Question 1515 Marks
Evaluvate the following intregals:
$\int\frac{5\text{x}+3}{\sqrt{\text{x}^2+4\text{x}+10}}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{5\text{x}+3}{\sqrt{\text{x}^2+4\text{x}+10}}\ \text{dx}$
Consider,
$5\text{x}+3=\text{A}\frac{\text{d}}{\text{dx}}(\text{x}^2+4\text{x}+10)+\text{B}$
$\Rightarrow5\text{x}+3=\text{A}(2\text{x}+4)+\text{B}$
$\Rightarrow5\text{x}+3=(2\text{A})\text{x}+4\text{A}+\text{B}$
Equating coefficient of like terms
$2\text{A}=5$
$\Rightarrow\text{A}=\frac{5}{2}$
And
$4\text{A}+\text{B}=3$
$\Rightarrow4\times\frac{5}{2}+\text{B}=3$
$\Rightarrow\text{B}=-7$
$\therefore\text{I}=\frac{5}{2}\int\frac{(2\text{x}+4)\text{dx}}{\sqrt{\text{x}^2+4\text{x}+10}}-7\int\frac{\text{dx}}{\sqrt{\text{x}^2+4\text{x}+10}}$
$=\frac{5}{2}\int\frac{(2\text{x}+4)\text{dx}}{\sqrt{\text{x}^2+4\text{x}+10}}-7\int\frac{\text{dx}}{\sqrt{\text{x}^2+4\text{x}+4-4+10}}$
$=\frac{5}{2}\int\frac{(2\text{x}+4)\text{dx}}{\sqrt{\text{x}^2+4\text{x}+10}}-7\int\frac{\text{dx}}{\sqrt{(\text{x}+2})^2+(\sqrt{6})^2}$
Putting, $\text{x}^2+4\text{x}+10=\text{t}$
$\Rightarrow(2\text{x}+4)\text{dx}=\text{dt}$
$\text{I}=\frac{5}{2}\int\frac{\text{dt}}{\sqrt{\text{t}}}-7\log\big|(\text{x}+2)^2+6\big|+\text{C}$
$=\frac{5}{2}\int\text{t}^{-\frac{1}{2}}\text{dt}-7\log\big|\text{x}+2+\sqrt{\text{x}^2+4\text{x}+10}\big|+\text{C}$
$=\frac{5}{2}\times2\sqrt{\text{t}}-7\log\big|\text{x}+2+\sqrt{\text{x}^2+4\text{x}+10}\big|+\text{C}$
$=5\sqrt{\text{x}^2+4\text{x}+10}-7\log\big|\text{x}+2+\sqrt{\text{x}^2+4\text{x}+10}\big|+\text{C}$
View full question & answer→Question 1525 Marks
Evaluate the follwing intregals:
$\int\frac{1}{\text{x}(\text{x}^4-1)}\ \text{dx}$
AnswerWe have,
$\text{I}=\int\frac{\text{dx}}{\text{x}(\text{x}^4-1)}$
$=\int\frac{\text{x}^3\text{dx}}{\text{x}^4(\text{x}^4-1)}$
Putting $\text{x}^4=\text{t}$
$\Rightarrow4\text{x}^3\text{dx}=\text{dt}$
$\Rightarrow\text{x}^3\text{dx}=\frac{\text{dt}}{4}$
$\therefore\text{I}=\frac{1}{4}\int\frac{\text{dt}}{\text{t}(\text{t}-1)}$
Let $\frac{1}{\text{t}(\text{t}-1)}=\frac{\text{A}}{\text{t}}+\frac{\text{B}}{\text{t}-1}$
$\rightarrow\frac{1}{\text{t}(\text{t}-1)}=\frac{\text{A}(\text{t}-1)+\text{B}\text{t}}{(\text{t}-1)}$
$\Rightarrow1=\text{A}(\text{t}-1)+\text{Bt}$
Putting t - 1 = 0
⇒ t = 1
$\therefore$ 1 = A × 0 + B (1)
⇒ B = 1
Putting t = 0
$\therefore$ 1 = A (0 - 1) + B × 0
⇒ A = -1
$\therefore$ $\text{I}=-\frac{1}{4}\int\frac{\text{dt}}{\text{t}}+\frac{1}{4}\int\frac{\text{dt}}{\text{t}-1}$
$=-\frac{1}{4}\log|\text{t}|+\frac{1}{4}\log|\text{t}-1|+\text{C}$
$=\frac{1}{4}\log\Big|\frac{\text{t}-1}{\text{t}}\Big|+\text{C}$
$=\frac{1}{4}\log\Big|\frac{\text{x}^2-1}{\text{x}^4}\Big|+\text{C}$
View full question & answer→Question 1535 Marks
Evaluate the following intregals:
$\int\frac{\text{x}^3}{(\text{x}-1)(\text{x}-2)(\text{x}-3)}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{\text{x}^3}{(\text{x}-1)(\text{x}-2)(\text{x}-3)}\ \text{dx}$
$=\int1+\frac{6\text{x}^2-9\text{x}+6}{(\text{x}-1)(\text{x}-2)(\text{x}-3)}\ \text{dx}$
Let $\frac{6\text{x}^2-9\text{x}+6}{(\text{x}-1)(\text{x}-2)(\text{x}-3)}=\frac{\text{A}}{\text{x}-1}+\frac{\text{B}}{\text{x}-2}+\frac{\text{C}}{\text{x}-3}$
$\text{x}\Rightarrow6\text{x}^2-11+6=\text{A}(\text{x}-2)(\text{x}-3)\\+\text{B}(\text{x}-1)(\text{x}-3)+\text{C}(\text{x}-1)(\text{x}-2)$
put x = 1
$\Rightarrow1=2\text{A}\Rightarrow\text{A}=\frac{1}{2}$
put x = 2
$\Rightarrow8=-\text{B}\Rightarrow\text{B}=-8$
put x = 3
$\Rightarrow27=2\text{C}\Rightarrow\text{C}=\frac{27}{2}$
Thus,
$\text{I}=\int\text{dx}+\frac{1}{2}\int\frac{\text{dx}}{\text{x}-1}-8\int\frac{\text{dx}}{\text{x}-2}+\frac{27}{2}\int\frac{\text{dx}}{\text{x}-3}$
$=\text{x}+\frac{1}{2}\log|\text{x}-1|-8\log|\text{x}-2|+\frac{27}{2}\log|\text{x}-3|+\text{C}$
Hence,
$\text{I}=\text{x}+\frac{1}{2}\log|\text{x}-1|-8\log|\text{x}-2|+\frac{27}{2}\log|\text{x}-3|+\text{C}$
View full question & answer→Question 1545 Marks
Evaluate the following integrals:
$\int \frac{\text{x}^2}{(\text{x}-1)\sqrt{\text{x}+2}}\text{ dx}$
AnswerWe have,
$\text{I}=\int \frac{\text{x}^2}{(\text{x}-1)\sqrt{\text{x}+2}}\text{ dx}$
Putting $\text{x}+2=\text{t}^2$
$\text{x}=\text{t}^2-2$
Diff both sides
$\text{dx}=2\text{t dt} $
$\text{I}=\int\frac{(\text{t}^2-2)^2}{(\text{t}^2-2-1)\text{t}}2\text{t dt}$
$=2\int\frac{(\text{t}^2-2)^2\text{dt}}{\text{t}^2-3}$
$=2\int\frac{(\text{t}^4-4\text{t}^2+4)}{\text{t}^2-3}\text{ dt}$
Dividing numerator by denominator, we get
$\therefore\ \text{I}=2\int\Big(\text{t}^2-1+\frac{1}{\text{t}^2-3}\Big)\text{ dt}$
$=2\int\text{t}^2\text{ dt}-2\int\text{ dt}+2\int\frac{\text{dt}}{\text{t}^2-(\sqrt{3})^2}$
$=2\Big[\frac{\text{t}^3}{3}\Big]-2\text{t}+2\times\frac{1}{2\sqrt{3}}\log\Big|\frac{\text{t}-\sqrt{3}}{\text{t}+\sqrt{3}}\Big|+\text{C}$
$=\frac{2}{3}(\sqrt{\text{x}+2})^3-2\sqrt{\text{x}+2}+\frac{1}{\sqrt{3}}\log\bigg|\frac{\sqrt{\text{x}+2}-\sqrt{3}}{\sqrt{\text{x}+2}+\sqrt{3}}\bigg|+\text{C}$
$=\frac{2}{3}(\text{x}+2)^{\frac{3}{2}}-2\sqrt{\text{x}+2}+\frac{1}{\sqrt{3}}\log\bigg|\frac{\sqrt{\text{x}+2}-\sqrt{3}}{\sqrt{\text{x}+2}+\sqrt{3}}\bigg|+\text{C}$ View full question & answer→Question 1555 Marks
Evaluate the following intregals:
$\int\frac{\text{x}+2}{\sqrt{\text{x}^2+2\text{x}-1}}\text{dx}$
AnswerLet $\text{I}=\int\frac{\text{x}+2}{\sqrt{\text{x}^2+2\text{x}-1}}\text{dx}$
consider,
$\text{x}+2=\text{A}\frac{\text{d}}{\text{dx}}(\text{x}^2+2\text{x}-1)+\text{B}$
$\Rightarrow\text{x}+2=\text{A}(2\text{x}+2)+\text{B}$
$\Rightarrow\text{x}+2=(2\text{A})\text{x}+2\text{A}+\text{B}$
Equating coefficient of like terms.
