55 Marks
Evaluate the following integrals:$\int\frac{1-3\text{x}}{3\text{x}^2+4\text{x}+2}\text{ dx}$
AnswerLet $\text{I}=\int\frac{1-3\text{x}}{3\text{x}^2+4\text{x}+2}\text{ dx}$
Let $1-3\text{x}=\lambda\frac{\text{d}}{\text{dx}}\big(3\text{x}^2+4\text{x}+2\big)+\mu$
$=\lambda(6\text{x}+4)+\mu$
$1-3\text{x}=(6\lambda)\text{x}+(4\lambda+\mu)$
Comparing the coefficients of like powers of x,
$6\lambda=3\Rightarrow\lambda=-\frac{1}{2}$
$4\lambda+\mu=1\Rightarrow4\Big(-\frac{1}{2}\Big)+\mu=1$
$\mu=3$
So, $\text{I}=\int\frac{-\frac{1}{2}(6\text{x}+4)+3}{3\text{x}^2+4\text{x}+2}\text{ dx}$
$\text{I}=-\frac{1}{2}\int\frac{6\text{x}+4}{3\text{x}^2+4\text{x}+2}\text{ dx}+3\int\frac{1}{3\text{x}^2+4\text{x}+2}\text{ dx}$
$\text{I}=-\frac{1}{2}\int\frac{6\text{x}+4}{3\text{x}^2+4\text{x}+2}\text{ dx}+\frac{3}{3}\int\frac{1}{\text{x}^2+\frac{4}{3}\text{x}+\frac{2}{3}}\text{ dx}$
$\text{I}=-\frac{1}{2}\int\frac{6\text{x}+4}{3\text{x}^2+4\text{x}+2}\text{ dx}+\int\frac{1}{\text{x}^2+2\text{x}\big(\frac{2}{3}\big)+\big(\frac{2}{3}\big)^2-\big(\frac{2}{3}\big)^2+\frac{2}{3}}\text{ dx}$
$\text{I}=-\frac{1}{2}\int\frac{6\text{x}+4}{3\text{x}^2+4\text{x}+2}\text{ dx}+\int\frac{1}{\big(\text{x}+\frac{2}{3}\big)^2+\frac{2}{9}}\text{ dx}$
$\text{I}=-\frac{1}{2}\int\frac{6\text{x}+4}{3\text{x}^2+4\text{x}+2}\text{ dx}+\int\frac{1}{\big(\text{x}+\frac{2}{3}\big)^2+\big(\frac{\sqrt2}{3}\big)^2}\text{ dx}$
$\text{I}=-\frac{1}{2}\log\big|3\text{x}^2+4\text{x}+2\big|+\frac{3}{\sqrt2}\tan^{-1}\bigg(\frac{\text{x}+\frac{2}{3}}{\frac{\sqrt2}{3}}\bigg)+\text{C}$ $\Big[\text{Since }\int\frac{1}{\text{x}^2+\text{a}^2}\text{ dx}=\frac{1}{\text{a}}\tan^{-1}\Big(\frac{\text{x}}{\text{a}}\Big)+\text{C}\Big]$
$\text{I}=\frac{1}{2}\log\big|3\text{x}^2+4\text{x}+2\big|+\frac{3}{\sqrt2}\tan^{-1}\Big(\frac{3\text{x}+2}{\sqrt2}\Big)+\text{C}$
View full question & answer→Question 65 Marks
Evaluate the following integrals:$\int\frac{\text{x}^2}{\text{x}^2+6\text{x}+12}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\text{x}^2\text{ dx}}{\text{x}^2+6\text{x}+12}$ Now,
Therefore,
$\frac{\text{x}^2}{\text{x}^2+6\text{x}+12}=1-\frac{(6\text{x}+12)}{\text{x}^2+6\text{x}+12}\ ....(1)$Let $6\text{x}+12=\text{A}\frac{\text{d}}{\text{dx}}\big(\text{x}^2+6\text{x}+12\big)+\text{B}$
$\Rightarrow6\text{x}+12=\text{A}(2\text{x}+6)+\text{B}$
$\Rightarrow6\text{x}+12=(2\text{A})\text{x}+6\text{A}+\text{B}$
Equating coefficients of like terms $2\text{A}=6$ $\text{A}=3$ $6\text{A}+\text{B}=12$ $18+\text{B}=12$ $\text{B}=-6$ $\therefore\ \frac{\text{x}^2}{\text{x}^2+6\text{x}+12}=1-\frac{3(2\text{x}+6)}{\text{x}^2+6\text{x}+12}$ $\text{I}=\int\frac{\text{x}^2\text{ dx}}{\text{x}^2+6\text{x}+12}$ $=\int\text{dx}-3\int\frac{(2\text{x}+6)\text{ dx}}{\text{x}^2+6\text{x}+12}+6\int\frac{\text{dx}}{\text{x}^2+6\text{x}+12}$ $=\int\text{dx}-3\int\frac{(2\text{x}+6)\text{ dx}}{\text{x}^2+6\text{x}+12}+6\int\frac{\text{dx}}{\text{x}^2+6\text{x}+9+3}$ $=\int\text{dx}-3\int\frac{(2\text{x}+6)\text{ dx}}{\text{x}^2+6\text{x}+12}+6\int\frac{\text{dx}}{(\text{x}+3)^2+\big(\sqrt3\big)^2}$ $=\text{x}-3\log\big|\text{x}^2+6\text{x}+12\big|+\frac{6}{\sqrt3}\tan^{-1}\Big(\frac{\text{x}+3}{\sqrt3}\Big)+\text{C}$ $=\text{x}-3\log\big|\text{x}^2+6\text{x}+12\big|+2\sqrt3\tan^{-1}\Big(\frac{\text{x}+3}{\sqrt3}\Big)+\text{C}$ View full question & answer→Question 75 Marks
Evaluate the following integrals:
$\int\cot^6\text{x}\text{ dx}$
Answer$\int\cot^6\text{x}\text{ dx}$
$=\int\cot^4\text{x}\cdot\big(\text{cosec}^2-1\big)\text{dx}$
$=\int\cot^4\text{x}\cdot\text{ cosec}^2\text{x}\text{ dx}-\int\cot^4\text{x}\text{dx}$
$=\int\cot^4\text{x}\cdot\text{ cosec}^2\text{x}\text{ dx}-\int\cot^2\text{x}\cdot\cot^2\text{x}\text{ dx}$
$=\int\cot^4\text{x}-\text{ cosec}^2\text{x}\text{ dx}-\int\big(\text{cosec}^2\text{x}-1\big)\cot^2\text{x}\text{ dx}$
$=\int\cot^4\text{x}\cdot\text{ cosec}^2\text{x}\text{ dx}-\int\cot^2\text{x}\cdot\text{cosec}^2\text{x}\text{ dx}+\int\cot^2\text{x}\text{ dx}$
$=\int\cot^4\text{x}\cdot\text{ cosec}^2\text{x}\text{ dx}-\int\cot^2\text{x}\cdot\text{cosec}^2\text{x}\text{ dx}+\int(\text{cosec}^2\text{x}-1)\text{ dx}$
Now, let $\text{I}_1=\int\cot^4\text{x}\cdot\text{cosec}^2\text{x}\text{ dx}-\int\cot^2\text{x}\cdot\text{cosec}^2\text{x}\text{ dx}$
And $\text{I}_2=\int(\text{cosec}^2\text{x}-1)\text{dx}$
First we integrate $I_1$
$\text{I}_1=\int\cot^4\text{x}\cdot\text{cosec}^2\text{x}\text{ dx}-\int\cot^2\text{x}\cdot\text{cosec}^2\text{x}\text{ dx}$
Let $\cot\text{x}=\text{t}$
$-\text{cosec}^2\text{x}\text{ dx}=\text{dt}$
$\text{cosec}^2\text{x}\text{ dx}=-\text{dt}$
$\text{I}_1=-\int\text{t}^4\text{dt}+\int\text{t}^2\text{dt}$
$=\frac{-\text{t}^5}{5}+\frac{\text{t}^3}{3}+\text{C}_1$
$=-\frac{\cot^5\text{x}}{5}+\frac{\cot^3\text{x}}{3}+\text{C}_1$
Now we integrate $I_2$
$\text{I}_2=\int(\text{cosec}^2\text{x}-1)\text{dx}$
$=-\cot\text{x}-\text{x}+\text{C}_1$
Now, $\int\cot^6\text{x}\text{ dx}=\text{I}_1+\text{I}_2$
$=-\frac{1}{5}\cot^5\text{x}+\frac{1}{3}\cot^3\text{x}-\cot\text{x}-\text{x}+\text{x}+\text{C}_1+\text{C}_2$
$=-\frac{1}{5}\cot^5\text{x}+\frac{1}{3}\cot^3\text{x}-\cot\text{x}-\text{x}+\text{C}$ $\big[\therefore\text{ C}=\text{C}_1+\text{C}_2\big]$
View full question & answer→Question 85 Marks
Evaluvate the following intregals:
$\int\frac{\text{x}+2}{\sqrt{\text{x}^2+2\text{x}+3}}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{\text{x}+2}{\sqrt{\text{x}^2+2\text{x}+3}}\ \text{dx}$
$\text{x}+2=\text{A}\frac{\text{d}}{\text{dx}}[\text{x}^2+2\text{x}+3]+\text{B}$
$\Rightarrow\text{x}+2=2\text{Ax}+2\text{A}+\text{B}$
Comparing the coefficient, we have,
$2\text{A}=1\text{ and }2\text{A}+\text{B}=2$
$\Rightarrow\text{A}=\frac{1}{2}$
Substituting the value of A in 2A + B = 2, we have,
$2\times\frac{1}{2}+\text{B}=2$
$\Rightarrow1+\text{B}=2$
$\Rightarrow\text{B}=2-1$
$\Rightarrow\text{B}=1$
Thus we have,
$\text{x}+2=\frac{1}{2}[2\text{x}+2]+1$
hence,
$\text{I}=\int\frac{\text{x}+2}{\sqrt{\text{x}^2+2\text{x}+3}}\text{dx}$
$=\int\frac{\big[\frac{1}{2}[2\text{x}+2]+1\big]}{\sqrt{\text{x}^2+2\text{x}+3}}\text{dx}$
$=\int\frac{\big[\frac{1}{2}[2\text{x}+2]\big]}{\sqrt{\text{x}^2+2\text{x}+3}}\text{dx}+\int\frac{\text{dx}}{\sqrt{\text{x}^2+2\text{x}+3}}$
$=\frac{1}{2}\int\frac{[2\text{x}+2]}{\sqrt{\text{x}^2+2\text{x}+3}}\text{dx}+\int\frac{\text{dx}}{\sqrt{\text{x}^2+2\text{x}+3}}$
Substituting t = x2 + 2x + 3 and dt = 2x + 2 in the first intrgrand, we have,
$\text{I}=\frac{1}{2}\int\frac{\text{dt}}{\sqrt{\text{t}}}+\int\frac{\text{dx}}{\sqrt{\text{x}^2+2\text{x}+3}}$
$=\frac{1}{2}\times2\sqrt{\text{t}}+\int\frac{\text{dx}}{\sqrt{\text{x}^2+2\text{x}+1+2}}+\text{C}$
$=\sqrt{\text{t}}+\int\frac{\text{dx}}{\sqrt{(\text{x}+1)^2+(\sqrt{2}})^2}+\text{C}$
$\text{I}=\sqrt{\text{x}^2+2\text{x}+3}+\log\big[|\text{x}+1|+\sqrt{(\text{x}+1)^2+(\sqrt{2}}^2\Big]+\text{C}$
$\text{I}=\sqrt{\text{x}^2+2\text{x}+3}+\log\Big[|\text{x}+1|+\sqrt{\text{x}^2+2\text{x}+3}\Big]+\text{C}$
View full question & answer→Question 95 Marks
Evaluate the following integrals:
$\int(2\text{x}+5)\sqrt{10-4\text{x}-3\text{x}^2}\text{dx}$
Answer$\text{I}=\int(2\text{x}+5)\sqrt{10-4\text{x}-3\text{x}^2}\text{dx}$Let $(2\text{x}+5)=\text{A}\frac{\text{d}}{\text{dx}}(10-4\text{x}-3\text{x}^2)+\text{B}$
$\Rightarrow(2\text{x}+5)=\text{A}(-4-6\text{x})+\text{B}$
$\Rightarrow(2\text{x}+5)=-6\text{A}\text{x}+(\text{B}-4\text{A})$
$\Rightarrow2=-6\text{A}\text{ and }(\text{B}-4\text{A})=5$
$\Rightarrow\text{A}=-\frac{1}{3}\text{ and }\text{B}=\frac{11}{3}$
$\Rightarrow(2\text{x}+5)=-\frac{1}{3}(-4-6\text{x})+\frac{11}{3}$
$\Rightarrow\text{I}=-\frac{1}{3}(-4-6\text{x})\sqrt{10-4\text{x}-3\text{x}^2}\text{dx}\\+\frac{11}{3}\int\sqrt{10-4\text{x}-3\text{x}^2}\text{dx}$
Let $\text{I}=-\frac{1}{3}\text{I}_1+\frac{11}{3}\text{I}_2\ \dots(1)$
Now,
$\text{I}_1=\int(-4-6\text{x})\sqrt{10-4\text{x}-3\text{x}^2}\text{dx}$
Let $(10-4\text{x}-3\text{x}^2)=\text{t},\text{ or, }(-4-6\text{x})\text{dx = dt}$
$\Rightarrow\text{I}_1=\int\sqrt{\text{t}}\text{dt}$
$=\frac{2}{3}\text{t}^{\frac{3}{2}}+\text{c}_1$
$\Rightarrow\text{I}_1=\frac{2}{3}(10-4\text{x}-3\text{x}^2)^{\frac{3}{2}}+\text{c}_1$
$\text{I}_2=\int\sqrt{(10-4\text{x}-3\text{x}^2)}\text{dx}$
$=\int\sqrt{3\Big(\frac{10}{3}-\frac{4}{3}\text{x}-\text{x}^2\Big)}\text{dx}$
$=\sqrt3\int\sqrt{\Big(\frac{26}{9}-\frac{4}{9}-\frac{4}{3}\text{x}-\text{x}^2\Big)}\text{dx}$
$=\sqrt3\int\sqrt{\Big[\Big(\frac{\sqrt{26}}{3}\Big)^2-\Big(\frac{4}{9}+\frac{4}{3}\text{x}-\text{x}^2\Big)\Big]}\text{dx}$
$=\sqrt3\int\sqrt{\Big[\Big(\frac{\sqrt{26}}{3}\Big)^2-\Big(\text{x}+\frac{2}{3}\Big)^2\Big]}\text{dx}$
$=\sqrt3\sin\Bigg(\frac{\text{x}+\frac{2}{3}}{\frac{\sqrt{26}}{3}}\Bigg)+\text{c}_2$
$=\sqrt3\sin\Big(\frac{3\text{x}+2}{\sqrt{26}}\Big)+\text{c}_2$
Using (1), we get
$\text{I}=-\frac{1}{3}\times\frac{2}{3}(10-4\text{x}-3\text{x}^2)^{\frac{3}{2}}\\+\frac{11}{3}\times\sqrt3\sin\Big(\frac{3\text{x}+2}{\sqrt{26}}\Big)+\text{C}$
$\therefore\ \text{I}=-\frac{2}{9}(10-4\text{x}-3\text{x}^2)^{\frac{3}{2}}+\frac{11\sqrt3}{3}\sin\Big(\frac{3\text{x}+2}{\sqrt{26}}\Big)+\text{C}$
View full question & answer→Question 105 Marks
Evaluate the following integrals:
$\int(\text{x}-2)\sqrt{2\text{x}^2-6\text{x}+5}\text{dx}$
AnswerLet $\text{I}=\int(\text{x}-2)\sqrt{2\text{x}^2-6\text{x}+5}\text{dx}$Also, $\text{x}-2=\lambda\frac{\text{d}}{\text{dx}}(2\text{x}^2-6\text{x}+5)+\mu$
$=(4\lambda)\text{x}+\mu-6\lambda$ Equating the co-efficient of like terms $4\lambda=1\Rightarrow\lambda=\frac{1}{4}$And