$2\text{A}=1$
$\Rightarrow\text{A}=\frac{1}{2}$
And
$2\text{A}+\text{B}=2$
$\Rightarrow2\times\frac{1}{2}+\text{B}=2$
$\Rightarrow\text{B}=1$
Then,
$\text{I}=\int\frac{\big[\frac{1}{2}(2\text{x}+2)+1}{\sqrt{\text{x}^2+2\text{x}+1}}\text{dx}$
$=\frac{1}{2}\int\frac{(2\text{x}+2)\text{dx}}{\sqrt{\text{x}^2+2\text{x}+1}}+\int\frac{\text{dx}}{\sqrt{\text{x}^2+2\text{x}-1}}$
Let $\text{x}^2+2\text{x}-1=\text{t}$
$\Rightarrow(2\text{x}+2)\text{dx}=\text{dt}$
$\therefore\text{I}=\frac{1}{2}\int\frac{\text{dt}}{\sqrt{\text{t}}}+\int\frac{\text{dx}}{\sqrt{\text{x}^2+2\text{x}+1-2}}$
$=\frac{1}{2}\int\text{t}^{-\frac{1}{2}}\text{dt}+\int\frac{\text{dx}}{\sqrt{\text{x}^2+2\text{x}+1-2}}$
$=\frac{1}{2}\bigg[\frac{\text{t}^{\frac{1}{2}}+1}{-\frac{1}{2}+1}\bigg]+\int\frac{\text{dx}}{\sqrt{(\text {x}+1)^2-(\sqrt{2})^2}}$
$=\sqrt{\text{t}}+\log\Big|\text{x}+1+\sqrt{(\text{x}+1)^2-(\sqrt{2})^2}\Big|+\text{C}$
$=\sqrt{\text{x}^2+2\text{x}-1}+\log\big|\text{x}+1+\sqrt{\text{x}^2+2\text{x}-1}\big|+\text{C}$
View full question & answer→Question 1565 Marks
Evaluate the following integrals:
$\int(\text{x}=1)\sqrt{\text{x}^2-\text{x}+1}\text{dx}$
AnswerLet $\text{I}=\int(\text{x}=1)\sqrt{\text{x}^2-\text{x}+1}\text{dx}\ \dots(1)$
Let $\text{x}+1=\lambda\frac{\text{d}}{\text{dx}}(\text{x}^2-\text{x}+1)+\mu$
$=\lambda(2\text{x}-1)+\mu$
Equating similar terms, we get,
$2\lambda=1\ \Rightarrow\ \lambda=\frac{1}{2}$
$-\lambda+\mu=1$
$\Rightarrow\mu=1+\lambda=1+\frac{1}{2}=\frac{3}{2}$
$\therefore\ \mu=\frac{3}{2}$
So,
$\text{I}=\int\Big(\frac{1}{2}(2\text{x}-1)+\frac{3}{2}\Big)\sqrt{\text{x}^2-\text{x}+1}\text{dx}$
$=\frac{1}{2}\int(2\text{x}-1)\sqrt{\text{x}^2-\text{x}+1}\text{dx}+\frac{3}{2}\int\sqrt{\text{x}^2-\text{x}+1}\text{dx}$
Let $\text{x}^2-\text{x}+1=\text{t}$
$\Rightarrow(2\text{x}-1)\text{dx}=\text{dt}$
$=\frac{1}{2}\int\sqrt{\text{t}}\text{dt}+\frac{3}{2}\int\sqrt{\Big(\text{x}-\frac{1}{2}\Big)^2+\Big(\frac{\sqrt3}{2}\Big)^2}\text{dx}$
$=\frac{1}{2}\frac{\text{t}^{\frac{3}{2}}}{\frac{3}{2}}+\frac{3}{2}\begin{Bmatrix}\frac{\big(\text{x}-\frac{1}{2}\big)}{2}\sqrt{\text{x}^2-\text{x} +1}\\+\frac{3}{8}\log\Big|\Big(\text{x}-\frac{1}{2}\Big)+\sqrt{\text{x}^2-\text{x}+1}\Big|\end{Bmatrix}$
$\Rightarrow\text{I}=\frac{1}{3}\text{t}^{\frac{3}{2}}+\frac{3}{8}(2\text{x}-1)\sqrt{\text{x}^2-\text{x}+1}+\frac{9}{16}\\\times\log\Big|\Big(\text{x}-\frac{1}{2}\Big)+\sqrt{\text{x}^2-\text{x}+1}\Big|+\text{C}$
Hence,
$\Rightarrow\text{I}=\frac{1}{3}(\text{x}^2-\text{x}+1)^{\frac{3}{2}}+\frac{3}{8}(2\text{x}-1)\sqrt{\text{x}^2-\text{x}+1}+\frac{9}{16}\\\times\log\Big|\Big(\text{x}-\frac{1}{2}\Big)+\sqrt{\text{x}^2-\text{x}+1}\Big|+\text{C}$
View full question & answer→Question 1575 Marks
Evaluate the following integrals:
$\int\frac{\log\big(1+\frac{1}{\text{x}}\big)}{\text{x}(1+\text{x})}\text{dx}$
AnswerLet I $=\int\frac{\log\big(1+\frac{1}{\text{x}}\big)}{\text{x}(1+\text{x})}\text{dx}\ .....(1)$
Let $\log\Big(1+\frac{1}{\text{x}}\Big)=\text{t}$ then,
$\text{d}\Big[\log\Big(1+\frac{1}{\text{x}}\Big)\Big]=\text{dt}$
$\Rightarrow\frac{1}{1+\frac{1}{\text{x}}}\times\frac{-1}{\text{x}^2}\text{dx}=\text{dt}$
$\Rightarrow\frac{1}{\frac{\text{x}+1}{\text{x}}}\times\frac{-1}{\text{x}^2}\text{dx}=\text{dt}$
$\Rightarrow\frac{-\text{x}}{\text{x}^2(\text{x}+1)}\text{dx}=-\text{dt}$
$\Rightarrow\frac{\text{dx}}{\text{x}(\text{x}+1)}=-\text{dt}$
Putting $\log\Big(1+\frac{1}{\text{x}}\Big)=\text{t}$ and $\frac{\text{dx}}{\text{x}(\text{x}+1)}=-\text{dt}$ in equation (1), we get
$\text{I}=\int\text{tx}-\text{dt}$
$=-\frac{\text{t}^2}{2}+\text{C}$
$=-\frac{1}{2}\Big[\log\Big(1+\frac{1}{\text{x}}\Big)\Big]^2+\text{C}$
$\therefore\text{I}=-\frac{1}{2}\Big[\log\Big(1+\frac{1}{\text{x}}\Big)\Big]^2+\text{C}$
View full question & answer→Question 1585 Marks
Evaluate the following integrals:
$\int\frac{\text{dx}}{(\text{x}^2+1)(\text{x}^2+4)}$
AnswerLet $\frac{1}{(\text{x}^2+1)(\text{x}^2+4)}=\frac{\text{Ax}+\text{B}}{(\text{x}^2+1)}+\frac{\text{Cx}+\text{D}}{\text{x}^2+4}$
$\Rightarrow1=(\text{Ax}+\text{B})(\text{x}^2+4)+(\text{Cx}+\text{D})(\text{x}^2+1)$
$=(\text{A}+\text{C})\text{x}^3+(\text{B}+\text{D})\text{x}^2+(4\text{A}+\text{C})\text{x}+4\text{B}+\text{D}$
Equating similar terms, we get,
A + C = 0, B + D = 0, 4A + C = 0, 4B + D = 1
Solving, we get, $\text{A}=0,\text{B}=\frac{1}{3},\text{C}=0,\text{D}=-\frac{1}{3}$
Thus,
$\text{I}=\int\frac{\frac{1}{3}\text{dx}}{(\text{x}^2+1)}-\int\frac{\frac{1}{3}\text{dx}}{(\text{x}^2+4)}$
$=\frac{1}{3}\tan^{-1}\text{x}-\frac{1}{6}\tan^{-1}\Big(\frac{\text{x}}{2}\Big)+\text{C}$ $\big[\because\int\frac{\text{dx}}{\text{x}^2+\text{a}^2}=\frac{1}{\text{a}}\tan^{-1}\frac{\text{x}}{\text{a}}\big]$
$\therefore\text{I}=\frac{1}{3}\tan^{-1}\Big(\frac{\text{x}}{2}\Big)+\text{C}$
View full question & answer→Question 1595 Marks
Evaluate the following intregals:
$\int\frac{2\text{x}-3}{(\text{x}^2-1)(2\text{x}-3)}\ \text{dx}$
Answer$\int\frac{2\text{x}-3}{(\text{x}^2-1)(2\text{x}-3)}\ \text{dx}$
$=\int\frac{(2\text{x}-3)}{(\text{x}-1)(\text{x}+1)(2\text{x}+3)}\ \text{dx}$
Let $\int\frac{(2\text{x}-3)}{(\text{x}-1)(\text{x}+1)(2\text{x}+3)}=\frac{\text{A}}{\text{x}-1}+\frac{\text{B}}{\text{x}+1}+\frac{\text{C}}{2\text{x}+ 3}$
$\Rightarrow\int\frac{(2\text{x}-3)}{(\text{x}-1)(\text{x}+1)(2\text{x}+3)}=\frac{\text{A}(\text{x}+1)(2\text{x}+3)+\text{B}(\text{x}+1)(2\text {x}-3)+\text{C}(\text{x}^2-1)}{(\text{x}-1))(\text{x}+1)(2\text{X}+3)}$
$\Rightarrow2\text{x}-3=\text{A}(\text{x}+1)(2\text{x}+3)\\+\text{B}(\text{x}-1)(2\text{x}+3)+\text{C}(\text{x}+1)(\text{x}-1)\ ...(1)$
Putting x + 1 = 0 or x = -1 in eq (1)
⇒ -2 - 3 = B (-1 -1) (-2 + 3)
⇒ -5 = B (-2) (1)
$\Rightarrow\text{B}=\frac{5}{2}$
Putting x - 1 = 0 or x = 1 in eq (1)
⇒ 2 - 3 = A (1 + 1) (2 + 3)
⇒ -1 = A (2) (5)
$\Rightarrow\text{A}=\frac{-1}{10}$
Putting 2x + 3 = 0 or $\text{x}=\frac{-3}{2}$ in eq (1)
$\Rightarrow2\times-\frac{3}{2}-3=\text{A}\times0+\text{B}\times0+\text{C}\Big(-\frac{3}{2}+1\Big)\Big(\frac{-3}{2}-1\Big)$
$\Rightarrow-6=\text{C}\Big(-\frac{1}{2}\Big)\Big(\frac{-5}{2}\Big)$
$\Rightarrow\text{C}=-\frac{24}{5}$
$\therefore\frac{2\text{x}-3}{(\text{x}-1)(\text{X}+1)(2\text{X}+3)}=\frac{-1}{10(\text{x}-1)}+\frac{5}{2(\text{x}+1)}-\frac{24}{5(2\text{x}+3)}$
$\Rightarrow\int\frac{(2\text{x}-3)}{(\text{x}-1)(\text{x}+1)(2\text{x}+3)}\ \text{dx}=\frac{-1}{10}\int\frac{1}{\text{x}-1}\text{dx}\\+\frac{5}{2}\int\frac{1}{\text{x}+1}\text{dx}-\frac{24}{5}\int\frac{1}{2\text{X}+3}\ \text{dx}$
$=\frac{-1}{10}\ln|\text{x}-1|+\frac{5}{2}\ln|\text{x}+1|-\frac{24}{5}\ln\frac{|2\text{x}+3|}{3}+\text{C}$
$=-\frac{1}{10}\ln|\text{x}-1|+\frac{5}{2}\ln|\text{x}+1|-\frac{12}{5}\ln|2\text{x}+3|+\text{C}$
View full question & answer→Question 1605 Marks
Evaluate the following integrals:
$\int\frac{1}{\text{x}(\text{x}^6+1)}\text{dx}$
AnswerLet $\text{I}=\int\frac{1}{\text{x}(\text{x}^6+1)}\text{dx}$
$=\int\frac{\text{x}^5}{\text{x}^6(\text{x}^6+1)}\text{dx}$
Let $\text{x}^6=\text{t}$
$\Rightarrow6\text{x}^5\text{dx = dt}$
$\Rightarrow\text{x}^5\text{dx}=\frac{\text{dt}}{6}$
$\text{I}=\frac{1}{6}\int\frac{\text{dt}}{\text{t}(\text{t}+1)}$
$=\frac{1}{6}\int\frac{\text{dt}}{\text{t}^2+\text{t}}$
$=\frac{1}{6}\int\frac{\text{dt}}{\text{t}^2+2\text{t}\big(\frac{1}{2}\big)+\big(\frac{1}{2}\big)^2-\big(\frac{1}{2}\big)^2}$
$=\frac{1}{6}\int\frac{\text{dt}}{\big(\text{t}+\frac{1}{2}\big)^2-\big(\frac{1}{2}\big)^2}$
Let $\text{t}+\frac{1}{2}=\text{u}$
$\Rightarrow\text{dt = du}$
$\text{I}=\frac{1}{6}\int\frac{\text{du}}{\text{u}^2-\big(\frac{1}{2}\big)^2}$
$=\frac{1}{6}\times\frac{1}{2\big(\frac{1}{2}\big)}\log\Bigg|\frac{\text{u}-\frac{1}{2}}{\text{u}+\frac{1}{2}}\Bigg|+\text{C}$ $\bigg[\text{Since}\int\frac{1}{\text{x}^2-\text{a}^2}\text{dx}=\frac{1}{2\text{a}}\log\Big|\frac{\text{x}-\text{a}}{\text{x}+\text{a}}\Big|+\text{C}\bigg]$
$\text{I}=\frac{1}{6}\log\Bigg|\frac{\text{t}+\frac{1}{2}-\frac{1}{2}}{\text{t}+\frac{1}{2}+\frac{1}{2}}\Bigg|+\text{C}$
$\text{I}=\frac{1}{6}\log\bigg|\frac{\text{x}^6}{\text{x}^6+1}\bigg|+\text{C}$
View full question & answer→Question 1615 Marks
Evaluate the following intregals: $\int\frac{\text{x}^2+\text{x}+1}{(\text{x}^2+1)(\text{x}+2)}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\text{x}^2+\text{x}+1}{(\text{x}^2+1)(\text{x}+2)}\text{ dx}$
We express
$\frac{\text{x}^2+\text{x}+1}{(\text{x}^2+1)(\text{x}+2)}=\frac{\text{A}}{\text{x}+2}+\frac{\text{Bx}+\text{C}}{\text{x}^2+1}$
$\Rightarrow\text{x}^2+\text{x}+1=\text{A}(\text{x}^2+1)+(\text{Bx}+\text{C})(\text{x}+2)$
Equating the coefficient of $x^2, x$ and constant, we get
$1=\text{A}+\text{B}$ and $1=2\text{B}+\text{C}$ and $1=\text{A}+2\text{C}$
$\text{or}\text{ A } =\frac{3}{5}$ and $\text{B}=\frac{2}{5}$ and $\text{C}=\frac{1}{5}$
$ \therefore\text{I}=\int\bigg(\frac{\frac{3}{5}}{\text{x}+2}+\frac{\frac{2}{5}\text{x}+\frac{1}{5}}{\text{x}^2+1}\bigg)\text{dx}$
$=\frac{3}{5}\int\frac{1}{\text{x}+2}\ \text{dx}+\frac{2}{5}\int\frac{\text{x}}{\text{x}+1}\ \text{dx}+\frac{1}{5}\int\frac{1}{\text{x}^2+1}\ \text{dx}$
$=\frac{3}{5}\text{I}_1+\frac{2}{5}\text{I}_2+\frac{1}{5}\text{I}_3\ \dots(1)$
Let $x + 2 = u$
On Differentiating both sides, we get
$\text{dx}=\text{du}$
$\text{I}_1=\int\frac{1}{\text{u}}\text{du}$