$\mu-6\lambda=-2$ $\Rightarrow\mu-6\times\frac{1}{4}=-2$ $\Rightarrow\mu=-2+\frac{3}{2}=-\frac{1}{2}$ $\therefore\ \text{I}=\int\Big[\frac{1}{4}(4\text{x}-6)-\frac{1}{2}\Big]\sqrt{2\text{x}^2-6\text{x}+5}\text{dx}$ $=\frac{1}{4}\int(4\text{x}-6)\sqrt{2\text{x}^2-6\text{x}+5}\text{dx}\\-\frac{1}{2}\int\sqrt{2\text{x}^2-6\text{x}+5}\text{dx}$ Let $2\text{x}^2-6\text{x}+5=\text{t}$ $\Rightarrow(4\text{x}-6)\text{dx = dt}$ $\therefore\ \text{I}=\frac{1}{4}\int\text{t}^{\frac{1}{2}}\text{dt}-\frac{1}{2}\int\sqrt{2\Big(\text{x}^2-3\text{x}+\frac{5}{2}\Big)}\text{dx}$ $=\frac{1}{4}\int\text{t}^{\frac{1}{2}}-\frac{\sqrt2}{2}\int\sqrt{\text{x}^2-3\text{x}\Big(\frac{3}{2}\Big)^2-\Big(\frac{3}{2}\Big)^2+\frac{5}{2}}\text{dx}$ $=\frac{1}{4}\Bigg[\frac{\text{t}^{\frac{3}{2}}}{\frac{3}{2}}\Bigg]-\frac{1}{\sqrt2}\int\sqrt{\Big(\text{x}-\frac{3}{2}\Big)^2-\frac{9}{4}+\frac{5}{2}}\text{dx}$ $=\frac{1}{6}\text{t}^{\frac{3}{2}}-\frac{1}{\sqrt2}\int\sqrt{\Big(\text{x}-\frac{3}{2}\Big)^2-\frac{9+10}{4}}\text{dx}$ $=\frac{1}{6}\text{t}^{\frac{3}{2}}-\frac{1}{\sqrt2}\int\sqrt{\Big(\text{x}-\frac{3}{2}\Big)^2+\Big(\frac{1}{2}\Big)^2}\text{dx}$ $=\frac{1}{6}\big(2\text{x}^2-6\text{x}+5)^{\frac{3}{2}}-\frac{1}{\sqrt2}\bigg[\bigg(\frac{\text{x}-\frac{3}{2}}{2}\bigg)\sqrt{\Big(\text{x}-\frac{3}{2}\Big)^2+\Big(\frac{1}{2}\Big)^2}\\+\frac{1}{8}\log\bigg|\Big(\text{x}-\frac{3}{2}\Big)+\sqrt{\text{x}^2-3\text{x}+\frac{5}{2}}\bigg|\bigg]+\text{C}$ $=\frac{1}{6}\big(2\text{x}^2-6\text{x}+5)^{\frac{3}{2}}-\frac{1}{\sqrt2}\bigg[\frac{2\text{x}-3}{4}\sqrt{\text{x}^2-3\text{x}+\frac{5}{2}}\\+\frac{1}{8}\log\bigg|\frac{2\text{x}-3}{2}+\sqrt{\text{x}^2-3\text{x}+\frac{5}{2}}\bigg|\bigg]+\text{C}$
View full question & answer→Question 115 Marks
Evaluate the following integrals:$\int\frac{\text{x}^2+1}{\text{x}^2-5\text{x}+6}\text{ dx}$
AnswerLet $\text{I}=\int\Big(\frac{\text{x}^2+1}{\text{x}^2-5\text{x}+6}\Big)\text{dx}$ Dividing Numerator by Denominator
$\frac{\text{x}^2+1}{\text{x}^2-5\text{x}+6}=1+\Big(\frac{5\text{x}-5}{\text{x}^2-5\text{x}+6}\Big)\ ....(1)$ Also $\frac{5\text{x}-5}{\text{x}^2-5\text{x}+6}=\frac{5\text{x}-5}{(\text{x}-2)(\text{x}-3)}$ Let $\Rightarrow\frac{5\text{x}-5}{(\text{x}-2)(\text{x}-3)}=\frac{\text{A}}{\text{x}-2}+\frac{\text{B}}{\text{x}-3}$ $\Rightarrow\frac{5\text{x}-5}{(\text{x}-2)(\text{x}-3)}=\frac{\text{A}(\text{x}-3)+\text{B}(\text{x}-2)}{(\text{x}-2)(\text{x}-3)}$ $\Rightarrow5\text{x}-5=\text{A}(\text{x}-3)+\text{B}(\text{x}-2)$ Let $\text{x}=3$ $5\times3-5=\text{}A\times0+\text{B}(3-2)$ $10=\text{B}$ Let $\text{x}=2$ $5\times2-5=\text{}A(2-3)+\text{B}\times0$ $\text{A}=-5$ $\Rightarrow\frac{5\text{x}-5}{(\text{x}-2)(\text{x}-3)}=\frac{-5}{\text{x}-2}+\frac{10}{\text{x}-3}\ ... ..(2)$ From (1) and (2) $\text{I}=\int\text{dx}-5\int\frac{\text{dx}}{\text{x}-2}+10\int\frac{\text{dx}}{\text{x}-3}$ $=\text{x}-5\log|\text{x}-2|+10\log|\text{x}-3|+\text{C}$ View full question & answer→Question 125 Marks
Evaluate the following integrals:$\int\frac{2\text{x}}{2+\text{x}-\text{x}^2}\text{ dx}$
Answer$\int\frac{2\text{x}\text{ dx}}{(2+\text{x}-\text{x}^2)}$ $2\text{x}=\text{A}\frac{\text{d}}{\text{dx}}\big(2+\text{x}-\text{x}^2\big)+\text{B}$ $2\text{x}=\text{A}(0+1-2\text{x})+\text{B}$ $2\text{x}=(-2\text{A})\text{x}+\text{A}+\text{B}$ Comparing the coefficients of like power of x, $-2\text{A}=2$ $\text{A}=-1$ $\text{A}+\text{B}=0$ $-1+\text{B}=0$ $\text{B}=1$ Now, $\int\frac{2\text{x}\text{ dx}}{(2+\text{x}-\text{x}^2)}$ $=\int\Big(\frac{-1(1-2\text{x})+1}{-\text{x}^2+\text{x}+2}\Big)\text{dx}$ $=-\int\Big(\frac{1-2\text{x}}{-\text{x}^2+\text{x}+2}\Big)\text{dx}+\int\frac{\text{dx}}{-\text{x}^2+\text{x}+2}$ $=-\text{I}_1+\text{I}_2\ ....(1)$ (say) where $\text{I}_1=\int\Big(\frac{1-2\text{x}}{-\text{x}^2+\text{x}+2}\Big)\text{dx}$ $\text{I}_2=\int\frac{\text{dx}}{-\text{x}^2+\text{x}+2}$ $\text{I}_1=\int\Big(\frac{1-2\text{x}}{-\text{x}^2+\text{x}+2}\Big)\text{dx}$ Let $-\text{x}^2+\text{x}+2=\text{t}$ $\Rightarrow(1-2\text{x})\text{dx}=\text{dt}$ $\text{I}_1=\int\frac{\text{dt}}{\text{t}}$ $\text{I}_1=\log|\text{t}|+\text{C}_1\ ....(2)$ $\text{I}_2=\int\frac{\text{dx}}{-\text{x}^2+\text{x}+2}$ $\text{I}_2=\int\frac{-\text{dx}}{\text{x}^2+\text{x}+2}$ $\text{I}_2=\int\frac{-\text{dx}}{\text{x}^2-\text{x}+\big(\frac{1}{2}\big)^2-\big(\frac{1}{2}\big)^2-2}$ $\text{I}_2=\int\frac{-\text{dx}}{\big(\text{x}-\frac{1}{2}\big)^2-\big(\frac{3}{2}\big)^2}$ $\text{I}_2=-\frac{1}{2\times\frac{3}{2}}\log\Bigg|\frac{\text{x}-\frac{1}{2}-\frac{3}{2}}{\text{x}-\frac{1}{2}+\frac{3}{2}}\Bigg|+\text{C}_2$ $\text{I}_2=-\frac{1}{3}\log\Big|\frac{\text{x}-2}{\text{x}+1}\Big|+\text{C}_2\ ....(3)$ From (1) (2) and (3) $\int\Big(\frac{2\text{x}}{2+\text{x}+\text{x}^2}\Big)\text{ dx}=-\log\big|2+\text{x}-\text{x}^2\big|-\frac{1}{3}\log\Big|\frac{\text{x}-2}{\text{x}+1}\Big|+\text{C}_1+\text{C}_2$ $=-\log\big|2+\text{x}-\text{x}^2\big|+\frac{1}{3}\log\Big|\frac{1+\text{x}}{\text{x}-2}\Big|+\text{C}$Where, $\text{C}=\text{C}_1+\text{C}_2$
View full question & answer→Question 135 Marks
Evaluate the following intregals:
$\int\frac{\cos\text{x}}{(1-\sin\text{x})^3(2+\sin\text{x})}\ \text{dx}$
AnswerLet
$\sin\text{x}=\text{t}$
$\Rightarrow\cos\text{x}=\text{dt}$
$\therefore\int\frac{\cos\text{x}}{(1+\sin\text{x})^3(2+\sin\text{x})}=\int\frac{1}{(1-\text{t})^3(2+\text{t})}\ \text{dt}$
Let $\text{f}(\text{t})=\frac{1}{(1-\text{t})^3(2+\text{t})}$
Then suppose
$\frac{1}{(1-\text{t})^3(2+\text{t})}=\frac{\text{A}}{1-\text{t}}+\frac{\text{B}}{(1-\text{t})^2}+\frac{\text{C}}{(1-\text{t})^3}+\frac{\text{D}}{(2+\text{t})}$
$\Rightarrow1=\text{A}(1-\text{t})^2(2+\text{t})+\text{B}(1-\text{t})(2+\text{t})\\+\text{C}(2+\text{t})+\text{D}(1-\text{t})^3$
Put t = 1
1 = 27D
$\Rightarrow\text{D}=\frac{1}{27}$
Similarly, we can find that $\text{A}=\frac{-1}{27}$ and $\text{B}=\frac{+1}{9}$
$\therefore\int\frac{1}{(1-\text{t})^3(2+\text{t})}\ \text{dt}=\frac{-1}{27}\int\frac{1}{1-\text{t}}\ \text{dt}+\frac{1}{9}\int\frac{\text{dt}}{(1-\text{t})^2}\\+\frac{1}{3}\int\frac{\text{dt}}{(1-\text{t})^3}+\frac{1}{27}\int\frac{\text{dt}}{2+\text{t}}$
$=\frac{-1}{27}\log|1-\text{t}|+\frac{1}{9(1-\text{t})}+\frac{1}{6(1-\text{t})^2}+\frac{1}{27}\log|2+\text{t}|+\text{C}$
Putting $\text{t}=\sin\text{x}$ we get
$\int\frac{\cos\text{x}}{(1-\sin\text{x})^3(2+\sin\text{x})}\ \text{dx}$
$=\frac{-1}{27}\log|1-\sin\text{x}|+\frac{1}{9(1-\sin\text{x})}\\+\frac{1}{6(1-\sin\text{x})^2}+\frac{1}{27}\log|2+\sin\text{x}|+\text{C}$
View full question & answer→Question 145 Marks
Evaluate the following intregals:
$\int\frac{\text{x}^4}{(\text{x}-1)(\text{x}^2+1)}\ \text{dx}$
Answerwe have
$\text{I}=\int\frac{\text{x}^4\text{dx}}{(\text{x}-1)(\text{x}^2+1)}\ \text{dx}$
$=\int\Big[\frac{\text{x}^4-1+1}{(\text{x}-1)(\text{x}^2+1)}\Big]\text{dx}$
$=\int\frac{(\text{x}^4-1)\text{dx}}{(\text{x}-1)(\text{x}^2+1)}+\int\frac{\text{dx}}{(\text{x}-1)(\text{x}^2+1)}$
$=\int\frac{(\text{x}^2-1)(\text{x}^2+1)\text{ dx}}{(\text{x}-1)(\text{x}^2+1)}+\int\frac{\text{dx}}{(\text{x}-1)(\text{x}^2+1)}$
$=\int\frac{(\text{x}-1)(\text{x}-1)\text{ dx}}{(\text{x}-1)}+\frac{\text{dx}}{(\text{x}-1)(\text{x}^2+1)}$
$=\int(\text{x}+1)\ \text{dx}+\int\frac{\text{dx}}{(\text{x}-1)(\text{x}^2+1)}\ ...(1)$
Let $\frac{1}{(\text{x}-1)(\text{x}^2+1)}=\frac{\text{A}}{\text{x}+1}+\frac{\text{Bx}+\text{C}}{\text{x}^2+1}$
$\Rightarrow\frac{1}{(\text{x}-1)(\text{x}^2+1)}=\frac{\text{A}(\text{x}^2+1)+(\text{Bx}+\text{C})(\text{x}-1)}{(\text{x}-1)(\text{x}^2+1)}$
$\Rightarrow 1 = Ax^2 + A + Bx^2 - Bx + Cx - C$
$\Rightarrow 1 = (A + B)x^2 +(C - B)x + A - C$
Equating coefficient of like terms
$A + B = 0 ...(1)$
$C - B = 0 ...(2)$
$A - C = 1 ...(3)$
Solving (1), (2), (3) we get
$\text{B}=-\frac{1}{2},\text{A}=\frac{1}{2},\text{C}=-\frac{1}{2}$
$\therefore\frac{1}{(\text{x}-1)(\text{x}^2+1)}=\frac{1}{2(\text{x}-1)}+\frac{-\frac{\text{x}}{2}-\frac{1}{2}}{\text{x}^2+1}$
$\Rightarrow\frac{1}{(\text{x}-1)(\text{x}^2+1)}=\frac{1}{2(\text{x}-1)}-\frac{1}{2}\Big(\frac{\text{x}}{\text{x}^2+1}\Big)-\frac{1}{2(\text{x}^2+1)}\ ...(2)$
From (1) and (2)
$\text{I}=\int(\text{x}+1)\ \text{dx}+\frac{1}{2}\int\frac{\text{dx}}{\text{x}-1}-\frac{1}{2}\int\frac{\text{x dx}}{\text{x}^2+1}-\frac{1}{2}\int\frac{\text{dx}}{\text{x}^2+1}$
Putting $\text{x}^2+1=\text{t}$
$\Rightarrow2\text{x dx}=\text{dt}$
$\Rightarrow\text{x dx}=\frac{\text{dt}}{2}$
$\therefore\text{I}=\int(\text{x}+1)\ \text{dx}+\frac{1}{2}\int\frac{\text{dx}}{\text{x}-1}-\frac{1}{4}\int\frac{\text{dt}}{\text{t}}-\frac{1}{2}\int\frac{\text{dx}}{\text{x}^2+1}$
$=\frac{\text{x}^2}{2}+\text{x}+\frac{1}{2}\log|\text{x}-1|-\frac{1}{4}\log|\text{t}|-\frac{1}{2}\tan^{-1}\text{x}+\text{C}$
View full question & answer→Question 155 Marks
Evaluate the following integrals:$\int\frac{\text{x}^3+\text{x}^2+2\text{x}+1}{\text{x}^2-\text{x}+1}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\text{x}^3+\text{x}^2+2\text{x}+1}{\text{x}^2-\text{x}+1}\text{ dx}$ $=\int\Big[\text{x}+2+\frac{3\text{x}-1}{\text{x}^2-\text{x}+1}\Big]\text{dx}$ $\text{I}=\frac{\text{x}^2}{2}+2\text{x}+\int\frac{3\text{x}-1}{\text{x}^2-\text{x}+1}\text{ dx}+\text{C}_1\ ....(1)$ Let $\text{I}_1=\int\frac{3\text{x}-1}{\text{x}^2-\text{x}+1}\text{ dx}$ Let $3\text{x}-1=\lambda\frac{\text{d}}{\text{dx}}\big(\text{x}^2-\text{x}+1\big)+\mu$ $=\lambda(2\text{x}-1)+\mu$ $3\text{x}-1=(2\lambda)\text{x}-\lambda+\mu$Comparing the coefficients of like powers of x,
$3=2\lambda\Rightarrow\lambda=\frac{3}{2}$ $-\lambda+\mu=-1\Rightarrow-\Big(\frac{3}{2}\Big)+\mu=-1$ $\mu=\frac{1}{2}$ So, $\text{I}_1=\int\frac{\frac{3}{2}(2\text{x}-1)+\frac{1}{2}}{\text{x}^2-\text{x}+1}\text{ dx}$ $\text{I}_1=\frac{3}{2}\int\frac{(2\text{x}-1)}{\text{x}^2-\text{x}+1}\text{ dx}+\frac{1}{2}\int\frac{1}{\text{x}^2-2\text{x}\big(\frac{1}{2}\big)+\big(\frac{1}{2}\big)^2-\big(\frac{1}{2}\big)^2+1}\text{ dx}$ $\text{I}_1=\frac{3}{2}\int\frac{2\text{x}-1}{\text{x}^2-\text{x}+1}\text{ dx}+\frac{1}{2}\int\frac{1}{\big(\text{x}-\frac{1}{2}\big)^2-\Big(\frac{\sqrt3}{2}\Big)^2}\text{ dx}$ $\text{I}_1=\frac{3}{2}\log\big|\text{x}^2-\text{x}+1\big|+\frac{1}{2}\times\frac{2}{\sqrt3}\tan^{-1}\bigg(\frac{\text{x}+\frac{1}{2}}{\frac{\sqrt3}{2}}\bigg)+\text{C}_2$ $\Big[\text{since},\int\frac{1}{\text{x}^2+\text{a}^2}\text{ dx}=\frac{1}{\text{a}}\tan^{-1}\Big(\frac{\text{x}}{\text{a}}\Big)+\text{C}\Big]$ $\text{I}_1=\frac{3}{2}\log\big|\text{x}^2-\text{x}+1\big|+\frac{1}{\sqrt3}\tan^{-1}\Big(\frac{2\text{x}+1}{\sqrt3}\Big)+\text{C}_2\ .....(2)$ Using equation (1) and (2) $\text{I}=\frac{\text{x}^2}{2}+2\text{x}+\frac{3}{2}\log\big|\text{x}^2-\text{x}+1\big|+\frac{1}{\sqrt3}\tan^{-1}\Big(\frac{2\text{x}+1}{\sqrt3}\Big)+\text{C}$
View full question & answer→Question 165 Marks
Evaluate the following integrals:$\int\frac{\text{x}-1}{3\text{x}^2-4\text{x}+3}\text{dx}$
AnswerLet $\text{I}=\int\frac{\text{x}^2}{\text{x}^2+7\text{x}+10}\text{dx}$
$=\int\begin{Bmatrix}1-\frac{7\text{x}+10}{\text{x}^2+7\text{x}+10}\end{Bmatrix}\text{dx}$
$\text{I}=\text{x}-\int\frac{7\text{x}+10}{\text{x}^2+7\text{x}+10}\text{dx}+\text{c}_1\dots\text{(i})$
Let $\text{I}_1=\int\frac{7\text{x}+10}{\text{x}^2+7\text{x}+10}\text{dx}$