$=\log|\text{u}|+\text{C}_1$
$=\log|\text{x}+2|+\text{C}_1\ \dots(2)$
And, $\text{I}_2=\int\frac{\text{x}}{\text{x}+1}\text{ dx}$
Let $(\text{x}^2+1)=\text{u}$
On differentiating both sides, we get
$2\text{x}\ \text{dx}=\text{du}$
$\therefore\text{I}_2=\frac{1}{2}\int\frac{1}{\text{u}}\text{ du}$
$=\frac{1}{2}\log|\text{u}|+\text{C}_2$
$=\frac{1}{2}\log|\text{x}^2+1|+\text{C}_2\ \dots(3)$
And $\text{I}_3=\int\frac{1}{\text{x}^2+1}\ \text{dx}$
$=\tan^{-1}\text{x}+\text{C}_3\ \dots(4)$
From $(1), (2), (3)$ and $(4)$ we get
$\therefore\text{I}=\frac{3}{5}(\log|\text{x}+2|+\text{C}_1)+\frac{2}{5}(\frac{1}{2}\log|\text{x}^2+1|+\text{C}_2)$
$+\frac{1}{5}\tan^{-1}\text{x}+\text{C}$
$=\frac{3}{5}\log|\text{x}+2|+\frac{1}{5}\log|\text{x}^2+1|+\frac{1}{5}\tan^{-1}\text{x}+\text{C}$
Hence, $\int\frac{\text{x}^2+\text{x}+1}{(\text{x}^2+1)(\text{x}+2)}\ \text{dx}=\frac{3}{5}\log|\text{x}+2|+\frac{1}{5}\tan^{-1}\text{x}+\text{C}$
View full question & answer→Question 1625 Marks
Evaluate the following integrals:
$\int\frac{\text{x}}{3\text{x}^4-18\text{x}^2+11}\text{dx}$
Answer$\int\frac{\text{x dx}}{3\text{x}^4-18\text{x}^2+11}$
Let $\text{x}^2=\text{t}$
$\Rightarrow2\text{x}\text{ dx = dt}$
$\Rightarrow\text{x dx}=\frac{\text{dt}}{2}$
Now, $\int\frac{\text{x dx}}{3\text{x}^4-18\text{x}^2+11}$
$=\frac{1}2{}\int\frac{\text{dt}}{3\text{t}^2-18\text{t}+11}$
$=\frac{1}{3\times2}\int\frac{\text{dt}}{\text{t}^2-6\text{t}+\frac{11}{3}}$
$=\frac{1}{6}\int\frac{\text{dt}}{\text{t}^2-6\text{t}+9-9+\frac{11}{3}}$
$=\frac{1}{6}\int\frac{\text{dt}}{(\text{t}-3)^2-\frac{16}{3}}$
$=\frac{1}{6}\int\frac{\text{dt}}{(\text{t}-3)^2-\Big(\frac{4}{\sqrt{3}}\Big)^2}$
$=\frac{1}{6}\times\frac{1}{2\times\frac{4}{\sqrt{3}}}\log\Bigg|\frac{\text{t}-3-\frac{4}{\sqrt{3}}}{\text{t}-3+\frac{4}{\sqrt{3}}}\Bigg|+\text{C}$
$=\frac{\sqrt{3}}{48}\log\Bigg|\frac{\text{x}^2-3-\frac{4}{\sqrt{3}}}{\text{x}^2-3+\frac{4}{\sqrt{3}}}\Bigg|+\text{C}$
View full question & answer→Question 1635 Marks
$\int\frac{3\text{x}+5}{\sqrt{7\text{x}+9}}\text{dx}$
Answer$\text{Let I}=\int\Big(\frac{3\text{x}+5}{\sqrt{7\text{x}+9}}\Big)\text{dx}$
$\text{Putting}\ 7\text{x}+9=\text{t}$
$\Rightarrow\text{x}=\frac{\text{t}-9}{7}\ \&\ 7\text{dx}=\text{dt}$
$\Rightarrow\text{dx}=\frac{\text{dt}}{7}$
$\therefore\text{I}=\int\Bigg(\frac{3\big(\frac{\text{t}-9}{7}\big)+5}{\sqrt{\text{t}}}\Bigg)\text{dt}$
$=\int\Big(\frac{3}{7}\frac{\text{t}}{\sqrt{\text{t}}}-\frac{27}{7\sqrt{\text{t}}}+\frac{5}{\sqrt{\text{t}}}\Big)\frac{\text{dt}}{7}$
$=\frac{3}{7\times7}\int\text{t}^\frac{1}{2}\text{dt}-\frac{27}{7\times7}\int\text{t}^{-\frac{1}{2}}\text{dt}+\frac{5}{7}\int^{-\frac{1}{2}}\text{dt}$
$=\frac{3}{7\times7}\bigg[\frac{\text{t}^{\frac{1}{2}+1}}{\frac{1}{2}+1}\bigg]-\frac{27}{7\times7}\bigg[\frac{\text{t}^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\bigg]+\frac{5}{7}\bigg[\frac{\text{t}^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\bigg]+\text{C}$
$=\frac{2}{7\times7}\text{t}^\frac{3}{2}-\frac{27}{7\times7}2\text{t}^\frac{1}{2}+\frac{10\sqrt{t}}{7}+\text{C}$
$=\frac{2}{7\times7}(7\text{x}+9)^\frac{3}{2}-\frac{54}{7\times7}(7\text{x}+9)^\frac{1}{2}+\frac{10}{7}\sqrt{7\text{x}+9}+\text{C}$ $[\because\text{t}=7\text{x}+9]$
$=\frac{2}{7\times7}(7\text{x}+9)^\frac{3}{2}+\Big(10-\frac{54}{7}\Big)\frac{\sqrt{7\text{x}+9}}{7}+\text{C}$
$=\frac{2}{7\times7}(7\text{x}+9)^\frac{3}{2}+\Big(\frac{70-54}{7}\Big)\frac{\sqrt{7\text{x}+9}}{7}+\text{C}$
$=\frac{2}{7\times7}(7\text{x}+9)^\frac{3}{2}+\frac{16}{7\times7}\frac{\sqrt{7\text{x}+9}}{7}+\text{C}$
$=\frac{2}{7\times7}\Big[(7\text{x}+9)^\frac{1}{2}[7\text{x}+9+8]\Big]+\text{C}$
$=\frac{2}{49}\Big[(7\text{x}+9)^\frac{1}{2}[7\text{x}+17]\Big]+\text{C}$
View full question & answer→Question 1645 Marks
Evaluate the following integrals:
$\int\frac{\text{x}^2+1}{\text{x}^4-\text{x}^2+1}\ \text{dx}$
Answerlet $\text{I}=\int\frac{\text{x}^2+1}{\text{x}^4-\text{x}^2+1}\ \text{dx}$
Dividing numerator and denominator bt $x^2$
$\therefore\text{I}=\frac{\Big(1+\frac{1}{\text{x}^2}\Big)}{\text{x}^2-1+\frac{1}{\text{x}^2}}\ \text{dx}$
$=\int\frac{\Big(1+\frac{1}{\text{x}^2}\Big)\text{dx}}{\Big(\text{x}-\frac{1}{\text{x}}\Big)^2+1}$
let $\Big(\text{x}-\frac{1}{\text{x}^2}\Big)\text{dx}=\text{dt}$
$\Rightarrow\text{I}=\int\frac{\text{dt}}{\text{t}^2+1}$
$=\tan^{-1}\text{t}+\text{C}$
$\therefore\text{I}=\tan^{-1}\Big(\frac{\text{x}^2-1}{\text{x}}\Big)+\text{C}$
View full question & answer→Question 1655 Marks
Evaluate the following integrals:
$\int\frac{\text{x}}{(\text{x}-3)\sqrt{\text{x}+1}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\text{x}}{(\text{x}-3)\sqrt{\text{x}+1}}\text{ dx}$
$=\int\frac{(\text{x}-3)+3}{(\text{x}-3)\sqrt{\text{x}+1}}\text{ dx}$
$\text{I}=\int\frac{\text{dx}}{\sqrt{\text{x}+1}}+3\int\frac{\text{dx}}{(\text{x}-3)\sqrt{\text{x}+1}}\ ...(\text{i})$
Now, $\int\frac{\text{dx}}{\sqrt{\text{x}+1}}=2\sqrt{\text{x}+1}+\text{C}_1$
and, $\int\frac{\text{dx}}{\sqrt{\text{x}+2}}=2\sqrt{\text{x}+2}+\text{C}_2$
$\int\frac{\text{dx}}{(\text{x}-3)\sqrt{\text{x}+1}}$
Let $\text{x}+1=\text{t}^2$
$\text{dx}=2\text{t dt}$
$\int\frac{\text{dx}}{(\text{x})-3\sqrt{\text{x}+1}}=2\int\frac{\text{t dt}}{(\text{t}^2-4)\text{t}}$
$=2\Big|\frac{\text{dt}}{\text{t}^2-4}\Big|$
$=\frac{2}{2\times2}\log\Big|\frac{\text{t}-2}{\text{t}+2}\Big|+\text{C}_2$
$\therefore\ \int\frac{\text{dx}}{(\text{x}-3)\sqrt{\text{x}+1}}=\frac{1}{2}\log\bigg|\frac{\sqrt{\text{x}+1}-2}{\sqrt{\text{x}}+1+2}\bigg|+\text{C}_2$
Thus, from (i)
$\text{I}=2\sqrt{\text{x}+1}+\frac{3}{2}\log\bigg|\frac{\sqrt{\text{x}+1}-2}{\sqrt{\text{x}+1}+2}\bigg|+\text{C} [$When $C = C_1 + C_2]$
View full question & answer→Question 1665 Marks
Evaluate the following integrals:
$\int\frac{\text{x}^2+1}{\text{x}^2+7\text{x}^2+1}\ \text{dx}$
AnswerWe have,
$\text{I}=\int\Big(\frac{\text{x}^2+1}{\text{x}^4+7\text{x}^2+1}\Big)\text{dx}$
Dividing numerator and denominator by $x^2$
$\text{I}=\int\Bigg(\frac{1+\frac{1}{\text{x}^2}}{\text{x}^2+7+\frac{1}{\text{x}^2}}\Bigg)\text{dx}$
$=\int\frac{\Big(1+\frac{1}{\text{x}^2}\Big)\text{dx}}{\text{x}^2+\frac{1}{\text{x}^2}-2+9}$
$\Rightarrow\int\frac{\Big(1+\frac{1}{\text{x}^2}\Big)\text{dx}}{\Big(\text{x}-\frac{1}{\text{x}^2}\Big)^2+3^2}$
Putting $\text{x}-\frac{1}{\text{x}}=\text{t}$
$\Rightarrow\Big(1+\frac{1}{\text{x}^2}\Big)\text{dx}=\text{dt}$
$\therefore\text{I}=\int\frac{\text{dt}}{\text{t}^2+3^2}$
$=\frac{1}{3}\tan^{-1}\Big(\frac{\text{t}}{3}\Big)+\text{C}$
$=\frac{1}{3}\tan^{-1}\Big(\frac{\text{x}-\frac{1}{\text{x}}}{3}\Big)+\text{C}$
$=\frac{1}{3}\tan^{-1}\Big(\frac{\text{x}^2-1}{3\text{x}}\Big)+\text{C}$
View full question & answer→Question 1675 Marks
Evaluate the following integrals:
$\int\frac{1}{(\text{x}^2+2\text{x}+10)^2}\text{ dx}$
AnswerLet $\int\frac{1}{(\text{x}^2+2\text{x}+10)^2}\text{ dx}$
$=\int\frac{1}{\big[(\text{x}+1)^2+3^2\big]}\text{ dx}$
Let $\text{x}+1=3\tan\theta$
On differentiating both sides, we get
$\text{dx}=3\sec^2\theta\text{ d}\theta$
$\therefore\ \text{I}=\int\frac{1}{\big[3^2\tan^2\theta+3^2\big]^2}3\sec^2\theta\text{ d}\theta$
$=\frac{1}{27}\int\frac{\sec^2\theta}{\sec^4\theta}\text{ d}\theta$
$=\frac{1}{27}\int\frac{1}{\sec^2\theta}\text{ d}\theta$
$=\frac{1}{27}\int\cos^2\theta\text{ d}\theta$
$=\frac{1}{54}\int(1+\cos2\theta)\text{d}\theta$
$=\frac{1}{54}\Big(\theta+\frac{\sin2\theta}{2}\Big)+\text{ C}$
$=\frac{1}{54}\Big(\theta+\frac{\tan\theta}{1+\tan^2\theta}\Big)+\text{C}$
$=\frac{1}{54}\begin{pmatrix}\tan^{-1}\frac{\text{x}+1}{3}+\frac{\tan\Big(\tan^{-1}\frac{\text{x}+1}{3}\Big)}{1+\tan^{2}\Big(\tan^{-1}\frac{\text{x}+1}{3}\Big)}\end{pmatrix}+\text{C}$
$=\frac{1}{54}\begin{pmatrix}\tan^{-1}\frac{\text{x}+1}{3}+\frac{\frac{\text{x}+1}{3}}{1+\Big(\frac{\text{x}+1}{3}\Big)^2}\end{pmatrix}+\text{C}$
$=\frac{1}{54}\Bigg(\tan^{-1}\frac{\text{x}+1}{3}+\frac{\frac{\text{x}+1}{3}}{\frac{\text{x}^2+2\text{x}+10}{9}}\Bigg)+\text{C}$
$=\frac{1}{54}\bigg(\tan^{-1}\frac{\text{x}+1}{3}+\frac{3(\text{x}+1)}{\text{x}^2+2\text{x}+10}\bigg)+\text{C}$
View full question & answer→Question 1685 Marks
Evaluate the following integrals:
$\int\frac{\cos^5\text{x}}{\sin\text{x}}\text{ dx}$
Answer $\int\frac{\cos^5\text{x}}{\sin\text{x}}\text{ dx}$ $=\int\frac{\cos^4\text{x}\cos\text{x}}{\sin\text{x}}\text{ dx}$ $=\int\frac{(\cos^2\text{x})^2\cos\text{x}}{\sin\text{x}}\text{ dx}$ $=\int\frac{(1-\sin^2\text{x})^2\times\cos\text{x}}{\sin\text{x}}\text{ dx}$ $=\int\frac{(1-\sin^4\text{x}-2\sin^2\text{x})}{\sin\text{x}}\cos\text{x}\text{ dx}$Let $\sin\text{x}=\text{t}$
$\Rightarrow\cos\text{x}\text{ dx}=\text{dt}$
Now, $\int\frac{(1-\sin^4\text{x}-2\sin^2\text{x})}{\sin\text{x}}\cos\text{x}\text{ dx}$
$=\int\frac{(1+\text{t}^4-2\text{t}^2)}{\text{t}}\text{ dt}$
$=\int\Big(\frac{1}{\text{t}}+\text{t}^3-2\text{t}\Big)\text{dt}$
$=\log|\text{t}|+\frac{\text{t}^4}{4}-\frac{2\text{t}^2}{2}+\text{C}$
$=\log|\text{t}|+\frac{\text{t}^4}{4}-\text{t}^2+\text{C}$
$=\log|\sin\text{x}|+\frac{\sin^4\text{x}}{4}-\sin^2\text{x}+\text{C}$
View full question & answer→Question 1695 Marks
Evalute the following integrals:
$\int\frac{\sin2\text{x}}{\text{a}^2+\text{b}^2\sin^2\text{x}}\text{dx}$
AnswerLet $\int\frac{\sin2\text{x}}{\text{a}^2+\text{b}^2\sin^2\text{x}}\text{dx}\ .....(\text{i})$