Let $7\text{x}+10=\lambda\frac{\text{d}}{\text{dx}}\Big(\text{x}^2+7\text{x}+10{}\Big)+\mu$
$=\lambda(2\text{x}+7)+\mu$
$7\text{x}+10=(2\lambda)\text{x}+7\lambda+\mu$
Comparing the coefficients of like powers of x,
$7=2\lambda$
$\Rightarrow\lambda=\frac72$
$7\lambda+\mu=10$
$\Rightarrow7\Big(\frac72\Big)+\mu=10$
$\mu=-\frac{29}{2}$
View full question & answer→Question 175 Marks
Evaluate the following integrals:
$\int(2\text{x}-5)\sqrt{2+3\text{x}-\text{x}^2}\text{dx}$
AnswerLet $\text{I}=\int(2\text{x}-5)\sqrt{2+3\text{x}-\text{x}^2}\text{dx}$
Also, $2\text{x}-5=\lambda\frac{\text{d}}{\text{dx}}(2+3\text{x}-\text{x}^2)+\mu$
$\Rightarrow2\text{x}-5=\lambda(-2\text{x}+3)+\mu$
$\Rightarrow2\text{x}-5=(-2\lambda)\text{x}=3\lambda+\mu$
Equating co-efficients of like terms
$-2\lambda=2$
$\Rightarrow\lambda=-1$
And
$3\lambda+\mu=-5$
$\Rightarrow3(-1)+\mu=-5$
$\Rightarrow\mu=-5+3$
$\Rightarrow\mu=-2$
$\therefore\ 2\text{x}-5=-1(-2\text{x}+3)-2$
Hence, $\text{I}=\int[-(-2\text{x}+3)-2]\sqrt{2+3\text{x}-\text{x}^2}\text{dx}$
$=-\int(-2\text{x}+3)\sqrt{2+3\text{x}-\text{x}^2}\text{dx}-2\int\sqrt{2+3\text{x}-\text{x}^2}\text{dx}$
$=-\text{I}_1-2\text{I}_2\ \dots(1)$
$\text{I}_1=\int(-2\text{x}+3)\sqrt{2+3\text{x}-\text{x}^2}\text{dx}$
Let $2+3\text{x}-\text{x}^2=\text{t}$
$\Rightarrow(-2\text{x}+3)\text{dx}=\text{dt}$
$\therefore\ \text{I}_1=\int\text{t}^{\frac{1}{2}}\text{dt}$
$=\frac{\text{t}^{\frac{1}{2}+1}}{\frac{1}{2}+1}$
$=\frac{2}{3}\text{t}^{\frac{3}{2}}$
$=\frac{2}{3}\big(2+3\text{x}-\text{x}^2\big)^{\frac{3}{2}}\ \dots(2)$
And $\text{I}_2=\int\sqrt{2+3\text{x}-\text{x}^2}\text{dx}$
$\text{I}_2=\int\sqrt{2-(\text{x}^2-3\text{x})}\text{dx}$
$=\int\sqrt{2-\Big[\text{x}^2-3\text{x}+\Big(\frac{3}{2}\Big)^2-\Big(\frac{3}{2}\Big)^2\Big]}\text{dx}$
$=\int\sqrt{2+\frac{9}{4}-\Big(\text{x}-\frac{3}{2}\Big)^2}\text{dx}$
$=\int\sqrt{\Big(\frac{\sqrt{17}}{2}\Big)^2-\Big(\text{x}-\frac{3}{2}\Big)^2}\text{dx}$
$=\frac{\text{x}-\frac{3}{2}}{2}\sqrt{\Big(\frac{\sqrt{17}}{2}\Big)^2-\Big(\text{x}-\frac{3}{2}\Big)^2}+\frac{\Big(\frac{\sqrt{17}}{2}\Big)^2}{2}\sin^{-1}\Bigg(\frac{\text{x}-\frac{3}{2}}{\frac{\sqrt{17}}{2}}\Bigg)$
$=\frac{2\text{x}-3}{4}\sqrt{2+3\text{x}-\text{x}^2}+\frac{17}{8}\sin^{-1}\Big(\frac{2\text{x}-3}{\sqrt{17}}\Big)\ \dots(3)$
From eq. (1), (2) and (3) we have
$\text{I}=-\frac{2}{3}\big(2+3\text{x}-\text{x}^2\big)^{\frac{3}{2}}-\frac{(2\text{x}-3)}{2}\sqrt{2+3\text{x}-\text{x}^2}\\-\frac{17}{4}\sin^{-1}\Big(\frac{2\text{x}-3}{\sqrt{17}}\Big)+\text{C}$
View full question & answer→Question 185 Marks
Evaluate the following intregals:
$\int\frac{3}{(1-\text{x})(1+\text{x}^2)}\ \text{dx}$
AnswerWe have,
$\text{I}=\int\frac{3\text{dx}}{(1-\text{x})(1+\text{x}^2)}$
$=3\int\frac{\text{dx}}{(1-\text{x})(1+\text{x}^2)}$
Let $\frac{1}{(1-\text{x})(1+\text{x})^2}=\frac{\text{A}}{1-\text{x}}+\frac{\text{Bx}+\text{C}}{\text{x}^2+1}$
$\Rightarrow\frac{1}{(1-\text{x})(\text{x}^2+1)}=\frac{\text{A}(\text{x}^2+1)+(\text{Bx}+\text{C})(1-\text{x})}{(1-\text{x})(\text{x}^2+1)}$
$\Rightarrow1=\text{Ax}^2+\text{A}+\text{Bx}-\text{Bx}^2+\text{C}-\text{Cx}$
$\Rightarrow1=(\text{A}-\text{B})\text{x}^2+(\text{B}-\text{C})\text{x}+\text{A}+\text{C}$
Equating coefficient of like terms.
$\text{A}-\text{B}=0\ ...(1)$
$\text{B}-\text{C}=0\ ...(2)$
$\text{A}+\text{C}=1\ ...(3)$
Solving (1), (2) and (3), we get
$\text{A}=\frac{1}{2},\text{B}=\frac{1}{2},\text{C}=\frac{1}{2}$
$\therefore\frac{1}{(1-\text{x})(\text{x}^2+1)}=\frac{1}{2(1-\text{x})}+\frac{\frac{\text{x}}{2}+\frac{1}{2}}{\text{x}^2+1}$
$\int\frac{3\text{dx}}{(1-\text{x})(\text{x}^2+1)}=\frac{3}{2}\int\frac{\text{dx}}{1-\text{x}}+\frac{3}{2}\int\frac{\text{x dx}}{\text{x}^2+1}+\frac{3}{2}\int\frac{\text{dx}}{\text{x}^2+1}$
Putting $\text{x}^2+1=\text{t}$
$\Rightarrow\text{x dx}=\frac{\text{dt}}{2}$
$\therefore\text{I}=\frac{3}{2}\int\frac{\text{dx}}{1-\text{x}}+\frac{3}{4}\int\frac{\text{dt}}{\text{t}}+\frac{3}{2}\int\frac{\text{dx}}{\text{x}^2+1}$
$=\frac{3}{2}\frac{\log|1-\text{x}|}{-1}+\frac{3}{4}\log|\text{t}|+\frac{3}{2}\times\tan^{-1}\text{x}+\text{C}$
$=\frac{-3}{2}\log|1-\text{x}|+\frac{3}{4}\log|1+\text{x}^2|+\frac{3}{2}\tan^{-1}\text{x}+\text{C}$
$=\frac{-3}{4}\times2\log|1-\text{x}|+\frac{3}{4}\log|1+\text{x}^2|+\frac{3}{2}(2\tan^{-1}\text{x})+\text{C}$
$=\frac{3}{4}\big[\log|1+\text{x}^2|-\log|(1-\text{x}^2)|\big]+\frac{3}{4}(2\tan^{-1}\text{x})+\text{C}$
$=\frac{3}{4}\Big[\log\Big|\frac{1+\text{x}^2}{(1-\text{x})^2}\Big|+2\tan^{-1}(\text{x})\Big]+\text{C}$
View full question & answer→Question 195 Marks
Evaluvate the following intregals:
$\int\frac{3+2\cos\text{x}+4\sin\text{x}}{2\sin\text{x}+\cos\text{x}+3}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{3+2\cos\text{x}+4\sin\text{x}}{2\sin\text{x}+\cos\text{x}+3}\ \text{dx}$
Let $3+2\cos\text{x}+4\sin\text{x}=\lambda\frac{\text{d}}{\text{dx}}(2\sin\text{x}+\cos\text{x}+3)+\mu(2\sin\text{x}+\cos\text{x}+3)+\text{v}$
$3+2\cos\text{x}+4\sin\text{x}=\lambda(2\cos\text{x}-\sin\text{x})+\mu(2\sin\text{x}+\cos\text{x}+3)+\text{v}$
$3+2\cos\text{x}+4\sin\text{x}=(-\lambda+2\mu)\sin\text{x}+(2\lambda+\mu)\cos\text{x}+3\mu+\text{v}$
Comparing the coefficient of $\sin\text{x}\ \&\cos\text{x}$ on the both the sides,
$-\lambda+2\mu\ =4\dots\dots(1)$
$2\lambda + \mu = 2 \ \dots\dots(2)$
$2\mu + \text{v} = 3 \ \dots\dots(3)$
Solving Equation (1), (2) and (3), we get
$\lambda=0,\mu=2,\text{v}=-3$
$\text{I}==\int\frac{2(2\sin\text{x}+\cos\text{x}+3)-3}{(2\sin\text{x}+\cos\text{x}+3)}\ \text{dx}$
$=2\int\text{dx}-2\int\frac{1}{2\sin\text{x}+\cos\text{x}+3}\ \text{dx}$
$\text{I}=2\text{x}-3\text{I}_1+\text{C}_1\dots\dots(4)$
Let $\text{I}_1=\int\frac{1}{2\sin\text{x}+\cos\text{x}+3}\ \text{dx}$
$\sin\text{x}=\frac{2\tan\frac{\text{x}}{2}}{\Big(1+\tan^2\frac{\text{x}}{2}\Big)},\cos\text{x}=\frac{1-\tan^2\frac{\text{x}}{2}}{\Big(1+\tan^2\frac{\text{x}}{2}\Big)}$
$\text{I}_1=\int\frac{1}{2\Bigg(\frac{2\tan\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}\Bigg)+\Bigg(\frac{1-\tan^2\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}\Bigg)+3}\ \text{dx}$
$=\int\frac{\Big(1+\tan^2\frac{\text{x}}{2}\Big)}{4\tan\frac{\text{x}}{2}+1-\tan^2\frac{\text{x}}{2}+3\Big(1+\tan^2\frac{\text{x}}{2}\Big)}\ \text{dx}$
$=\int\frac{\sec^2\frac{\text{x}}{2}}{2\tan^2\frac{\text{x}}{2}+4\tan\frac{\text{x}}{2}+4}\ \text{dx}$
Let $\tan\frac{\text{x}}{2}=\text{t}$
$\frac{1}{2}\sec^2\frac{\text{x}}{2}=\text{dt}$
$\text{I}_1=\int\frac{2\text{dt}}{2\text{t}^2+4\text{t}+4}$
$\frac{2}{2}\int\frac{\text{dt}}{\text{t}^2+2\text{t}+2}$
$=\int\frac{\text{dt}}{\text{t}^2+2\text{t}+1-1+2}$
$=\int\frac{\text{dt}}{(\text{t}+1)^2+1}$
$=\tan^{-1}\Big(\tan\frac{\text{x}}{2}+1)+\text{C}_2$
Now, using equation (1),
$\text{I}=2\times-3\tan^{-1}\Big(\tan\frac{\text{x}}{2}+1\Big)+\text{C}$
View full question & answer→Question 205 Marks
Evaluate the following intregals:
$\int\frac{4\text{x}^2+3}{(\text{x}^2+2)(\text{x}^2+3)(\text{x}^2+4)}\ \text{dx}$
AnswerWe have,
$\int\frac{(4\text{x}^2+3)\text{dx}}{(\text{x}^2+2)(\text{x}^2+3)(\text{x}^2+4)}\ $
Putting $x^2 = t$
Then,
$\frac{4\text{x}^2+3}{(\text{x}^2+2)(\text{x}^2+3)(\text{x}^2+4)}=\frac{4\text{t}^2+3}{(\text{t}+2)(\text{t}+3)(\text{t}+4)}$
Let $\frac{4\text{x}^2+3}{(\text{x}^2+2)(\text{x}^2+3)(\text{x}^2+4)}=\frac{\text{A}}{\text{t}+2}+\frac{\text{B}}{\text{t}+3}+\frac{\text{C}}{\text{t}+4}$
$\Rightarrow\frac{4\text{x}^2+3}{(\text{x}^2+2)(\text{x}^2+3)(\text{x}^2+4)}=\frac{\text{A}(\text{t}+3)(\text{t}+4)+\text{C}(\text{t}+2)(\text{t}+3)}{(\text{t}+2)(\text{t}+3)(\text{t}+4)}$
$\Rightarrow 4t^2 + 3 = A(t + 3)(t+ 4 ) + B(t + 2)(t + 4) + C(t + 2)(t + 3)$
putting $t + 3 = 0$
$⇒ t = -3$
$\therefore$ $4 \times (-3)^2 + 3 = B(-3 + 2)(-3 + 4)$
$\Rightarrow 39 = B(-1)$
$\Rightarrow B = -39$
Putting$ t + 2 = 0$
$⇒ t = -2$
$\therefore$ $4(-2)^2 + 3 = A(-2 + 3)(-2 + 4)$
$\therefore$ $19 = A \times 1 \times 2$
$\Rightarrow\text{A}=\frac{19}{2}$
Let $t + 4 = 0$
$\Rightarrow t = -4$
$\therefore$ $4 \times (-4)^2 + 3 = C(-4 + 2)(-4 + 3)$
$\Rightarrow 67 = C(-2)(-1)$
$\Rightarrow\text{C}=\frac{67}{2}$
$\therefore\frac{4\text{t}^2+3}{(\text{t}+2)(\text{t}+3)(\text{t}+4)}=\frac{19}{2(\text{t}+2)}-\frac{39}{\text{x}^2+3}-\frac{39}{(2\text{t}+4)}$
$\Rightarrow\frac{4\text{t}^2+3}{(\text{x}^2+2)(\text{x}^2+3)(\text{x}^2+4)}=\frac{19}{2(\text{x}^2+2)}-\frac{39}{\text{x}^2+3}+\frac{67}{2(\text{x}^2+4)}$
$\therefore\text{I}=\frac{19}{2}\int\frac{\text{dx}}{\text{x}^2+(\sqrt{2})^2}-39\int\frac{\text{dx}}{\text{x}^2+(\sqrt{3})^2}-\frac{67}{2}\int\frac{\text{dx}}{\text{x}^2+2^2}$
$=\frac{19}{2}\times\frac{1}{\sqrt{2}}\tan^{-1}\Big(\frac{\text{x}}{\sqrt{2}}\Big)-\frac{39}{\sqrt{3}}\tan^{-1}\Big(\frac{\text{x}}{\sqrt{3}}\Big)-67\times\frac{1}2{}\tan^{-1}\Big(\frac{\text{x}}{2}\Big)+\text{C}$
$=\frac{19}{2\sqrt{2}}\tan^{-1}\Big(\frac{\text{x}}{\sqrt{2}}\Big)-\frac{39}{\sqrt{3}}\tan^{-1}\Big(\frac{\text{x}}{\sqrt{3}}\Big)-\frac{67}{4}\tan^{-1}\Big(\frac{\text{x}}{2}\Big)+\text{C}$
View full question & answer→Question 215 Marks
Evaluate the following integrals:
$\int(3\text{x}+1)\sqrt{4-3\text{x}-2\text{x}^2}\text{dx}$
Answer$\text{I}=\int(3\text{x}+1)\sqrt{4-3\text{x}-2\text{x}^2}\text{dx}$Let $(3\text{x}+1)=\text{A}\frac{\text{d}}{\text{dx}}(4-3\text{x}-2\text{x}^2)+\text{B}$
$\Rightarrow(3\text{x}+1)=\text{A}(-3-4\text{x})+\text{B}$
$\Rightarrow(3\text{x}+1)=-4\text{A}\text{x}+(\text{B}-3\text{A})$
$\Rightarrow3=-4\text{A}\text{ and }(\text{B}-3\text{A})=1$
$\Rightarrow\text{A}=-\frac{3}{4}\text{ and }\text{B}=-\frac{5}{4}$
$\Rightarrow(3\text{x}+1)=-\frac{3}{4}(-3-4\text{x})-\frac{5}{4}$
$\Rightarrow\text{I}=-\frac{3}{4}(-3-4\text{x})\sqrt{4-3\text{x}-2\text{x}^2}\text{dx}\\-\frac{5}{4}\int\sqrt{4-3\text{x}-2\text{x}^2}$
Let $\text{I}=-\frac{3}{4}\text{I}_1-\frac{5}{4}\text{I}_2\ \dots(1)$
Now,
$\text{I}_1=\int(-3-4\text{x})\sqrt{4-3\text{x}-2\text{x}^2}\text{dx}$
Let $(4-3\text{x}-2\text{x}^2)=\text{t},\text{ or, }(-3-4\text{x})\text{dx = dt}$
$\Rightarrow\text{I}_1=\int\sqrt{\text{t}}\text{dt}$
$=\frac{2}{3}\text{t}^{\frac{3}{2}}+\text{c}_1$
$\Rightarrow\text{I}_1=\frac{2}{3}(4-3\text{x}-2\text{x}^2)^{\frac{3}{2}}+\text{c}_1$
$\text{I}_2=\int\sqrt{4-3\text{x}-2\text{x}^2}\text{dx}$
$=\int\sqrt{2\Big(2-\frac{3}{2}\text{x}-\text{x}^2\Big)}\text{dx}$
$=\sqrt2\int\sqrt{\Big(\frac{17}{4}-\frac{9}{4}-\frac{3}{2}\text{x}-\text{x}^2\Big)}\text{dx}$