Let $\text{a}^2+\text{b}^2\sin^2\text{x}=\text{t}$ then,
$\text{d}\big(\text{a}^2+\text{b}^2\sin^2\text{x}\big)=\text{dt}$
$=\text{b}^2(2\sin\text{x}\cos\text{x})\text{dx}=\text{dt}$
$\Rightarrow\text{dx}=\frac{\text{dt}}{\text{b}^2(2\sin\text{x}\cos\text{x})}$
$=\frac{\text{dt}}{\text{b}^2\sin2\text{x}}$
Putting $\text{a}^2+\text{b}^2\sin^2\text{x}=\text{t}$ and $\text{dx}=\frac{\text{dt}}{\text{b}^2\sin2\text{x}}$ in equation (i), we get,
$\text{I}=\int\frac{\sin2\text{x}}{\text{t}}\times\frac{\text{dt}}{\text{b}^2\sin2\text{x}}$
$=\frac{1}{\text{b}^2}\int\frac{\text{dt}}{\text{t}}$
$=\frac{1}{\text{b}^2}\log|\text{t}|+\text{C}$
$=\frac{1}{\text{b}^2}\log|\text{a}^2+\text{b}^2\sin^2\text{x}|+\text{C}$
$\Rightarrow\text{I}=\frac{1}{\text{b}^2}\log|\text{a}^2+\text{b}^2\sin^2\text{x}|+\text{C}$
View full question & answer→Question 1705 Marks
Evaluate the following integrals:
$\int\sin^5\text{x}\text{ dx}$
AnswerLet $\text{I}=\int\sin^5\text{x}\text{ dx}$ Then
$\text{I}=\int\sin^3\text{x }\sin^2\text{x}\text{ dx}$
$=\int\sin^3\text{x}(1-\cos^2\text{x})\text{ dx} $
$=\int\big(\sin^3\text{x}-\sin^3\text{x}\cos^2\text{x}\big)\text{ dx}$
$=\int\big[\sin\text{x}(1-\cos^2\text{x})-\sin^3\text{x}\cos^2\text{x}\big]\text{dx}$
$=\int\big(\sin\text{x}-\sin\text{x}\cos^2\text{x}-\sin^3\text{x}\cos^2\text{x}\big)\text{dx}$
$\text{I}=\int\sin\text{x}\text{ dx}-\int\sin\text{x}\cos^2\text{x}\text{ dx}-\int\sin^3\text{x}\cos^2\text{x}\text{dx}$
Putting $\cos\text{x}=\text{t}$ and $-\sin\text{x}\text{ dx}=\text{dt}$ is $2^{nd}$ and $3^{rd}$ integrals, we get
$\text{I}=\int\sin\text{x}\text{ dx}-\int\text{t}^2(-\text{dt})+\int\sin^2\text{x}\text{t}^2\text{ dt}$
$=\int\sin\text{x}\text{ dx}+\int\text{t}^2\text{dt}+\int\big(1-\cos^2\text{x}\big)\text{t}^2\text{dt}$
$=\int\sin\text{x}\text{ dx}+\int\text{t}^2\text{dt}+\int(1-\text{t}^2)\text{t}^2\text{dt}$
$=-\cos\text{x}+\frac{\text{t}^3}{3}+\frac{\text{t}^3}{3}-\frac{\text{t}^5}{5}+\text{C}$
$=-\cos\text{x}+\frac{2}{3}\text{t}^3-\frac{1}{5}\text{t}^5+\text{C}$
$=-\cos\text{x}+\frac{2}{3}(\cos^3\text{x})-\frac{1}{5}\big(\cos^5\text{x})+\text{C}$
$\therefore\ \text{I}=-\big[\cos\text{x}-\frac{2}{3}\cos^3\text{x}+\frac{1}{5}\cos^5\text{x}\big]+\text{C}$
View full question & answer→Question 1715 Marks
Evaluate the follwing intregals:
$\int\frac{\text{x}^2}{(\text{x}-1)(\text{x}^2+1)}\ \text{dx}$
Answer$\text{I}=\int\frac{\text{x}^2}{(\text{x}-1)(\text{x}^2+1)}\ \text{dx}$
$\frac{\text{x}^2}{(\text{x}-1)^2(\text{x}^2+1)}=\frac{\text{A}}{(\text{x}-1)}+\frac{\text{Bx}+\text{C}}{\text{x}^2+1}$
$=\frac{\text{A}(\text{x}^2+1)+(\text{Bx}+\text{C})(\text{x}-1)}{(\text{x}-1)(\text{x}^2+1)}$
$\Rightarrow\frac{\text{x}^2}{(\text{x}+1)(\text{x}^2+1)}=\frac{(\text{A}+\text{B})\text{x}^2+(\text{C}-\text{B})\text{x}+(\text{A}-\text{C})}{(\text{x}-1)(\text{x}^2+1)}$
Comapairing coefficient, we get
$\text{A}+\text{B}=\text{C}=\frac{1}{2}$
$\therefore\text{I}=\frac{1}{2}\int\frac{1}{(\text{x}-1)\ \text{dx}}+\frac{1}{2}\int\frac{\text{x}}{\text{x}^2+1}\ \text{dx}+\frac{1}{2}\int\frac{1}{\text{x}^2+1}\ \text{dx}$
$=\frac{1}{2}\ln|\text{x}-1|+\frac{1}{4}\ln|\text{x}^2+1|+\frac{1}{2}\tan^{-1}\text{(x)}+\text{C}$
View full question & answer→Question 1725 Marks
Evaluate the following integral:
$\int\frac{1}{\sqrt{(2-\text{x})^2-1}}\text{ dx}$
Answer$\int\frac{1}{\sqrt{(2-\text{x})^2-1}}\text{ dx}$
Let $2-\text{x}=\text{t}$
$-\text{dx}=\text{dt}$
$\text{dx}=-\text{dt}$
Now, $\int\frac{1}{\sqrt{(2-\text{x})^2-1}}\text{ dx}$
$=\int\frac{-\text{dt}}{\sqrt{\text{t}^2-1}}$
$=-\log\big|\text{t}+\sqrt{\text{t}^2-1}\big|+\text{C}$
$=-\log\big|(2-\text{x})+\sqrt{(2-\text{x})^2-1}\big|+\text{C}$
View full question & answer→Question 1735 Marks
Evaluate the following intregals:
$\int\frac{1}{1-\sin\text{x}+\cos\text{x}}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{1}{1-\sin\text{x}+\cos\text{x}}\ \text{dx}$
Putting $\sin\text{x}=\frac{2\tan\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}},\cos\text{x}=\frac{1-\tan^2\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}$
$\text{I}=\int\frac{1}{1-\frac{2\tan\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}+\frac{1-\tan^2\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}}$
$=\int\frac{1+\tan^2\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}-2\tan\frac{\text{x}}{2}+1-\tan^2\frac{\text{x}}{2}}\ \text{dx}$
$=\int\frac{\sec^2\frac{\text{x}}{2}}{2-2\tan\frac{\text{x}}{2}}\ \text{dx}$
Let $\tan\frac{\text{x}}{2}=\text{t}$
$\frac{1}{2}\sec^2\frac{\text{x}}{2}\text{dx}=\text{dt}$