$=\sqrt2\int\sqrt{\Big[\Big(\frac{\sqrt{17}}{2}\Big)^2-\Big(\frac{9}{4}+\frac{3}{2}\text{x}+\text{x}^2\Big)\Big]}\text{dx}$
$=\sqrt2\int\sqrt{\Big[\Big(\frac{\sqrt{17}}{2}\Big)^2-\Big(\text{x}+\frac{3}{2}\Big)^2\Big]}\text{dx}$
$=\sqrt2\sin\Bigg(\frac{\text{x}+\frac{3}{2}}{\frac{\sqrt{17}}{2}}\Bigg)+\text{c}_2$
$=\sqrt2\sin\Big(\frac{2\text{x}+3}{\sqrt{17}}\Big)+\text{c}_2$
Using (1), we get
$\text{I}=-\frac{3}{4}\times\frac{2}{3}(4-3\text{x}-2\text{x}^2)^{\frac{3}{2}}\\-\frac{5}{4}\times\sqrt2\sin\Big(\frac{2\text{x}+3}{\sqrt{17}}\Big)+\text{C}$
$\therefore\ \text{I}=-\frac{1}{2}(4-3\text{x}-2\text{x}^2)^{\frac{3}{2}}-\frac{5\sqrt2}{4}\sin\Big(\frac{2\text{x}+3}{\sqrt{17}}\Big)+\text{C}$
View full question & answer→Question 225 Marks
Evaluate the following integrals:$\int\frac{\text{x}+2}{2\text{x}^2+6\text{x}+5}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\text{x}+2}{2\text{x}^2+6\text{x}+5}\text{ dx}$
Let $\text{x}+2=\lambda\frac{\text{d}}{\text{dx}}\big(2\text{x}^2+6\text{x}+5\big)+\mu$
$=\lambda(4\text{x}+6)+\mu$
$\text{x}+2=(4\lambda)\text{x}+(6\lambda+\mu)$
Comparing the coefficients of like powers of x,
$4\lambda=1\Rightarrow\lambda=\frac{1}{4}$
$6\lambda+\mu=2\Rightarrow6\Big(\frac{1}{4}\Big)+\mu=2$
$\mu=\frac{1}{2}$
So, $\text{I}=\int\frac{\frac{1}{4}(4\text{x}+6)+\frac{1}{2}}{2\text{x}^2+6\text{x}+5}\text{ dx}$
$\text{I}=\frac{1}{4}\int\frac{4\text{x}+6}{2\text{x}^2+6\text{x}+5}\text{ dx}+\frac{1}{2}\int\frac{1}{2\text{x}^2+6\text{x}+5}\text{ dx}$
$\text{I}=\frac{1}{4}\int\frac{4\text{x}+6}{2\text{x}^2+6\text{x}+5}\text{ dx}+\frac{1}{2}\int\frac{1}{\text{x}^2+3\text{x}+\frac{5}{2}}\text{ dx}$
$\text{I}=\frac{1}{4}\int\frac{4\text{x}+6}{2\text{x}^2+6\text{x}+5}\text{ dx}+\frac{1}{4}\int\frac{1}{\text{x}^2+2\text{x}\big(\frac{3}{2}\big)+\big(\frac{3}{2}\big)^2-\big(\frac{3}{2}\big)^3+\frac{5}{2}}\text{ dx}$
$\text{I}=\frac{1}{4}\int\frac{4\text{x}+6}{2\text{x}^2+6\text{x}+5}\text{ dx}+\frac{1}{4}\int\frac{1}{\big(\text{x}+\frac{3}{2}\big)^2+\frac{1}{4}}\text{ dx}$
$\text{I}=\int\frac{4\text{x}+6}{2\text{x}^2+6\text{x}+5}\text{ dx}+\frac{1}{4}\int\frac{1}{\big(\text{x}+\frac{3}{2}\big)^2+\big(\frac{1}{2}\big)^2}\text{ dx}$
$\text{I}=\frac{1}{4}\int\frac{4\text{x}+6}{2\text{x}^2+6\text{x}+5}\text{ dx}+\frac{1}{4}\times\frac{1}{\frac{1}{2}}\tan^{-1}\bigg(\frac{\text{x}+\frac{3}{2}}{\frac{1}{2}}\bigg)+\text{C dx}$ $\Big[\text{Since }\int\frac{1}{\text{x}^2+\text{a}^2}\text{ dx}=\frac{1}{\text{a}}\tan^{-1}\Big(\frac{\text{x}}{\text{a}}\Big)+\text{C}\Big]$
$\text{I}=\frac{1}{4}\log\big|2\text{x}^2+6\text{x}+5\big|+\frac{1}{2}\tan^{-1}(2\text{x}+3)+\text{C}$
View full question & answer→Question 235 Marks
Evaluate the following integrals:
$\int\frac{1}{\sin^3\text{x}\cos^5\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\frac{1}{\sin^3\text{x}\cos^5\text{x}}\text{dx}\ ...(\text{i})$
Then, $\text{I}=\int\sin^{-3}\text{x}\cos^{-5}\text{x}\text{ dx}$
Since -3 - 5 = -8, which is given integer. So, we divide both numerator and denominator by $\cos^8\text{x}$
$\therefore\ \text{I}=\int\frac{\frac{1}{\cos^8\text{x}}}{\frac{\sin^3\text{x}\cos^5\text{x}}{\cos^8\text{x}}}\text{ dx}$
$=\int\frac{\sec^8\text{x}}{\tan^3\text{x}}\text{ dx}$
$=\int\frac{(\sec^2\text{x})^3}{\tan^3\text{x}}\sec^2\text{x}\text{ dx}$
$=\int\frac{(1+\tan^2\text{x})^3}{\tan^3\text{x}}\sec^2\text{x}\text{ dx}$
$\text{I}=\int\frac{\big(1+\tan^6\text{x}+3\tan^4\text{x}+3\tan^2\text{x}\big)\sec^2\text{x}}{\tan^3\text{x}}\text{ dx}\ ...(\text{ii})$
Let $\text{t}=\tan\text{x}$ Then
$\text{d}(\tan\text{x})=\text{dt}$
$\sec^2\text{x}\text{ dx}=\text{dt}$
$\therefore\ \text{I}\int\frac{\big(1+\text{t}^6+3\text{t}^4+3\text{t}^2\big)}{\text{t}^3}\text{ dt}$
$=\int\big(\text{t}^{-3}+\text{t}^3+3\text{t}+3\text{t}^{-1}\big)\text{dt}$
$=-\frac{\text{t}^{-2}}{2}+\frac{\text{t}^{4}}{4}+\frac{3}{2}\text{t}^2+3\log\text{t}+\text{C}$
$=-\frac{1}{2\text{t}^{2}}+\frac{\text{t}^{4}}{4}+\frac{3}{2}\text{t}^2+3\log\text{t}+\text{C}$
$=-\frac{1}{2}\times\frac{1}{\tan^2\text{x}}+\frac{\tan^4\text{x}}{4}+\frac{3}{2}\times\tan^2\text{x}+3\log|\tan\text{x}|+\text{C}$
$\therefore\ \text{I}=\frac{-1}{2\tan^2\text{x}}+3\log|\tan\text{x}|+\frac{3}{2}\tan^2\text{x}+\frac{1}{4}\times\tan^4\text{x}+\text{C}$
View full question & answer→Question 245 Marks
Evaluate the following integrals:
$\int\frac{1}{\sin\text{x}+\sin2\text{x}}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{1}{\sin\text{x}+\sin2\text{x}}\ \text{dx}$
$=\int\frac{\text{dx}}{\sin\text{x}+2\sin\text{x}\cos\text{x}}$
$=\int\frac{\sin\text{x dx}}{(1-\cos^2\text{x})+2(1-\cos^2\text{x})\cos\text{x}}$
Let $\cos\text{x}=\text{t}\Rightarrow-\sin\text{x dx}=\text{dt}$
$\therefore\text{I}=\int\frac{\text{dt}}{(\text{t}^2-1)+2(\text{t}^2-1)\text{t}} $
$=\int\frac{\text{dt}}{(\text{t}^2-1)(1+2\text{t})}$
Let $\int\frac{1}{(\text{t}^2-1)(1+2\text{t})}=\frac{\text{A}}{\text{t}-1}+\frac{\text{B}}{\text{t}+1}+\frac{\text{C}}{1+2\text{t}}$
Put t = 1
⇒ 1 = 6A $\Rightarrow\text{A}=\frac{1}{6}$
Put t = -1
⇒ 1 = 2B $\Rightarrow\text{B}=\frac{1}{2}$
put $\text{t}=-\frac{1}{2}$
$\Rightarrow1=-\frac{3}{4}\text{C}\Rightarrow\text{C}=-\frac{4}{3}$
Thus,
$\text{I}=\frac{1}{6}\int\frac{\text{dt}}{\text{t}-1}+\frac{1}{2}\int\frac{\text{dt}}{\text{t}+1}-\frac{4}{3}\int\frac{\text{dt}}{1+2\text{t}}$
$=\frac{1}{6}\log|\text{t}-1|+\frac{1}{2}\log|\text{t}+1|-\frac{2}{3}\log|1+2\text{t}|+\text{C}$
Hence,
$\text{I}=\frac{1}{6}\log|\cos\text{x}-1|+\frac{1}{2}\log|\cos\text{x}+1|\\-\frac{2}{3}\log|1+2\cos\text{x}|+\text{C}$
View full question & answer→Question 255 Marks
Evaluate the following integrals:$\int\frac{(3\sin\text{x}-2)\cos\text{x}}{5-\cos^2\text{x}-4\sin\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{(3\sin\text{x}-2)\cos\text{x}}{5-\cos^2\text{x}-4\sin\text{x}}\text{ dx}$ $\therefore\ \text{I}=\int\frac{(3\sin\text{x}-2)\cos\text{x}}{5-(1-\sin^2\text{x})-4\sin\text{x}}\text{ dx}$ $\Rightarrow\text{I}=\int\frac{(3\sin\text{x}-2)\cos\text{x}}{5-1+\sin^2\text{x}-4\sin\text{x}}\text{ dx}$ Substitute $\sin\text{x}=\text{t}$ $\Rightarrow\cos\text{x}\text{ dx}=\text{dt}$ Thus, $\text{I}=\int\frac{(3\text{t}-2)}{4+\text{t}^2-4\text{t}}\text{ dt}$ $\text{I}=\int\frac{(3\text{t}-2)}{\text{t}^2-4\text{t}+4}\text{ dt}$ $\text{I}=\int\frac{(3\text{t}-2)}{(\text{t}-2)^2}\text{ dt}$ Now let us separate the integrand into the simplest form using partial fractions. $\frac{(3\text{t}-2)}{(\text{t}-2)^2}=\frac{\text{A}}{(\text{t}-2)}+\frac{\text{B}}{(\text{t}-2)^2}$ $=\frac{\text{A}(\text{t}-2)+\text{B}}{(\text{t}-2)^2}$ $=\frac{\text{At}-2\text{A}+\text{B}}{(\text{t}-2)^2}$ $\Rightarrow3\text{t}-2=\text{At}-2\text{A}+\text{B}$ Comparing the coefficients, we have, $\text{A}=3$and
$-2\text{A}+\text{B}=-2$ Substituting the value of A = 3 in the above equation, we have, $\Rightarrow-2\times3+\text{B}=-2$ $\Rightarrow-6+\text{B}=-2$ $\Rightarrow\text{B}=6-2$ $\Rightarrow\text{B}=4$Thus, $\text{I}=\int\frac{(3\text{t}-2)}{(\text{t}-2)^2}\text{ dt}$ becomes,
$\text{I}=\int\frac{3}{(\text{t}-2)^2}\text{ dt}+\int\frac{4}{(\text{t}-2)^2}\text{ dt}$ $=3\log|\text{t}-2|-4\Big(\frac{1}{\text{t}-2}\Big)+\text{C}$ $=3\log|2-\text{t}|+4\Big(\frac{1}{2-\text{t}}\Big)+\text{C}$ Now, substituting $\text{t}=\sin\text{x},$ we have, $=3\log|2-\sin\text{x}|+4\Big(\frac{1}{2-\sin\text{x}}\Big)+\text{C}$
View full question & answer→Question 265 Marks
Evaluate the following integrals:$\int\frac{(1-\text{x}^2)}{\text{x}(1-2\text{x})}\text{ dx}$
AnswerLet $\text{I}=\int\frac{1-\text{x}^2}{\text{x}(1-2\text{x})}\text{ dx}$ $=\int\frac{1-\text{x}^2}{\text{x}-2\text{x}^2}\text{ dx}$ $=\int\frac{1-\text{x}^2}{2\text{x}^2-\text{x}}\text{ dx}$ $=\int\bigg[\frac{1}{2}+\frac{\frac{\text{x}}{2}-1}{2\text{x}^2-\text{x}}\bigg]\text{dx}$ $\text{I}=\frac{1}{2}\text{x}+\int\frac{\frac{\text{x}}{2}-1}{2\text{x}^2-\text{x}}\text{ dx}+\text{C}_1\ ....(1)$ $\text{I}_1=\int\frac{2\text{x}+1}{\text{x}^2-\text{x}}\text{ dx}$ Let $\frac{\text{x}}{2}-1=\lambda\frac{\text{d}}{\text{dx}}\big(2\text{x}^2-\text{x}\big)+\mu$ $=\lambda(4\text{x}-1)+\mu$ $\frac{\text{x}}{2}-1=(4 \lambda)\text{x}-\lambda+\mu$Comparing the coefficients of like powers of x,
$\frac{1}{2}=4\lambda\Rightarrow\lambda=\frac{1}{8}$ $-\lambda+\mu=-1\Rightarrow-\Big(\frac{1}{8}\Big)+\mu=-1$ $\mu=-\frac{7}{8}$ So, $\text{I}_1=\int\frac{\frac{1}{8}(4\text{x}-1)-\frac{7}{8}}{2\text{x}^2-\text{x}}\text{ dx}$ $\text{I}=\frac{1}{8}\int\frac{4\text{x}-1}{2\text{x}^2-\text{x}}\text{ dx}-\frac{7}{8}\int\frac{1}{2\big(\text{x}^2-\frac{\text{x}}{2}\big)}\text{ dx}$ $\text{I}=\frac{1}{8}\int\frac{4\text{x}-1}{2\text{x}^2-\text{x}}\text{ dx}-\frac{7}{16}\int\frac{1}{\text{x}^2-2\text{x}\big(\frac{1}{4}\big)+\big(\frac{1}{4}\big)^2-\big(\frac{1}{4}\big)^2}\text{ dx}$ $\text{I}=\frac{1}{8}\int\frac{4\text{x}-1}{2\text{x}^2-\text{x}}\text{ dx}-\frac{7}{16}\int\frac{1}{\big(\text{x}-\frac{1}{4}\big)^2-\big(\frac{1}{4}\big)^2}\text{ dx}$ $\text{I}=\frac{1}{8}\log\big|2\text{x}^2+\text{x}\big|-\frac{7}{16}\times\frac{1}{2\big(\frac{1}{24}\big)}\log\bigg|\frac{\text{x}-\frac{1}{4}-\frac{1}{4}}{\text{x}-\frac{1}{4}+\frac{1}{4}}\bigg|+\text{C}_2$ $\Big[\text{since},\int\frac{1}{\text{x}^2-\text{a}^2}\text{ dx}=\frac{1}{2\text{a}}\log\Big|\frac{\text{x}-\text{a}}{\text{x}+\text{a}}\Big|+\text{C}\Big]$ $\text{I}=\frac{1}{8}\log|\text{x}|+\frac{1}{8}\log|2\text{x}-1|-\frac{7}{8}\log|1-2\text{x}|+\frac{7}{8}\log2+\frac{7}{8}\log|\text{x}|+\text{C}_2$ $\text{I}_1=\log\big|\text{x}\big|-\frac{3}{4}\log|1-2\text{x}|+\text{C}_3\ ....(2)$ $\Big[\text{Say},\text{C}_3=\text{C}_2+\frac{7}{8}\log2\Big]$ Using equation (1) and (2) $\text{I}=\frac{1}{2}\text{x}+\log|\text{x}|-\frac{3}{4}\log|1-2\text{x}|+\text{C}$
View full question & answer→Question 275 Marks
Evaluate the following integrals:
$\int\frac{1}{\sin^4\text{x}\cos^2\text{x}}\text{dx}$
Answer$\int\frac{1}{\sin^4\text{x}\cos^2\text{x}}\text{dx}$
Dividing numerator & denominator by $\sin^2\text{x}$
$=\int\frac{\frac{1}{\sin^2\text{x}}}{\sin^4\text{x }\cdot\ \cot^2\text{x}}\text{dx}$
$=\int\frac{\text{cosec}^6\text{x}}{\cot^2}\text{dx}$
$=\int\frac{\text{cosec}^4\text{x }\cdot\text{ cosec}^2\text{x}\text{ dx}}{\cot^2\text{x}}$
$=\int\frac{(1+\cot^2\text{x})^2\text{cosec}^2\text{x}\text{ dx}}{\cot^2\text{x}}$
Let $\cot\text{x}=\text{t}$