$=\frac{2}{2}\int\frac{\text{dt}}{1-\text{t}}$
$=-\log|1-\text{t}|+\text{C}$
$\text{I}=-\log\big|1-\tan\frac{\text{x}}{2}\big|+\text{C}$
View full question & answer→Question 1745 Marks
Evalute the following integrals:
$\int\frac{\sin2\text{x}}{\text{a}\cos^2\text{x}+\text{b}\sin^2\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\frac{\sin2\text{x}}{\text{a}\cos^2\text{x}+\text{b}\sin^2\text{x}}\text{dx}$ $=\int\frac{\sin2\text{x}}{\text{a}(1-\sin^2\text{x})+\text{b}\sin^2\text{x}}\text{dx}$ $=\int\frac{\sin2\text{x}}{(\text{b}-\text{a})\sin^2\text{x}+\text{a}}\text{dx}$ Putting $\sin^2\text{x}=\text{t}$ $\Rightarrow2\sin\text{x}\cos\text{x}=\frac{\text{dt}}{\text{dx}}$ $\Rightarrow\sin2\text{x}=\frac{\text{dt}}{\text{dx}}$ $\Rightarrow\sin2\text{x }\text{dx}=\text{dt}$ $\therefore\text{I}=\int\frac{1}{(\text{b}-\text{a})\text{t}+\text{a}}\text{dt}$ $=\frac{1}{(\text{b}-\text{a})}\text{ln}|(\text{b}-\text{a})\text{t}+\text{a}|+\text{C}$ $\Big[\because\int\frac{1}{\text{ax}+\text{b}}\text{dx}=\frac{1}{\text{a}}\text{In}|\text{ax}+\text{b}|+\text{C}\Big]$ $=\frac{1}{(\text{b}-\text{a})}\text{ln}|(\text{b}-\text{a})\sin^2\text{x}+\text{a}|+\text{C}\ \big[\because\text{t}=\sin^2\text{x}\big]$ $=\frac{1}{(\text{b}-\text{a})}\text{ln}\big|\text{b}\sin^2\text{x}+\text{a}(1-\sin^2\text{x})\big|+\text{C}$$=\frac{1}{(\text{b}-\text{a})}\text{ln}\big|\text{b}\sin^2\text{x}+\text{a}\cos^2\text{x}\big|+\text{C}$
View full question & answer→Question 1755 Marks
Evalute the following integrals:
$\int\frac{\text{cosec x}}{\log\tan\frac{\text{x}}{2}}\text{dx}$
AnswerLet $\int\frac{\text{cosec x}}{\log\tan\frac{\text{x}}{2}}\text{dx}\ .....(\text{i})$
Let $\log\tan\frac{\text{x}}{2}=\text{t}$ then,
$\text{d}\Big[\log\tan\frac{\text{x}}{2}\Big]=\text{dt}$
$\Rightarrow\text{cosec x dx}=\text{dt}$
$\Rightarrow\text{dx}=\frac{\text{dt}}{\text{cosec x}}$
Putting $\log\tan\frac{\text{x}}{2}=\text{t}$ and $\text{dx}=\frac{\text{dt}}{\text{cosec x}}$ in equation (i), we get
$\text{I}=\int\frac{\text{cosec x}}{\text{t}}\times\frac{\text{dt}}{\text{cosec x}}$
$=\int\frac{\text{dt}}{\text{t}}$
$=\log|\text{t}|+\text{C}$
$=\log\Big|\log\tan\frac{\text{x}}{2}\Big|+\text{C}$
$\therefore\text{I}=\log\Big|\log\tan\frac{\text{x}}{2}\Big|+\text{C}$
View full question & answer→Question 1765 Marks
Evaluate the following integrals:
$\int4\text{x}^3\sqrt{5-\text{x}^2}\text{ dx}$
Answer$\int4\text{x}^3\sqrt{5-\text{x}^2}\text{ dx}$
$=4\int\text{x}^2\times\text{x}\sqrt{5-\text{x}^2}\text{ dx}$
Let $5-\text{x}^2=\text{t}$
$\Rightarrow\text{x}^2=5-\text{t}$
$\Rightarrow2\text{x}=-\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{x dx}=-\frac{\text{dt}}{2}$
Now, $4\int\text{x}^2\times\text{x}\sqrt{5-\text{x}^2}\text{ dx}$
$=\frac{4}{-2}\int(5-\text{x})\sqrt{\text{t}}\text{ dt}$
$=-2\int5\text{t}^{\frac{1}{2}}+2\int\text{t}^{\frac{3}{2}}\text{ dt}$
$=-10\Bigg[\frac{\text{t}^{\frac{1}{2}+1}}{\frac{1}{2}+1}\Bigg]+2\Bigg[\frac{\text{t}^{\frac{3}{2}+1}}{\frac{3}{2}+1}\Bigg]+\text{C}$
$=-\frac{20}{3}\text{t}^{\frac{3}{2}}+\frac{4}{5}\text{t}^{\frac{5}{2}}+\text{C}$
$=-\frac{20}{3}\big(5-\text{x}^2\big)^{\frac{3}{2}}+\frac{4}{5}\big(5-\text{x}^2\big)^{\frac{5}{2}}+\text{C}$
View full question & answer→Question 1775 Marks
Evaluate the following intregals:
$\int\frac{1}{\text{x}[6(\log\text{x})^2+7\log\text{x}+2]}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{\text{dx}}{\text{x}[6(\log\text{x})^2+7\log\text{x}+2]}$ $=\int\frac{1}{\text{x}(2\log\text{x}+1)(3\log\text{x}+2)}\text{ dx}$ Now,Let $\frac{1}{\text{x}(2\log\text{x}+1)(3\log\text{x}+2)}=\frac{\text{A}}{\text{x}(2\log\text{x}+1)}+\frac{\text{B}}{\text{x}(3\log\text{x}+2)}$
$\Rightarrow1=\text{A}(3\log\text{x}+2)+\text{B}(2\log\text{x}+1)$ Put $\text{x}=10^{-\frac12{}{}}$ $\Rightarrow1=\frac{1}{2}\text{A}\Rightarrow\text{A}=2$ $-\frac{2}{3}$ Put $\text{x}=10)^{-\frac23}$ $\Rightarrow1=-\frac{1}{3}\text{B}\Rightarrow\text{B}=-3$ $\therefore\text{I}=\int\frac{2\text{dx}}{\text{x}(2\log\text{x}+1)}-\int\frac{3\text{dx}}{\text{x}(3\log\text{x}+2)}$ $=\log|2\log\text{x}+1|-\log|3\log\text{x}+2|\text{C}$ $\therefore\text{I}=\log\Big|\frac{2\log\text{x}+1}{3\log\text{x}+2}\Big|+\text{C}$
View full question & answer→Question 1785 Marks
Evaluate the following integrals:
$\int\sin^7\text{x}\text{ dx}$
AnswerLet $\text{I}=\int\sin^7\text{x}\text{ dx}$ Then
$\text{I}=\int\sin^6\text{x }\sin\text{x}\text{ dx}$
$=\int\big(\sin^2\text{x}\big)^3\sin\text{x}\text{ dx}$
$=\int\big(1-\cos^2\text{x}\big)^3\sin\text{x}\text{ dx}$
$=\int\big(1-\cos^6\text{x}+3\cos^4\text{x}-3\cos^2\text{x}\big)\sin\text{x}\text{ dx}$