$-\text{cosec}^2\text{x}=\frac{\text{dt}}{\text{dx}}$
$-\text{cosec}^2\text{x}\text{ dx}=\text{dt}$
Now, $\int\frac{(1+\cot^2\text{x})^2\text{cosec}^2\text{x}\text{ dx}}{\cot^2\text{x}}$
$=\int\Big(\frac{1+\text{t}^2}{\text{t}}\Big)^2(-\text{dt})$
$=-\int\frac{\big(1+\text{t}^4+2\text{t}^2\big)}{\text{t}^2}\text{dt}$
$=-\int(\text{t}^{-2}+\text{t}^2+2)\text{dt}$
$=-\Big[\frac{\text{t}^{-2+1}}{-2+1}+\frac{\text{t}^3}{3}+2\text{t}\Big]+\text{C}$
$=-\Big[-\frac{1}{\text{t}}+\frac{\text{t}^3}{3}+2\text{t}\Big]+\text{C}$
$=-\frac{1}{3}\text{t}^3-2\text{t}+\frac{1}{\text{t}}+\text{C}$
$=-\frac{1}{3}\cot^3\text{x}-2\cot\text{x}+\frac{1}{\cot\text{x}}+\text{C}$
$=-\frac{1}{3}\cot^3\text{x}-2\cot\text{x}+\tan\text{x}+\text{C}$
View full question & answer→Question 285 Marks
Evaluate the following intregals:
$\int\frac{\text{x}+2}{\sqrt{\text{x}^2-1}}\text{dx}$
AnswerLet $\text{x}+2=\text{A}\frac{\text{d}}{\text{dx}}(\text{x}^2-1)+\text {B}\ ...(1)$
$\Rightarrow\text{x}+2=\text{A}(2\text{x})+\text{B}$
Equating the coefficient of x and constant term on both sides, we obtain
$2\text{A}=1\Rightarrow\text{A}=\frac{1}{2}$
$\text{B}=2$
From (1), we obtain
$(\text{x}+2)=\frac{1}{2}(2\text{x})+2$
Then, $\int\frac{\text{x}+2}{\sqrt{\text{x}^2-1}}\text{dx}=\int\frac{\frac{1}{2}(2\text{x})+2}{\sqrt{\text{x}^2-1}}\text{dx}$
$=\frac{1}{2}\int\frac{2\text{x}}{\sqrt{\text{x}^2-1}}\text{dx}+\int\frac{2}{\sqrt{\text{x}^2-1}}\text{dx}\ ...(2)$
$\text{In }\frac{1}{2}\int\frac{2\text{x}}{\sqrt{\text{x}^2-1}}\text{dx}, $
Let $\text{x}^2-1=\text{t}\Rightarrow2\text{x}\text{ dx}=\text{dt}$
In $\frac{1}{2}\int\frac{2\text{x}}{\sqrt{\text{x}^2-1}}\text{dx}$
let $\text{x}^2-1=\text{t}\Rightarrow2\text{x} \text{ dx}=\text{dt}$
$\frac{1}{2}\int\frac{2\text{x}}{\sqrt{\text{x}^2-1}}\text{dx}=\frac{1}{2}\int\frac{\text{dt}}{\sqrt{\text{t}}}$
$=\frac{1}{2}\big[2\sqrt{\text{t}}\big]$
$=\sqrt{\text{t}}$
$=\sqrt{\text{x}^2-1}$
Then, $\int\frac{2}{\sqrt{\text{x}^2-1}}\text{dx}=2\int\frac{1}{\sqrt{\text{x}^2-1}}\text{dx}=2\log\big|\text{x}+\sqrt{\text{x }+1}\big|$
From equation (2), we obtain
$\int\frac{\text{x}+2}{\sqrt{\text{x}^2-1}}\text{dx}=\sqrt{\text{x}^2-1}+2\log\big|\text{x}+\sqrt{\text{x}^2-1}\big|+\text{C}$
View full question & answer→Question 295 Marks
$\int\frac{\text{x}}{\sqrt{\text{x}+\text{a}}-\sqrt{\text{x}+\text{b}}}\text{dx}$
Answer$\int\frac{\text{x}}{\sqrt{\text{x}+\text{a}}-\sqrt{\text{x}+\text{b}}}\text{dx}$
$=\int\frac{\text{x}}{\sqrt{\text{x}+\text{a}}-\sqrt{\text{x}+\text{b}}}\times\frac{\sqrt{\text{x}+\text{a}}+\sqrt{\text{x}+\text{b}}}{\sqrt{\text{x}+\text{a}}+\sqrt{\text{x}+\text{b}}}\text{dx}$
$=\int\frac{\text{x}(\sqrt{\text{x}+\text{a}}+\sqrt{\text{x}+\text{b}})}{(\sqrt{\text{x}+\text{a}})^2-(\sqrt{\text{x}+\text{b}})^2}\text{dx}$
$=\int\frac{\text{x}(\sqrt{\text{x}+\text{a}}+\sqrt{\text{x}+\text{b}})}{\text{x}+\text{a}-\text{x}-\text{b}}\text{dx}$
$=\frac{1}{\text{a}-\text{b}}\int\text{x}(\sqrt{\text{x}+\text{a}}+\sqrt{\text{x}+\text{b}})\text{dx}$
$=\frac{1}{\text{a}-\text{b}}\big[\int\text{x}(\sqrt{\text{x}+\text{a}})\text{dx}+\int\text{x}(\sqrt{\text{x}+\text{b}})\text{dx}\big]$
$=\frac{1}{\text{a}-\text{b}}\big[\int(\text{x}+\text{a}-\text{a})(\sqrt{\text{x}+\text{a}})\text{dx}+\int(\text{x}+\text{b}-\text{b})(\sqrt{x+\text{b}})\text{dx}$
$=\frac{1}{\text{a}-\text{b}}\big[\int(\text{x}+\text{a})(\sqrt{\text{x}+\text{a}})\text{dx}-\text{a}\int(\sqrt{\text{x}+\text{a}})\text{dx}\\+\int(\text{x}+\text{b})(\sqrt{\text{x}+\text{b}})\text{dx}-\text{b}\int(\sqrt{\text{x}+\text{b}})\text{dx}\big]$
$=\frac{1}{\text{a}-\text{b}}\big[\int(\text{x}+\text{a})^\frac{3}{2}\text{dx}-\text{a}\int(\text{x}+\text{a})^\frac{1}{2}\text{dx}+\int(\text{x}+\text{b})^\frac{3}{2}\text{dx}-\text{b}\int(\text{x}+\text{b})^\frac{1}{2}\text{dx}\big]$
$=\frac{1}{\text{a}-\text{b}}\Big[\frac{(\text{x}+\text{a})^\frac{5}{2}}{\frac{5}{2}}-\text{a}\frac{(\text{x}+\text{a})^\frac{3}{2}}{\frac{3}{2}}+\frac{(\text{x}+\text{b})^\frac{5}{2}}{\frac{5}{2}}-\text{b}\frac{(\text{x}+\text{b})^\frac{3}{2}}{\frac{3}{2}}+\text{c}$ where, c is an arbitrary constant.
$=\frac{1}{\text{a}-\text{b}}\Big[\frac{2}{5}(\text{x}+\text{a})^\frac{5}{2}-\frac{2\text{a}}{3}(\text{x}+\text{a})^\frac{3}{2}+\frac{2}{5}(\text{x}+\text{b})^\frac{5}{2}-\frac{2\text{b}}{3}(\text{x}+\text{b})^\frac{3}{2}\Big]+\text{c}$ where, c is an arbitrary constant.
Hence, $\int\frac{\text{x}}{\sqrt{\text{x}+\text{a}}-\sqrt{\text{x}+\text{b}}}\text{dx}=$ $\frac{1}{\text{a}-\text{b}}\Big[\frac{2}{5}(\text{x}+\text{a})^\frac{5}{2}-\frac{2\text{a}}{3}(\text{x}+\text{a})^\frac{3}{2}+\frac{2}{5}(\text{x}+\text{b})^\frac{5}{2}-\frac{2\text{b}}{3}(\text{x}+\text{b})^\frac{3}{2}\Big]+\text{c}$ where, c is an arbitrary constant.
View full question & answer→Question 305 Marks
Evaluate the following integrals:$\int\frac{\sin^{-1}\text{x}}{\text{x}^2}\text{dx}$
AnswerLet $\text{I}=\int\frac{\sin^{-1}\text{x}}{\text{x}^2}\text{dx}$
$=\int\Big(\frac{1}{\text{x}^2}\Big)(\sin^{-1}\text{x})\text{dx}$
$\text{I}=\Big[\sin^{-1}\text{x}\int\frac{1}{\text{x}^2}\text{dx}-\int\Big(\frac{1}{\sqrt{1-\text{x}^2}}\int\frac{1}{\text{x}^2}\text{dx}\Big)\text{dx}\Big]$
$=\sin^{-1}\text{x}\big(-\frac{1}{\text{x}}\big)-\int\frac{1}{\sqrt{1-\text{x}^2}}\Big(-\frac{1}{\text{x}}\Big)\text{dx}$
$\text{I} =-\frac{1}{\text{x}}\sin^{-1}\text{x}+\int\frac{1}{\text{x}\sqrt{1-\text{x}^2}}\text{dx}$
$\text{I}=-\frac{1}{\text{x}}\sin^{-1}\text{x}+\text{I}_1 \dots(1)$
Where,
$\text{I}_1=\int\frac{1}{\text{x}\sqrt{1-\text{x}^2}}\text{dx}$
Let $1-\text{x}^2=\text{t}^2$
$-2\text{x dx}=2\text{t dt}$
$\text{I}_1=\int\frac{\text{x}}{\text{x}^2\sqrt{1-\text{x}^2}}\text{dx}$
$=-\int\frac{\text{tdt}}{(1-\text{t}^2)\sqrt{\text{t}}}$
$=-\int\frac{\text{dt}}{(1-\text{t}^2)}$
$=\int\frac{1}{\text{t}^2-1}\text{dt}$
$=\frac{1}{2}\log\Big|\frac{\text{t}-1}{\text{t+1}}\Big|$
$=\frac{1}2{\log}\Big|\frac{\sqrt{1-\text{x}^2}-1}{\sqrt{1-\text{x}^2}+1}\Big|+\text{C}_1$
Now
$\text{I}=-\frac{\sin^{-1}\text{x}}{\text{x}}+\frac{1}{2}\log\bigg|\Big(\frac{\sqrt{1-\text{x}^2}-1}{\sqrt{1-\text{x}^2+1}}\Big)\Big(\frac{\sqrt{1-\text{x}^2}-1}{\sqrt{1-\text{x}^2-1}}\Big)\bigg|+\text{C}$
$=-\frac{\sin^{-1}\text{x}}{\text{x}}+\frac{1}2{}\log\bigg|\frac{(\sqrt{1-\text{x}^2}-1)^2}{1-\text{x}^2-1}\bigg|+\text{C}$
$=-\frac{\sin^{-1}\text{x}}{\text{x}}+\frac{1}{2}\log\bigg|\frac{(\sqrt{1-\text{x}^2}-1)^2}{-\text{x}^2}\bigg|+\text{C}$
$=-\frac{\sin^{-1}\text{x}}{\text{x}}+\log\bigg|\frac{\sqrt{1-\text{x}^2}-1}{-\text{x}}\bigg|+\text{C}$
$\text{I}=-\frac{\sin^{-1}\text{x}}{\text{x}}+\log\bigg|\frac{1-\sqrt{1-\text{x}}^2}{\text{x}}\bigg|+\text{C}$
View full question & answer→Question 315 Marks
Evaluvate the following intregals:
$\int\frac{1}{\text{p}+\text{q}\tan\text{x}}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{1}{\text{p}+\text{q}\tan\text{x}}\ \text{dx}$
$=\int\frac{1}{\text{P}+\frac{q\sin\text{x}}{\cos\text{x}}}\ \text{dx}$
$=\int\frac{\cos\text{x}}{q\sin\text{x}+\text{p}\cos\text{x}}\ \text{dx}$
Let $\cos\text{x}=\text{A}(q\sin\text{x}+\text{p}\cos\text{x})+\text{B}(q\cos\text{x}-\text{p}\sin\text{x})$
$\Rightarrow\cos\text{x}=(\text{Ap}+\text{Bq})\cos\text{x}+(\text{Aq}-\text{Bp})\sin\text{x}$
Compairing coefficient of like terms
$\text{Ap}+\text{Bq}=0\dots(1)$
$\text{Aq}+\text{Bp}=1\dots(2)$
Multipiying eq (1) by p and eq (2) by q and then adding
$\Rightarrow\text{Ap}^2+\text{Bpq}=\text{p}$
$\Rightarrow\text{Aq}^2+\text{Bpq}=0$
$\Rightarrow\text{A}=\frac{\text{p}}{\text{p}^2\text{q}^2}$
Putting value of A in eq (1)
$\frac{\text{p}^2}{\text{p}^2+\text{q}^2}+\text{Bq}=1$
$\Rightarrow\text{Bq}=1-\frac{\text{p}^2}{\text{p}^2+\text{q}^2}$
$\Rightarrow\text{Bq}=\frac{\text{p}^2+\text{q}^2-\text{p}^2}{\text{p}^2+\text{q}^2}$
$\Rightarrow\text{B}=\frac{\text{q}}{\text{p}^2+\text{q}^2}$
$\therefore\text{I}=\int\Big[\frac{\text{p}^2}{\text{p}^2+\text{q}^2}\times\frac{(\text{q}\sin\text{x}+\text{p}\cos\text{x})}{(q\sin\text{x}+\text{p}\cos\text{x})}+\frac{\text{q}}{\text{p}^2+\text{q}^2}\times\frac{(q\cos\text{x}-\text{p}\sin\text{x})}{(q\sin\text{x}+\text{p}\cos\text{x})}\Big]\text{dx}$
$=\frac{\text{q}}{\text{p}^2+\text{q}^2}\int\text{dx}+\frac{\text{p}^2}{\text{p}^2+\text{q}^2}\int\Big(\frac{q\cos\text{x}-\text{p}\sin\text{x}}{q\sin\text{x}+\text{p}\cos\text{x}}\Big)\text{dx}$
Putting $\text{q}\sin\text{x}+\text{p}\cos\text{x}=\text{t}$
$\Rightarrow(\text{q}\cos\text{x}-\text{p}\sin\text{x})\text{ dx}=\text{dt}$
$\therefore\text{I}=\frac{\text{p}}{\text{q}+\text{q}^2}\int\text{dx}+\frac{\text{q}}{\text{p}^2+\text{q}^2}\int\frac{1}{\text{t}}\text{dt}$
$=\frac{\text{p}}{\text{q}+\text{q}^2}\text{x}+\frac{\text{q}}{\text{p}^2+\text{q}^2}\ln|\text{q}\sin\text{x}+\text{p}\cos\text{x}|+\text{C}$
View full question & answer→Question 325 Marks
Evaluate the following integrals:
$\int\frac{1}{\sin\text{x}(3+2\cos\text{x})}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{1}{\sin\text{x}(3+2\cos\text{x})}\ \text{dx}$
$=\frac{\sin\text{x dx}}{\sin^2\text{x}(3+2\cos\text{x })}$
$=\frac{\sin\text{x dx}}{(1-\cos^2\text{x})(3+2\text{x})}$
Let $\cos\text{x}=\text{t}$
$\Rightarrow-\sin\text{x dx}=\text{dt}$
$\therefore\text{I}=\int\frac{\text{dt}}{(\text{t}^2-1)(3+2\text{t})}$
Now,
Let $\frac{1}{(\text{t}^2-1)(3+2\text{t})}=\frac{\text{A}}{\text{t}-1}+\frac{\text{B}}{\text{t}+1}+\frac{\text{C}}{3+2\text{t}}$
$\Rightarrow 1 = A(t + 1)(3 + 2t) + B(t - 1)(3 + 2t) + C (t^2- 1)$
Put $t = -1$
$\Rightarrow 1 = -2B$ $\Rightarrow\text{A}=\frac{1}{10}$
Put $\text{t}=-\frac{3}{2}$
$\Rightarrow1=\frac{5}{4}\text{C}\Rightarrow\text{C}=\frac{4}{5}$
Thus,
$\text{I}=\frac{1}{10}\int\frac{\text{dt}}{\text{t}-1}-\frac{1}{2}\int\frac{\text{dt}}{\text{t}+1}+\frac{5}{4}\int\frac{\text{dt}}{3+2\text{t}}$
$=\frac{1}{10}\log|\text{t}-1|=\frac{1}{2}\log|\text{t}+1|+\frac{2}{5}\log|3+2\text{t}|+\text{C}$
Hence,
$\text{I}=\frac{1}{10}\log|\cos\text{x}-1|-\frac{1}{2}\log|\cos\text{x}+1|+\frac{2}{5}\log|3+2\cos\text{x}|+\text{C}$
View full question & answer→Question 335 Marks
Evaluate the following integrals:
$\int\frac{(\text{x}^2+1)(\text{x}^2+2)}{(\text{x}^2+3)(\text{x}^2+4)}\ \text{dx}$
Answer$\text{f}(\text{x})=\frac{(\text{x}^2+1)(\text{x}^2+2)}{(\text{x}^2+3)(\text{x}^2+4)}$
Now,
$\frac{(\text{x}^2+1)(\text{x}^2+2)}{(\text{x}^2+3)(\text{x}^2+4)}$
$=\frac{\text{x}^4+3\text{x}^2+2}{\text{x}^4+7\text{x}^2+12}$
$=\frac{(\text{x}^4+7\text{x}^2+12)-4\text{x}^2-10}{\text{x}^4+7\text{x}^2+12}$
$=1-\frac{4\text{x}^2+10}{\text{x}^4+7\text{x}^2+12}$
Now,
$\frac{4\text{x}^2+10}{\text{x}^4+7\text{x}^2+12}=\frac{4\text{x}^2+10}{(\text{x}^2+3)(\text{x }^2+4)}$
let $\frac{4\text{x}^2+10}{(\text{x}^2+3)(\text{x}^2+4)}=\frac{\text{Ax}+\text{B}}{\text{x}^2+3}+\frac{\text{Cx}+\text{D}}{\text{x}^2+4}$
$\Rightarrow4\text{x}^2+10=(\text{Ax}+\text{B})(\text{x}^2+4)+(\text{Cx}+\text{D})(\text{x}^2+3)$
let x = 0, we get
10 = 4B + 3D ...(1)
If x = 1, we get
14 = 5 (A + B) + 4 (C + D) = 5A + 5B + 4C + 4D ...(2)
If x = -1, we get
14 = 5 (-A + B) + 4 (-C + D) = -5A + 5B + -4C + 4D ...(3)
Applying (2) and (3), we get
28 = 10B + 8D
⇒ 14 = 5B + 4D ...(4)
From (1) we get,
10 = 4B + 3D
Multiplying equations (4) by 3 and (1) by 4 and substracting, we get
42 - 40 = 15B - 16G
⇒ 2 = -B
or B = -2 ...(5)
putting value of (B) in (1), we get
10 = 4 (-2) + 3D
$\frac{10+8}{3}=\text{D}$
⇒ D = 6
Compairing coefficient of $x^3$ in
$4x^2 + 10 = (Ax + B)(x^2+ 4) + (Cx + 4)(x^2+3)$, we get,
$0 = A + C$ ...(8)
Compairing coefficient of x, we get
$0 = 4A + 3C$
$\Rightarrow A = C = 0$
$\therefore\text{f}(\text{x})=1-\frac{(-2)}{\text{x}^2+3}-\frac{6}{\text{x}^2+4}$
$=1+\frac{2}{\text{x}^2+3}-\frac{6}{\text{x}^2+4}$
$\therefore\int\text{f}(\text{x})\text{dx}=\int1+\frac{2}{\text{x}^2+3}-\frac{6}{\text{x}^2+4}\ \text{dx}$
$=\text{x}+\frac{2}{\sqrt{3}}\tan^{-1}\times\frac{\text{x}}{\sqrt{3}}-3\tan^{-1}\frac{\text{x}}{2}+\text{C}$
View full question & answer→Question 345 Marks
Evaluate the following integrals:$\int\frac{\text{x}^2+\text{x}+1}{\text{x}^2-\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\text{x}^2+\text{x}+1}{\text{x}^2-\text{x}}\text{ dx}$ $=\int\Big[1+\frac{2\text{x}+1}{\text{x}^2-\text{x}}\Big]\text{dx}$ $=\text{x}+\int\frac{2\text{x}+1}{\text{x}^2-\text{x}}\text{ dx}+\text{C}_1\ ....(1)$ $\text{I}_1=\int\frac{2\text{x}+1}{\text{x}^2-\text{x}}\text{ dx}$ Let $2\text{x}+1=\lambda\frac{\text{d}}{\text{dx}}\big(\text{x}^2-\text{x}\big)+\mu$ $=\lambda(2\text{x}-1)+\mu$ $2\text{x}+1=(2\lambda)\text{x}-\lambda+\mu$Comparing the coefficients of like powers of x,
$2=2\lambda\Rightarrow\lambda=1$ $-\lambda+\mu=1\Rightarrow\mu=2$ So, $\text{I}_1=\int\frac{(2\text{x}-1)+2}{\text{x}^2-\text{x}}\text{ dx}$ $\text{I}=\int\frac{2\text{x}-1}{\text{x}^2-\text{x}}\text{ dx}+2\int\frac{1}{\text{x}^2-\text{x}}\text{ dx}$ $\text{I}=\int\frac{2\text{x}-1}{\text{x}^2-\text{x}}\text{ dx}+2\int\frac{1}{\text{x}^2-2\text{x}\big(\frac{1}{2}\big)+\big(\frac{1}{2}\big)^2-\big(\frac{1}{2}\big)^2}\text{ dx}$ $\text{I}=\int\frac{2\text{x}-1}{\text{x}^2-\text{x}}\text{ dx}+2\int\frac{1}{\big(\text{x}-\frac{1}{2}\big)^2-\big(\frac{1}{2}\big)^2}\text{ dx}$ $\text{I}=\log\big|\text{x}^2+\text{x}\big|+2\times\frac{1}{2\big(\frac{1}{2}\big)}\log\bigg|\frac{\text{x}-\frac{1}{2}-\frac{1}{2}}{\text{x}-\frac{1}{2}+\frac{1}{2}}\bigg|+\text{C}_1$ $\Big[\text{since},\int\frac{1}{\text{x}^2-\text{a}^2}\text{ dx}=\frac{1}{2\text{a}}\log\Big|\frac{\text{x}-\text{a}}{\text{x}+\text{a}}\Big|+\text{C}\Big]$ $\text{I}_1=\log\big|\text{x}^2+\text{x}\big|+2\log\Big|\frac{\text{x}-1}{\text{x}}\Big|+\text{C}_2\ ....(2)$ Using equation (1) and (2) $\text{I}=\text{x}+\log\big|\text{x}^2+\text{x}\big|+2\log\Big|\frac{\text{x}-1}{\text{x}}\Big|+\text{C}$
View full question & answer→Question 355 Marks
Evaluate the following integrals:$\int\frac{1}{\cos\text{x}+\text{cosec x}}\text{dx}$
AnswerLet $\text{I}=\int\frac{1}{\cos\text{x}+\text{cosec x}}\text{dx}$
$=\int\frac{\sin\text{x}}{1+\cos\text{x}\sin\text{x}}\text{dx}$
$=\int\frac{2\sin\text{x}}{2+2\cos\text{x}\sin\text{x}}\text{dx}$
$=\int\frac{\sin\text{x}+\cos\text{x}+\sin\text{x}-\cos\text{x}}{2+2\cos\text{x}\sin\text{x}}\text{dx}$
$=\int\frac{\sin\text{x}+\cos\text{x}}{2+2\cos\text{x}\sin\text{x}}\text{dx}+\int\frac{\sin\text{x}-\cos\text{x}}{2+2\cos\text{x}\sin\text{x}}\text{dx}$
$=\int\frac{\sin\text{x}+\cos\text{x}}{3-\sin^2\text{x}-\cos^2\text{x}+2\cos\text{x}\sin\text{x}}\text{dx}+\int\frac{\sin\text{x}-\cos\text{x}}{1+\sin^2\text{x}+\cos^2\text{x}+2\cos\text{x}\sin\text{x}}\text{dx}$
$=\int\frac{\sin\text{x}+\cos\text{x}}{3-(\sin\text{x}-\cos\text{x})^2}\text{dx}+\int\frac{\sin\text{x}-\cos\text{x}}{1+(\sin\text{x}+\cos\text{x})^2}\text{dx}$
$=\text{I}_1+\text{I}_2 \dots(1)$
Where, $\text{I}_1=\int\frac{\sin\text{x}+\cos\text{x}}{3-(\sin\text{x}-\cos\text{x}^2)}\text{dx}$ and $\text{I}_2=\int\frac{\sin\text{x}-\cos\text{x}}{1+(\sin\text{x}+\cos\text{x})^2}\text{dx}$
Now,
$\text{I}_1=\int\frac{\sin\text{x}+\cos\text{x}}{3-(\sin\text{x}-\cos\text{x})^2}\text{dx}$
Let $(\sin\text{x}-\cos\text{x})=\text{t}$
On differentiating both sides, we get
$(\cos\text{x}+\sin\text{x})\text{dx = dt}$
$\therefore\text{I}_1=\int\frac{1}{3-(\text{t})^2}\text{dt}$
$=\frac{1}{2\sqrt{3}}\log\bigg|\frac{\sqrt{3}+\text{t}}{\sqrt{3}-\text{t}}\bigg|+\text{C}_1$
$=\frac{1}{2\sqrt{3}}\log\bigg|\frac{\sqrt{3}+\sin\text{x}-\cos\text{x}}{\sqrt{3}-(\sin\text{x}-\cos\text{x})}\bigg|+\text{C}_1 \dots(2)$
Now,
$\text{I}_2=\int\frac{\sin\text{x}-\cos\text{x}}{1+(\sin\text{x}+\cos\text{x})^2}\text{dx}$
Let $(\sin\text{x}+\cos\text{x})=\text{t}$
On differentiating both sides, we get
$(\cos\text{x}-\sin\text{x})\text{dx = dt}$
$\therefore\text{I}_2=-\int\frac{1}{1+(\text{t})^2}\text{dt}$
$=-\tan^{-1}\text{t}+\text{c}_2$
$=-\tan^{-1}(\sin\text{x}+\cos\text{x})+\text{c}_2 \dots(3)$
On substituting (2) and (3) in (1), we get
$\text{I}=\frac{1}{2\sqrt{3}}\log\bigg|\frac{\sqrt{3}+\sin\text{x}-\cos\text{x}}{\sqrt{3}-\sin\text{x}+\cos\text{x}}\bigg|-\tan^{-1}(\sin\text{x}+\cos\text{x})+\text{C}$
Hence, $\int\frac{1}{\cos\text{x}+\text{cosec x}}\text{dx}=\frac{1}{2\sqrt{3}}\log\bigg|\frac{\sqrt{3}+\sin\text{x}-\cos\text{x}}{\sqrt{3}-\sin\text{x}+\cos\text{x}}\bigg|-\tan^{-1}(\sin\text{x}+\cos\text{x})+\text{C}$
View full question & answer→Question 365 Marks
Evaluate the following integrals:$\int\frac{5\text{x}-2}{1+2\text{x}+3\text{x}^2}\text{ dx}$
AnswerLet $\text{I}=\int\frac{5\text{x}-2}{1+2\text{x}+3\text{x}^2}\text{ dx}$ Rewriting the numerator we have, $5\text{x}-2=\text{A}\frac{\text{d}}{\text{dx}}\big(1+2\text{x}+3\text{x}^2\big)+\text{B}$ $\Rightarrow5\text{x}-2=\text{A}(2+6\text{x})+\text{B}$ $\Rightarrow5\text{x}-2=6\text{xa}+2\text{A}+\text{B}$ Comparing the coefficient, we have, $6\text{A}=5$ and $2\text{A}+\text{B}=-2$ $\Rightarrow\text{A}=\frac{5}{6}$ Substituting the value of A in 2A + B = -2, we have, $2\times\frac{5}{6}+\text{B}=-2$ $\Rightarrow\frac{10}{6}+\text{B}=-2$ $\Rightarrow\text{B}=-2-\frac{10}{6}$ $\Rightarrow\text{B}=\frac{-12-10}{6}$ $\Rightarrow\text{B}=\frac{-22}{6}$ $\Rightarrow\text{B}=\frac{-11}{3}$ $5\text{x}-2=\frac{5}{6}(2+6\text{x})-\frac{11}{3}$Thus, $\text{I}=\int\frac{5\text{x}-2}{1+2\text{x}+3\text{x}^2}\text{ dx}$ becomes,
$\text{I}=\int\frac{\big[\frac{5}{6}(2+6\text{x})-\frac{11}{3}\big]}{3\text{x}^2+2\text{x}+1}\text{ dx}$ $=\frac{5}{6}\int\frac{(2+6\text{x})}{3\text{x}^2+2\text{x}+1}\text{ dx}-\frac{11}{3}\int\frac{\text{dx}}{3\text{x}^2+2\text{x}+1}$ $=\frac{5}{6}\log\big(3\text{x}^2+2\text{x}+1\big)-\frac{11}{3\times3}\int\frac{\text{dx}}{\text{x}^2+\frac{2}{3}\text{x}+\frac{1}{3}}+\text{C}$ $=\frac{5}{6}\log\big(3\text{x}^2+2\text{x}+1\big)-\frac{11}{9}\int\frac{\text{dx}}{\text{x}^2+\frac{2}{3}\text{x}+\big(\frac{4}{3}\big)^2+\frac{1}{3}-\big(\frac{4}{3}\big)^2}+\text{C}$ $=\frac{5}{6}\log\big(3\text{x}^2+2\text{x}+1\big)-\frac{11}{9}\int\frac{\text{dx}}{\big(\text{x}+\frac{1}{3}\big)^2+\Big(\frac{\sqrt2}{3}\Big)^2}+\text{C}$ $=\frac{5}{6}\log\big(3\text{x}^2+2\text{x}+1\big)-\frac{11}{9}\times\frac{1}{\frac{\sqrt2}{3}}\tan^{-1}\Bigg[\frac{\big(\text{x}+\frac{1}{3}\big)}{\frac{\sqrt2}{3}}\Bigg]+\text{C}$ $=\frac{5}{6}\log\big(3\text{x}^2+2\text{x}+1\big)-\frac{11}{9}\times\frac{3}{\sqrt2}\tan^{-1}\Bigg[\frac{\big(\frac{3\text{x}+1}{3}\big)}{\frac{\sqrt2}{3}}\Bigg]+\text{C}$ $=\frac{5}{6}\log\big(3\text{x}^2+2\text{x}+1\big)-\frac{11}{3\sqrt2}\tan^{-1}\Big[\frac{3\text{x}+1}{\sqrt2}\Big]+\text{C}$
View full question & answer→Question 375 Marks
Evalute the following integrals:
$\int\frac{1}{(\text{x}+1)^2(\text{x}^2+1)}\ \text{dx}$
AnswerWe have,
$\text{I}=\int\frac{1}{(\text{x}+1)^2(\text{x}^2+1)}\ \text{dx}$
Let $\frac{1}{(\text{x}+1)^2(\text{x}^2+1)}=\frac{\text{A}}{\text{x}+1}+\frac{\text{Bx}+\text{C}}{(\text{x}+1)^2}+\frac{\text{Cx}+\text{D}}{\text{x}^2+1}$
$\Rightarrow\frac{1}{(\text{x}+1)^2(\text{x}^2+1)}=\frac{\text{A}(\text{x}+1)(\text{x}^2+1)+\text{B}(\text{x}^2+1)(\text{Cx}+\text{D})(\text{x}+)^2}{(\text{x}+1)^2(\text{x}^2+1)}$
$\Rightarrow1=\text{A}(\text{x}^3+\text{x}+\text{x}^2+1)+\text{B}(\text{x}^2+1)+(\text{Cx}+\text{D})(\text{x}^2+2\text{x}+1)$
$\Rightarrow1=\text{A}(\text{x}^3+\text{x}^2+\text{x}+1)+\text{B}(\text{x}^2+1)+\text{Cx}^3+2\text{Cx}^2+\text{Cx}+\text{Dx}^2+2\text{Dx}+\text{D}$
$\Rightarrow1=(\text{A}+\text{C})\text{x}^3+(\text{A}+\text{B}+2\text{C}+\text{D})\text{x}^2+(\text{A}+\text{C}+2\text{D})\text{x}+\text{A}+\text{B}+\text{D}$
Equating coefficients of like terms
A + C = 0 ...(1)
A + B + 2C + D = 0 ...(2)
A + C + 2D = 0 ...(3)
A + B + D = 1 ...(4)
$\text{A}=\frac{1}{2},\text{B}=\frac{1}{2},\text{C}=-\frac{1}2{}$ and $\text{D}=0$
$\therefore\frac{1}{(\text{x}+1)^2(\text{x}^2+1)}=\frac{1}{2(\text{x}+1)}+\frac{1}{2(\text{x}+1)^2}-\frac{1}{2}\times\frac{\text{x}}{\text{x}^2+1}$
$\Rightarrow\int\frac{\text{dx}}{(\text{x}+1)^2(\text{x}^+1)}=\frac{1}{2}\int\frac{\text{dx}}{\text{x}+1}+\frac{1}{2}\int\frac{\text{dx}}{(\text{x}+1)^2}-\frac{1}{2}\int\frac{\text{x dx}}{\text{x}^2+1}$
Putting $\text{x}^2+1=\text{dt}$
$\Rightarrow2\text{x dx}=\text{dt}$
$\Rightarrow\text{x dx}=\frac{\text{dt}}{2}$
$\therefore\text{I}=\frac{1}{2}\int\frac{\text{dx}}{\text{x}+1}+\frac{1}{2}\int\frac{\text{dx}}{(\text{x}+1)^2}-\frac{1}{4}\int\frac{\text{dt}}{\text{t}}$
$=\frac{1}{2}\log|\text{x}+1|-\frac{1}{2(\text{x}+1)}-\frac{1}{4}\log|\text{t}|+\text{C}'$
$=\frac{1}{2}\log|\text{x}+1|-\frac{1}{2(\text{x}+1)}-\frac{1}{4}\log|\text{x}^2+1|+\text{C}'$
View full question & answer→Question 385 Marks
Evaluate the following integrals:$\int\frac{\text{x}^3-3\text{x}}{\text{x}^4+2\text{x}^2-4}\text{ dx}$
Answer$\text{I}=\int\frac{\text{x}^3-3\text{x}}{\text{x}^4+2\text{x}^2-4}\text{ dx}$ $=\int\frac{\text{x}(\text{x}^2-3)}{\text{x}^4+2\text{x}^2-4}\text{ dx}$ Let $\text{x}^2=\text{t},$ or, $2\text{x}\text{ dx}=\text{dt}$ $\text{I}=\frac{1}{2}\int\frac{(\text{t}-3)}{\text{t}^2+2\text{t}-4}\text{ dt}$ $=\frac{1}{4}\int\frac{2\text{t}-6}{\text{t}^2+2\text{t}-4}\text{ dt}$$=\frac{1}{4}\int\frac{2\text{t}+2-8}{\text{t}^2+2\text{t}-4}\text{ dt}$
$=\frac{1}{4}\int\Big(\frac{2\text{t}+2}{\text{t}^2+2\text{t}-4}-\frac{8}{\text{t}^2+2\text{t}-4}\Big)\text{dt}$
$=\frac{1}{4}\Big(\int\frac{2\text{t}+2}{\text{t}^2+2\text{t}-4}\text{ dt}-\int\frac{8}{\text{t}^2+2\text{t}-4}\text{ dt}\Big)$ $\Rightarrow\text{I}=\frac{1}{4}(\text{I}_1+\text{I}_2)\ ....(1)$ Now, $\text{I}_1=\int\frac{2\text{t}+2}{\text{t}^2+2\text{t}-4}\text{ dt}$ Let $\text{t}^2+2\text{t}-4=\text{u}$ or, $(2\text{t}+2)\text{dt}=\text{du}$ $\Rightarrow\text{I}_1=\int\frac{1}{\text{u}}\text{ du}=\int|\text{u}|+\text{C}_1$ $\Rightarrow\text{I}_1=\ln|\text{t}^2+2\text{t}-4|+\text{C}_1$ $\therefore\ \text{I}_1=\ln|\text{x}^4+2\text{x}^2-4|+\text{C}_1$ Now, $\text{I}_2=\int\frac{-8}{(\text{t}+1)^2-5}\text{ dt}$ $\Rightarrow\text{I}_2=\int\frac{8}{\big(\sqrt5\big)^2-(\text{t}+1)^2}\text{ dt}$ $\therefore\ \text{I}_2=\frac{8}{2\sqrt5}\ln\Big|\frac{\sqrt5+\text{x}^2+1}{\sqrt5-\text{x}^2-1}\Big|+\text{C}_2$ So, from (1), we get $\text{I}=\frac{1}{4}\Big[\ln|\text{x}^4+2\text{x}^2-4|+\frac{4}{\sqrt5}\ln\Big|\frac{\sqrt5+\text{x}^2+1}{\sqrt5-\text{x}^2-1}\Big|\Big]+\text{C}$ $\therefore\ \text{I}=\frac{1}{4}\ln|\text{x}^4+2\text{x}^2-4|+\frac{1}{\sqrt5}\ln\Big|\frac{\sqrt5+\text{x}^2+1}{\sqrt5-\text{x}^2-1}\Big|+\text{C}$
View full question & answer→Question 395 Marks
Evaluvate the following intregals
$\int\frac{1}{4+3\tan\text{x}}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{1}{4+3\tan\text{x}}\ \text{dx}$
$\text{I}=\int\frac{\cos\text{x}}{4\cos\text{x}+3\sin\text{x}}\ \text{dx}$
Let $\cos\text{x}=\lambda\frac{\text{d}}{\text{dx}}(4\cos\text{x}+3\cos\text{x})+\mu(4\cos\text{x}+3\sin\text{x})+\text{v}$
$\cos\text{x}=\lambda(-4\sin\text{x}+3\cos\text{x})+\mu(4\cos\text{x}+3\sin\text{x})+\text{v}$
$\cos\text{x}=(-4\lambda+3\mu)\sin\text{x}+(3\lambda+4\mu)\cos\text{x}+\text{v}$
Compairing the coefficient of $\sin\text{x}\ \&\cos\text{x}$ on the both the sides,
$-4\lambda+3\mu=0\ \dots\dots(1)$
$3\lambda+4\mu=1\ \dots\dots(2)$
$\text{v}=0\ \dots\dots(3)$
solving the equation (1), (2) and (3),
$\lambda=\frac{3}{25}$
$\mu=\frac{4}{25}$
$\text{v}=0$
$\text{I}=\int\frac{3}{25}\frac{(-4\sin\text{x}+3\cos\text{x})}{(4\cos\text{x}+3\sin\text{x})}\ \text{dx}+\frac{4}{25}\int\text{dx}$
$\text{I}=\frac{3}{25}\log|4\cos\text{x}+3\sin\text{x}|+\frac{4}{25}\text{x}+\text{C}$
View full question & answer→Question 405 Marks
Evaluate the following intregals:
$\int\frac{1}{5+7\cos\text{x}+\sin\text{x}}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{1}{5+7\cos\text{x}+\sin\text{x}}\ \text{dx}$
Putting $\sin\text{x}=\frac{2\tan\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}\text{ and }\cos\text{x}=\frac{1-\tan^2\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}$
Now,
$\text{I}=\int\frac{1}{5+\frac{7\Big(1-\tan\frac{\text{x}}{2}\Big)}{\Big(1+\tan^2\frac{\text{x}}{2}\Big)}+\Bigg(\frac{2\tan\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}\Bigg)}\text{ dx}$
$=\int\frac{\Big(1+\tan^2\frac{\text{x}}{2}\Big)}{5\Big(1+\tan^2\frac{\text{x}}{2}\Big)+7-7\tan^2\frac{\text{x}}{2}+2\tan\frac{\text{x}}{2}}\ \text{dx}$
$=\int\frac{\sec^2\frac{\text{x}}{2}}{-2\tan^2\frac{\text{x}}{2}+12+2\tan\frac{\text{x}}{2}}\ \text{dx}$
Let $\tan\frac{\text{x}}{2}=\text{t}$
$\frac{1}{2}\sec^2\frac{\text{x}}{2}\text{dx}=\text{dt}$
$\text{I}=\int\frac{2\text{dt}}{-2\text{t}^2+12+2\text{t}}$
$=-\int\frac{\text{dt}}{\text{t}^2-\text{t}-6}$
$=-\int\frac{\text{dt}}{\text{t}^2-2\text{t}\Big(\frac{1}{2}\Big)+\Big(\frac{1}{2}\Big)^2-\Big(\frac{1}{2}\Big)^2-6}$
$=-\int\frac{\text{dt}}{\Big(\text{t}-\frac{1}{2}\Big)^2-\Big(\frac{5}{2}\Big)^2}$
$=-\frac{1}{2\Big(\frac{5}{2}\Big)}\log\Bigg|\frac{\text{t}-\frac{1}{2}-\frac{5}{2}}{\text{t}-\frac{1}{2}+\frac{5}{2}}\Bigg|+\text{C}$
$=-\frac{1}{5}\log\Big|\frac{\text{t}-3}{\text{t}+2}\Big|+\text{C}$
$\text{I}=\frac{1}{5}\log\Bigg|\frac{\tan\frac{\text{x}}{2}+2}{\tan\frac{\text{x}}{2}-3}\Bigg|+\text{C}$
View full question & answer→Question 415 Marks
Evaluate the following integrals:
$\int\frac{\text{x}^2}{(\text{x}^2+1)(3\text{x}^2+4)}\ \text{dx}$
AnswerWe have,
$\text{I}=\int\frac{\text{x}^2\text{dx}}{(\text{x}^2+1)(3\text{x}^2+4)} $
Putting $x^2 = t$
Then, $\frac{\text{x}^2}{(\text{x}^2+1)(3\text{x}^2+4)}=\frac{\text{t}}{(\text{t}+1)(3\text{t}+4)}$
Let $\frac{\text{t}}{(\text{t}+1)(3\text{t}+4)}=\frac{\text{A}}{\text{t}+1}+\frac{\text{B}}{3\text{t}+4}$
$\Rightarrow\frac{\text{t}}{(\text{t}+1)(3\text{t}+4)}=\frac{\text{A}(3\text{t}+4)+\text{B}(\text{t}+1)}{(\text{t}+1)(3\text{t}+4)}$
$\Rightarrow\text{t}=\text{A}(3\text{t}+4)+\text{B}(\text{t}+1)$
putting $t + 1 = 0$
$⇒ t = -1$
$\therefore$ -$1 = A (-3 + 4) + 0$
$⇒ A = -1$
putting $3t + 4 = 0$
$\Rightarrow\text{t}=-\frac{4}{3}$
$\therefore-\frac{4}{3}=0+\text{B}\big(-\frac{4}3{}+1\big)$
$\Rightarrow-\frac{4}{3}=\text{B}\times\big(-\frac{1}{3}\big)$
$\Rightarrow\text{B}=4$
$\therefore\frac{\text{t}}{(\text{t}+1)(3\text{t}+4)}=-\frac{1}{\text{t}+1}+\frac{4}{3\text{t}+4}$
$\Rightarrow\frac{\text{x}^2}{(\text{x}^2+1)(3\text{x}^2+4)}=\frac{1}{\text{x}^2+1}+\frac{4}{3\text{x}^2+4}$
$$$\Rightarrow\frac{\text{x}^2}{(\text{x}^2+1)(3\text{x}^2+4)}=\frac{1}{\text{x}^2+1}+\frac{4}{3\big(\text{x}^2+\frac{4}{3}\big)}$
$\Rightarrow\frac{\text{x}^2}{(\text{x}^2+1)(3\text{x}^2+4)}=-\int\frac{\text{dx}}{\text{x}^2+1}+\frac{4}{3}\int\frac{\text{dx}}{\text{x}^2+\big(\frac{2}{\sqrt{3}}\big)^2}$
$=-\tan^{-1}(\text{x})+\frac{4}{3}\times\frac{\sqrt{3}}{2}\tan^{-1}\big(\frac{\sqrt{3}\text{x}}{2}\big)+\text{C}$
$=-\tan^{-1}(\text{x})+\frac{2}{\sqrt{3}}\tan^{-1}\Big(\frac{\sqrt{3}\text{x}}{2}\Big)+\text{C}$
View full question & answer→Question 425 Marks
Evaluate the following integrals:
$\int\frac{\text{x}^3}{(\text{x}^2+1)^3}\text{dx}$
AnswerLet I $=\int\frac{\text{x}^3}{(\text{x}^2+1)^3}\text{dx}\ ....(1)$
Let $1 + x^2 = t$ then,
$d(1 + x^2) = dt$
$\Rightarrow2\text{x dx}=\text{dt}$
$\Rightarrow\text{x dx}=\frac{\text{dt}}{2}$
Puttting $1 + x^2 = t$ and $\text{x dx}=\frac{\text{dt}}{2}$ in equation (1), we get
$\text{I}=\int\frac{\text{x}^2}{\text{t}^3}\times\frac{\text{dt}}{2}$
$=\frac{1}{2}\int\frac{(\text{t}-1)}{\text{t}^3}\text{dt}\ \ [\because1+\text{x}^2=\text{t}]$
$=\frac{1}{2}\int\Big[\Big(\frac{\text{t}}{\text{t}^3}-\frac{1}{\text{t}^3}\Big)\text{dt}\Big]$
$=\frac{1}{2}\int\big(\text{t}^{-2}-\text{t}^{-3}\big)\text{dt}$
$=\frac{1}{2}\Big[-1\text{t}^{-1}-\frac{\text{t}^{-2}}{-2}\Big]+\text{C}$
$=\frac{1}{2}\Big[-\frac{1}{\text{t}}+\frac{1}{2\text{t}^2}\Big]+\text{C}$
$=-\frac{1}{2\text{t}}+\frac{1}{4\text{t}^2}+\text{C}$
$=-\frac{1}{2(1+\text{x}^2)}+\frac{1}{4(1+\text{x}^2)^2}+\text{C}$
$=\frac{-2(1+\text{x}^2)+1}{4(1+\text{x}^2)^2}+\text{C}$
$=\frac{-2-2\text{x}^2+1}{4(1+\text{x}^2)^2}+\text{C}$
$=\frac{-2\text{x}^2-1}{4(1+\text{x}^2)^2}+\text{C}$
$=-\frac{(1+2\text{x}^2)}{4(\text{x}^2+1)^2}+\text{C}$
$\therefore\text{I}=-\frac{(1+2\text{x}^2)}{4(\text{x}^2+1)^2}+\text{C}$
View full question & answer→Question 435 Marks
Evaluate the following integrals:
$\int(4\text{x}+1)\sqrt{\text{x}^2-\text{x}-2}\text{dx}$
AnswerLet $\text{I}=\int(4\text{x}+1)\sqrt{\text{x}^2-\text{x}-2}\text{dx}$
Let $4\text{x}+1=\lambda\frac{\text{d}}{\text{dx}}(\text{x}^2-\text{x}-2)+\mu$
$=\lambda(2\text{x}-1)+\mu$
Equating similar terms, we get,
$2\lambda=4\Rightarrow\lambda=2$
$-\lambda+\mu=1\Rightarrow\mu=3$
So,
$\text{I}=\int(2(2\text{x}-1)+3)\sqrt{\text{x}^2-\text{x}-2}\text{dx}$
$=2\int(2\text{x}-1)\sqrt{\text{x}^2-\text{x}-2}\text{dx}+3\int\sqrt{\text{x}^2-\text{x}-2}\text{dx}$
Let $\text{x}^2-\text{x}-2=\text{t}$
$(2\text{x}-1)\text{dx = dt}$
$\therefore\ \text{I}=2\int\sqrt{\text{t}}\text{dt}+3\int\sqrt{\Big(\text{x}-\frac{1}{2}\Big)^2-\Big(\frac{3}{2}\Big)^2}\text{dx}$
$\Rightarrow\text{I}=2\frac{\text{t}^{\frac{3}{2}}}{\frac{3}{2}}+3\begin{Bmatrix}\frac{\Big(\text{x}-\frac{1}{2}\Big)}{2}\sqrt{\text{x}^2-\text{x}-2}\\-\frac{9}{8}\log\Big|\Big(\text{x}-\frac{1}{2}\Big)+\sqrt{\text{x}^2-\text{x}-2}\Big|\end{Bmatrix}+\text{C}$
Hence,
$\Rightarrow\text{I}=\frac{4}{3}(\text{x}^2-\text{x}-2)^{\frac{3}{2}}+\frac{3}{4}(2\text{x}-1)\sqrt{\text{x}^2-\text{x}-2}\\-\frac{27}{8}\log\Big|\Big(\text{x}-\frac{1}{2}\Big)+\sqrt{\text{x}^2-\text{x}-2}\Big|+\text{C}$
View full question & answer→Question 445 Marks
Evaluate the following integrals:
$\int\frac{1}{\text{x}^4+3\text{x}^2+1}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{1}{\text{x}^4+3\text{x}^2+1}\ \text{dx}$
Dividing numerator and denominator by $x^2$
$\therefore\text{I}=\int\frac{\frac{1}{\text{x}^2}}{\text{x}^2+3+\frac{1}{\text{x}^2}}\ \text{dx}$
$=\frac{1}{2}\int\frac{\Big(1+\frac{1}{\text{x}^2}\Big)-\Big(1-\frac{1}{\text{x}^2}\Big)}{\text{x}^2+3+\frac{1}{\text{x}^2}}\ \text{dx}$
$=\frac{1}{2}\int\frac{\Big(1+\frac{1}{\text{x}^2}\Big)}{\Big(\text{x}-\frac{1}{\text{x}}\Big)^2+5}\ \text{dx}-\frac{1}{2}\int\frac{\Big(1-\frac{1}{\text{x}^2}\Big)}{\Big(\text{x}+\frac{1}{\text{x}}\Big)^2+1}$
Let $\Big(\text{x}-\frac{1}{\text{x}}\Big)=\text{t}$
$\Rightarrow\Big(1+\frac{1}{\text{x}^2}\Big)\text{dx}=\text{dt}$
And $\text{x}+\frac{1}{\text{x}}=\text{z}$
$\Rightarrow\Big(1-\frac{1}{\text{x}^2}\Big)\text{dx}=\text{dz}$
$\therefore\text{}=\frac{1}{2}\int\frac{\text{dt}}{\text{t}^2+5}-\frac{1}{2}\int\frac{\text{dz}}{\text{z}^2+1}$
$=\frac{1}{2\sqrt{5}}\tan^{-1}\Big(\frac{\text{t}}{\sqrt{5}}\Big)-\frac{1}{2}\tan^{-1}(\text{z})+\text{C}$
Hence,
$\text{I}=\frac{1}{2\sqrt{5}}\tan^{-1}\Big(\frac{\text{x}^2-1}{\sqrt{5}\text{x}}\Big)-\frac{1}{2}\tan^{-1}\Big(\frac{\text{x}^2+1}{\text{x}}\Big)+\text{C}$
View full question & answer→Question 455 Marks
Evaluate the following integrals:
$\int\frac{1}{\text{x}^4+\text{x}^2+1}\ \text{dx}$
AnswerWe have
$\text{I}=\int\frac{1}{\text{x}^4+\text{x}^2+1}\ \text{dx}$
$=\frac{1}{2}\int\frac{2\text{dx}}{\text{x}^4+\text{x}^2+1}$
$\Rightarrow\int\bigg(\frac{(\text{x}^2+1)-(\text{x}^2-1)}{\text{x}^4+\text{x}^2+1}\bigg)\text{dx}$
$\Rightarrow\frac{1}{2}\int\Big(\frac{\text{x}^2+1}{\text{x}^2+\text{x}^2+1}\Big)\text{dx}-\frac{1}{2}\int\Big(\frac{\text{x}^2-1}{\text{x}^4+\text{x}^2+1}\Big)\text{dx}$
$\Rightarrow\frac{1}{2}\int\Big(\frac{\text{x}^2+1}{\text{x}^4+\text{x}^2+1}\Big)\text{dx}-\frac{1}{2}\int\Big(\frac{\text{x}^2-1}{\text{x}^4+\text{x}^2+1}\Big)\text{dx}$
Dividing numerator and denominator bt $x^2$
$\text{I}=\frac{1}{2}\int\Bigg(\frac{1+\frac{1}{\text{x}^2}}{\text{x}^2+1+\frac{1}{\text{x}^2}}\Bigg)\text{dx}-\frac{1}{2}\int\Bigg(\frac{1-\frac{1}{\text{x}^2}}{\text{x}^2+1+\frac{1}{\text{x}^2}}\Bigg)\text{dx}$
$=\frac{1}{2}\int\Bigg(\frac{1+\frac{1}{\text{x}^2}}{\text{x}^2+\frac{1}{\text{x}^2}-2+3}\Bigg)\text{dx}-\frac{1}{2}\int\frac{\Big(1-\frac{1}{\text{x}^2}\Big)\text{dx}}{\text{x}^2+\frac{1}{\text{x}^2}+2-1}$
$=\frac{1}{2}\int\frac{\Big(1+\frac{1}{\text{x}^2}\Big)\text{dx}}{\Big(\text{x}-\frac{1}{\text{x}}\Big)^2+(\sqrt{3})^2}-\frac{1}{2}\int\frac{\Big(1-\frac{1}{\text{x}^2}\Big)\text{dx}}{\Big(\text{x}+\frac{1}{\text{x}}\Big)-1^2}$
putting $\text{x}-\frac{1}{\text{x}}=\text{t}$
$\Rightarrow\Big(1-\frac{1}{\text{x}^2}\Big)\text{dx}=\text{dp}$
$\therefore\text{I}=\frac{1}{2}\int\frac{\text{dt}}{\text{t}^2+(\sqrt{3})^2}-\frac{1}{2}\int\frac{\text{dp}}{\text{p}^2-1^2}$
$=\frac{1}{2}\times\frac{1}{\sqrt{3}}\tan^{-1}\Big(\frac{\text{t}}{\sqrt{3}}\Big)-\frac{1}{2}\times\frac{1}{2\times1}\Big|\frac{\text{p}-1}{\text{p}+1}\Big|+\text{C}$
$=\frac{1}{2\sqrt{3}}\tan^{-1}\Big(\frac{\text{x}-\frac{1}{\text{x}}}{\sqrt{3}}\Big)-\frac{1}{4}\log\Bigg|\frac{\text{x}+\frac{1}{\text{x}}-1}{\text{x}+\frac{1}{\text{x}}+1}\Bigg|+\text{C}$
$=\frac{1}{2\sqrt{3}}\tan^{-1}\Big(\frac{\text{x}^2-1}{\text{x}\sqrt{3}}\Big)-\frac{1}{4}\log\Big|\frac{\text{x}^2-\text{x}+1}{\text{x}^2+\text{x}+1}\Big|+\text{C}$
View full question & answer→Question 465 Marks
Evaluate the following integrals:
$\int(2\text{x}+3)\sqrt{\text{x}^2+4\text{x}+3}\text{dx}$
AnswerLet $\text{I}=\int(2\text{x}+3)\sqrt{\text{x}^2+4\text{x}+3}\text{dx}$Let $(2\text{x}+3)=\lambda\frac{\text{d}}{\text{dx}}(\text{x}^2+4\text{x}+3)+\mu$
$=\lambda(2\text{x}+4)+\mu$ Equating similar terms, we get, $\lambda=1\text{ and }4\lambda+\mu=3$ $\Rightarrow\mu=-1$ So, $\text{I}=\int((2\text{x}+4)+(-1))\sqrt{\text{x}^2+4\text{x}+3}\text{dx}$ $=\int(2\text{x}+4)\sqrt{\text{x}^2+4\text{x}+3}\text{dx}-\int\sqrt{\text{x}^2+4\text{x}+3}\text{dx}$ Let $\text{x}^2+4\text{x}+3=\text{t}$ $\Rightarrow(2\text{x}+4)\text{dx = dt}$ $\therefore\ \text{I}=\int\sqrt{\text{t}}\text{dt}-\int\sqrt{(\text{x}+2)^2-1}\text{dx}$ $=\frac{3}{2}\text{t}^{\frac{3}{2}}-\frac{(\text{x}+2)}{2}\sqrt{\text{x}^2+4\text{x}+3}\\+\frac{1}{2}\log\Big|(\text{x}+2)+\sqrt{\text{x}^2+4\text{x}+3}\Big|+\text{C}$ Hence, $\text{I}=\frac{2}{3}(\text{x}^2+4\text{x}+3)^{\frac{3}{2}}-\Big(\frac{\text{x}+2}{2}\Big)\sqrt{\text{x}^2+4\text{x}+3}\\+\frac{1}{2}\log\Big|(\text{x}+2)+\sqrt{\text{x}^2+4\text{x}+3}\Big|+\text{C}$
View full question & answer→Question 475 Marks
Evaluate the following intregals:
$\int\frac{\text{ax}^2+\text{bx}+\text{c}}{(\text{x}-\text{a})(\text{x}-\text{b})(\text{x}-\text{c})}\ \text{dx},$ where a, b, c are distinct
AnswerWe have
$\text{I}=\int\frac{\text{ax}^2+\text{bx}+\text{c}}{(\text{x}-\text{a})(\text{x}-\text{b})(\text{x}-\text{c})}\ \text{dx}$
Let $\int\frac{\text{ax}^2+\text{bx}+\text{c}}{(\text{x}-\text{a})(\text{x}-\text{b})(\text{x}-\text{c})}=\frac{\text{A}}{\text{x}-\text{a}}+\frac{\text{B}}{(\text{x}-\text{b}) }+\frac{ \text{C}}{(\text{x}-\text{c})}$
$\Rightarrow\text{ax}^2+\text{bx}+\text{c}\\=\text{A}(\text{x}-\text{b})(\text{x}-\text{c})+\text{B}(\text{x}-\text{c})(\text{x}-\text{a})+\text{C}(\text{x}-\text{a})(\text{x}-\text{b}$
$\Rightarrow\text{ax}^2+\text{bx}+\text{c}\\=\text{A}[\text{x}^2-(\text{b}+\text{c})\text{x}+\text{bc}]+\text{B}[\text{x}^2-(\text{c}+\text{a})\text{x}+\text{ca}]\\+\text{C}[\text{x}^2-(\text{a}-\text{b})\text{x}+\text{ab}]$
$\Rightarrow\text{ax}^2+]\text{bx}+\text{c}=(\text{A}+\text{B}+\text{C})\text{x}^2-[\text{A}(\text{b}+\text{c})+\text{B}(\text{c}+\text{a}) \\+\text{C}(\text{a}+\text{b})]\text{x}+\text{Abc}+\text{Bca}+\text{Cab}$
Equation the coefficient on both sides, we get
$\text{a}=\text{A}+\text{B}+\text{C}\ ...(1)$
$\text{b}=-[\text{A}(\text{b}+\text{c})+\text{B}(\text{c}+\text{a})+\text{C}(\text{a}+\text{b})]\ ...(2)$
$\text{c}=\text{Abc}+\text{Bca}+\text{Cab}\ ...(3)$
Solving (1), (2), (3) we get
$\text{A}=\frac{\text{a}^2+\text{ab}+\text{c}}{(\text{a}-\text{b})(\text{a}-\text{c})}$
$\text{B}=\frac{\text{ab}^2+\text{b}^2+\text{c}^2}{(\text{b}-\text{a})(\text{b}-\text{c})}$
$\text{C}=\frac{\text{ac}^2+\text{bc}+\text{c}}{(\text{c}-\text{a})(\text{c}-\text{b})}$
$\therefore\text{I}=\int\Big[\frac{\text{a}^2+\text{ab}+\text{c}}{(\text{a}-\text{b})(\text{a}-\text{c})}\times\frac{1}{\text{x}-\text{a}}+\frac{\text{ab}^2+\text{b}^2+\text{c}}{(\text{b}-\text{a})(\text{b}-\text{c})}\times\frac{1}{\text{x}-\text{b}}\\+\frac{\text{ac}^2+\text{bc}^2+\text{c}}{(\text{c}-\text{a})(\text{c}-\text{b})}\times\frac{1}{\text{x}-\text{c}}\Big]\ \text{dx}$
$=\frac{\text{a}^2+\text{ab}+\text{c}}{(\text{a}-\text{b})(\text{a}-\text{c})}\log|\text{x}-\text{a}|+\frac{\text{a}^2+\text{ab}+\text{c}}{(\text{a}-\text{b})(\text{a}-\text{c})}\log|\text{x}-\text{b}|\\+\frac{\text{ac}^2+\text{bc}^2+\text{c}}{(\text{c}-\text{a})(\text{c}-\text{b})}\log|\text{x}-\text{c}|+\text{K}$
View full question & answer→Question 485 Marks
Evaluate the following intregals:
$\int\frac{\text{x}+1}{\sqrt{4+5\text{x}-\text{x}}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\text{x}+1}{\sqrt{4+5\text{x}-\text{x}^2}}\text{ dx}$
Let $\text{x}+1=\lambda\frac{\text{d}}{\text{dx}}(4+5\text{x}-\text{x}^2)+\mu$
$=\lambda(5-2\text{x})+\mu$
$\text{x}=(-2\lambda)\text{x}+5\lambda+\mu$
comparing the coefficient of like powers of x,
$-2\lambda=1\ \Rightarrow\lambda=-\frac{1}{2}$
$5\lambda+\mu=1\ \Rightarrow5\Big(-\frac{1}{2}\Big)+\mu=1$
$\mu=\frac{7}{2}$
So, $\text{I}=\int\frac{-\frac{1}{2}(5-2\text{x})+\frac{7}{2}}{\sqrt{4+\text{x}-\text{x}^2}}\ \text{dx}$
$=-\frac{1}{2}\int\frac{(5-2\text{x})}{\sqrt{4+5\text{x}-\text{x}^2}}\ \text{ dx}+\frac{7}{2}\int\frac{1}{\sqrt{-\big[\text{x}^2-2\text{x}\big(\frac{5}{2}\big)+\big(\frac{5}{2}\big)^2-\big(\frac{5}{2}\big)^2-4\big]}}\text{ dx}$
$\text{I}=-\frac{1}{2}\int\frac{5-2\text{x}}{\sqrt{4+5\text{x}-\text{x}^2}}\text{dx}+\frac{7}{2}\int\frac{1}{\sqrt{\Big[\Big(\frac{\sqrt{41}}{2}}\Big)^2-\Big(\text{x}-\frac52\Big)^2\Big]}\text{dx}$
$\text{I}=-\frac{1}{2}\int\frac{5-2\text{x}}{\sqrt{4+5\text{x}-\text{x}^2}}\text{dx}+\frac{7}{2}\int\frac{1}{\sqrt{-\Big[\big(\text{x}-\frac{\sqrt{41}}{2}\big)^2-\big(\text{x}-\frac{5}{2}^2}\Big]}\text{ dx}$
$\text{I}=\frac{1}{2}(2\sqrt{4+5\text{x}-\text{x}^2})+\frac{7}{2}\sin^{-1}\Bigg(\frac{\text{x}-\frac{5}{2}}{\frac{\sqrt{41}}{2}}\Bigg)+\text{c}$ $\big[\text{since},\int\frac{1}{\sqrt{\text{x}}}\text{dx}=2\sqrt{\text{x}}+\text{c},\int\frac{1}{\sqrt{\text{a}^2-\text{x}^2}}\text{dx}=\sin^{-1}\big(\frac{\text{x}}{\text{a}}\big)+\text{c}\big]$
$\text{I}=-\sqrt{4+5\text{x}-\text{x}^2}+\frac{7}{2}\sin^{-1}\Big(\frac{2\text{x}-5}{\sqrt{41}}\Big)+\text{c}$
View full question & answer→Question 495 Marks
Evaluate the following integrals:
$\int\frac{(\text{x}-1)^2}{\text{x}^4+\text{x}^2+1}\text{ dx}$
AnswerLet $\text{I}=\int\frac{(\text{x}-1)^2}{\text{x}^4+\text{x}^2+1}\text{ dx}$
$=\frac{\text{x}^2-2\text{x}+1}{\text{x}^4+1+\frac{1}{\text{x}^2}}\text{ dx}$
Dividing numerator and denominator by $x^2$
$\therefore\text{I}=\int\frac{1-\frac{2}{\text{x}}+\frac{1}{\text{x}^2}}{\text{x}^2+1+\frac{1}{\text{x}^2}}\ \text{dx}$
$=\int\frac{\Big(1+\frac{1}{\text{x}^2}\Big)}{\Big(\text{x}-\frac{1}{\text{x}}\Big)^2+3}\ \text{dx}-\int\frac{2\text{x}}{\text{x}^4+\text{x}^2+1}\ \text{dx}$
Let $\Big(\text{x}-\frac{1}{\text{x}}\Big)=\text{t}$
$\Rightarrow\Big(1+\frac{1}{\text{x}^2}\Big)\text{dx}=\text{dt}$ [for 1st part]
Let $\text{x}^2=\text{z}$
$\Rightarrow2\text{x dx}=\text{dz}$ [For 2nd part]
$\therefore\text{I}=\int\frac{\text{dt}}{\text{t}^2+3}-\int\frac{\text{dz}}{\text{z}^2+\text{z}+1}$
$=\int\frac{\text{dt}}{\text{t}^3+3}-\int\frac{\text{dz}}{\Big(\text{z}+\frac{1}{2}\Big)^2+\frac{3}{4}}$
$\Rightarrow\text{I}=\frac{1}{\sqrt{3}}\tan^{-1}\Big(\frac{\text{t}}{\sqrt{3}}\Big)-\frac{2}{\sqrt{3}}\tan^{-1}\Bigg(\frac{\text{z}+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\Bigg)+\text{C}$
$\Rightarrow\text{I}=\frac{1}{\sqrt{3}}\tan^{-1}\Big(\frac{\text{x}^2-1}{\sqrt{3}\text{x}}\Big)-\frac{2}{\sqrt{3}}\tan^{-1}\Big(\frac{2\text{z}+1}{\sqrt{3}}\Big)+\text{C}$
Hence,
$\text{I}=\frac{1}{\sqrt{3}}\tan^{-1}\Big(\frac{\text{x}^2-1}{\sqrt{3}\text{x}}\Big)-\frac{2}{\sqrt{3}}\tan^{-1}\Big(\frac{2\text{x}^2+1}{\sqrt{3}}\Big)+\text{C}$
View full question & answer→Question 505 Marks
Evaluate the following intregals:
$\int\frac{3\text{x}+1}{\sqrt{5-2\text{x}-\text{x}^2}}\text{ dx}$
Answer Let $\text{I}=\int\frac{3\text{x}+1}{\sqrt{5-2\text{x}-\text{x}^2}}\text{ dx}$
Let $3\text{x}+1=\lambda\frac{\text{d}}{\text{dx}}(5-2\text{x}-\text{x}^2)+\mu$
$=\lambda(-2-2\text{x})+\mu$
$3\text{x}+1=(-2\lambda)\text{x}-2\lambda+\mu$
Compairing the coefficient of like power of x,
$-2\lambda=3\ \Rightarrow\lambda=-\frac{3}{2}$
$-2\lambda+\mu=1\ \Rightarrow-2\Big(-\frac{3}{2}\Big)+\mu=1$
$\mu=-2$
So, $\text{I}=\int\frac{-\frac{3}{2}(-2-2\text{x})-2}{\sqrt{5-2\text{x}-\text{x}^2}}\text{ dx}$
$=-\frac{3}{2}\int\frac{(-2-2\text{x})}{\sqrt{5-2\text{x}-\text{x}^2}}\text{ dx}-2\int\frac{1}{\sqrt{-\big[\text{x}^2+2\text{x}-5\big]}}\text{ dx}$
$=-\frac{3}{2}\int\frac{(-2-2\text{x})}{\sqrt{5-2\text{x}-\text{x}^2}}\text{ dx}-2\int\frac{1}{\sqrt{-\big[\text{x}^2+2\text{x}+(1)^2-(1)^2-5\big]}}\text{ dx}$
$=-\frac{3}{2}\int\frac{(-2-2\text{x})}{\sqrt{5-2\text{x}-\text{x}^2}}\text{ dx}-2\int\frac{1}{\sqrt{-\big[(\text{x}+1)^2-(\sqrt{6})^2\big]}}\text{ dx}$
$=-\frac{3}{2}\int\frac{(-2-2\text{x})}{\sqrt{5-2\text{x}-\text{x}^2}}\text{ dx}-2\int\frac{1}{\sqrt{\big[(\sqrt{6})^2-(\text{x}+1)^2\big]}}\text{ dx}$
$\text{I}=-\frac{3}{2}\times2\sqrt{5-2\text{x}-\text{x}^2}-2\sin^{-1}\Big(\frac{\text{x}+1}{\sqrt{6}}\Big)+\text{c}$ $\big[\text{since}, \int\frac{1}{\sqrt{\text{x}}}\text{dx}=2\sqrt{\text{x}}+\text{c},\int\frac{1}{\sqrt{\text{a}^2-\text{x}^2}}\text{dx}=\sin^{-1}\big(\frac{\text{x}}{\text{a}}\big)+\text{c}\big]$
$\text{I}=-\frac{3}{2}\times2\sqrt{5-2\text{x}-\text{x}^2}-2\sin^{-1}\Big(\frac{\text{x}+1}{\sqrt{6}}\Big)+\text{C}$
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