$\text{I}=\int\sin\text{x}\text{ dx}-\int\cos^6\text{x }\sin\text{x}\text{ dx}+3\int\cos^4\text{x }\sin\text{x}\text{ dx}-3\int\cos^2\text{x }\sin\text{x}\text{ dx}$
Putting $\cos\text{x}=\text{t}$ and $-\sin\text{x}\text{dx}=\text{dt}$ in $2^{nd}, 3^{rd}$ and $4^{th}$ integral we get
$\text{I}=\int\sin\text{x}\text{ dx}-\int\text{t}^6(-\text{dt})+3\int\text{t}^4(-\text{dt})-3\int\text{t}^2(-\text{dt})$
$=-\cos\text{x}+\frac{\text{t}^7}{7}-\frac{3}{5}\text{t}^5+\frac{3}{3}\text{t}^3+\text{C}$
$=-\cos\text{x}+\frac{\cos^7\text{x}}{7}-\frac{3}{5}\cos^5\text{x}+\cos^3\text{x}+\text{C}$
$\therefore\ \text{I}=-\cos\text{x}+\cos^3\text{x}-\frac{3}{5}\cos^5\text{x}+\frac{1}{7}\cos^7\text{x}+\text{C}$
View full question & answer→Question 1795 Marks
Evaluate the following integrals:
$\int\frac{\text{x}}{\sqrt{\text{x}^2+\text{a}^2}+\sqrt{\text{x}^2-\text{a}^2}}\text{ dx}$
Answer$\int\frac{\text{x}}{\sqrt{\text{x}^2+\text{a}^2}+\sqrt{\text{x}^2-\text{a}^2}}\text{ dx}$ Let $\text{x}^2=\text{t}$ $\Rightarrow2\text{x}=\frac{\text{dt}}{\text{dx}}$ $\Rightarrow\text{x dx}=\frac{\text{dt}}{2}$ Now, $\int\frac{\text{x}}{\sqrt{\text{x}^2+\text{a}^2}+\sqrt{\text{x}^2-\text{a}^2}}\text{ dx}$$=\frac{1}{2}\int\frac{\text{dt}}{\sqrt{\text{t}+\text{a}^2}+\sqrt{\text{t}-\text{a}^2}}$
$=\frac{1}{2}\int\frac{\text{dt}}{\Big(\sqrt{\text{t}+\text{a}^2}+\sqrt{\text{t}-\text{a}^2}\Big)}\times\frac{\Big(\sqrt{\text{t}+\text{a}^2}-\sqrt{\text{t}-\text{a}^2}\Big)}{\Big(\sqrt{\text{t}+\text{a}^2}-\sqrt{\text{t}-\text{a}^2}\Big)}$
$=\frac{1}{2}\int\frac{\Big(\sqrt{\text{t}+\text{a}^2}-\sqrt{\text{t}-\text{a}^2}\Big)}{(\text{t}+\text{a}^2)-(\text{t}-\text{a}^2)}\text{ dt}$
$=\frac{1}{4\text{a}^2}\int\Big(\text{t}+\text{a}^2\Big)^{\frac12}\text{dt}-\frac{1}{4\text{a}^2}\big(\text{t}-\text{a}^2\big)^{\frac12}\text{dt}$
$=\frac{1}{4\text{a}^2}\begin{bmatrix}\frac{\big(\text{t}+\text{a}^2\big)^{\frac12+1}}{\frac{1}2+1}\end{bmatrix}-\frac{1}{4\text{a}^2}\begin{bmatrix}\frac{\big(\text{t}-\text{a}^2\big)^{\frac12+1}}{\frac12+1}\end{bmatrix}+\text{C}$
$=\frac{1}{6\text{a}^2}\begin{bmatrix}(\text{t}+\text{a}^2)^{\frac{3}{2}}-(\text{t}-\text{a}^2)^{\frac32}\end{bmatrix}+\text{C}$
$=\frac{1}{6\text{a}^2}\begin{bmatrix}(\text{x}^2+\text{a}^2)^{\frac32}-(\text{x}^2-\text{a}^2)^{\frac32}\end{bmatrix}+\text{C}$
View full question & answer→Question 1805 Marks
Evalute the following integrals:
$\int\frac{1}{\cos(\text{x}+\text{a})\cos(\text{x}+\text{b})}\text{dx}$
Answer$\int\frac{1}{\cos(\text{x}+\text{a})\cos(\text{x}+\text{b})}\text{dx}$
Multiplying and Dividing by $\sin\big[(\text{x}+\text{b})-(\text{x}+\text{a})\big],$ we get
$=\int\frac{1}{\sin\big[(\text{x}+\text{b})-(\text{x}+\text{a})\big]}\times\frac{\sin\big[(\text{x}+\text{b})-(\text{x}+\text{a})\big]}{\cos(\text{x}+\text{a})\cos(\text{x}+\text{b})}\text{dx}$
$=\int\frac{1}{\sin(\text{b}-\text{a})}\times\frac{\sin\big[(\text{x}+\text{b})-(\text{x}+\text{a})\big]}{\cos(\text{x}+\text{a})\cos(\text{x}+\text{b})}\text{dx}$
$=\frac{1}{\sin(\text{b}-\text{a})}\int\frac{\sin(\text{x}+\text{b})\cos(\text{x}+\text{a})-\sin(\text{x}+\text{a})\cos(\text{x}+\text{b})}{\cos(\text{x}+\text{a})\cos(\text{x}+\text{b})}\text{dx}$
$=\frac{1}{\sin(\text{b}-\text{a})}\Big[\int\frac{\sin(\text{x}+\text{b})}{\cos(\text{x}+\text{b})}\text{dx}-\int\frac{\sin(\text{x}+\text{a})}{\cos(\text{x}+\text{a})}\text{dx}\Big]$
$=\frac{1}{\sin(\text{b}-\text{a})}\Big[\int\tan(\text{x}+\text{b})\text{dx}-\int\tan(\text{x}+\text{a})\text{dx}\Big]$
$=\frac{1}{\sin(\text{b}-\text{a})}\big[\log(\sec(\text{x}+\text{b}))-\log(\sec(\text{x}+\text{a}))\big]+\text{C}$
$=\frac{1}{\sin(\text{b}-\text{a})}\bigg[\log\Big(\frac{\sec(\text{x}+\text{b})}{\sec(\text{x}+\text{a})}\Big)\bigg]+\text{C}$
Hence, $\int\frac{1}{\cos(\text{x}+\text{a})\cos(\text{x}+\text{b})}\text{dx}=\frac{1}{\sin(\text{b}-\text{a})}\bigg[\log\Big(\frac{\sec(\text{x}+\text{b})}{\sec(\text{x}+\text{a})}\Big)\bigg]+\text{C}$
View full question & answer→Question 1815 Marks
$\int(2\text{x}^2+3)\sqrt{\text{x+2}}\text{dx}$
AnswerLet $\text{I}=\int(2\text{x}^2+3)\sqrt{\text{x+2}}\text{dx}$
Substituting x + 2 = t and dx = dt, we get
$\text{I}=\int\big[2(\text{t}-2)^2+3\big]\sqrt{\text{t}}\text{dt}$
$=\int\big[2(\text{t}^2+4-4\text{t})+3\big]\sqrt{\text{t}}\text{dt}$
$=\int\big[2\text{t}^2+8-8\text{t}+3\big]\sqrt{\text{t}}\text{dt}$
$=\int\Big(2\text{t}^{\frac{5}{2}}+11\text{t}^{-\frac{1}{2}}-8\text{t}^{\frac{3}{2}}\Big)\text{dt}$
$=\frac{4}{7}\text{t}^{\frac{7}{2}}+\frac{22}{3}\text{t}^{\frac{3}{2}}-\frac{16}{5}\text{t}^{\frac{5}{2}}+\text{C}$
$=\frac{4}{7}(\text{x}+2)^\frac{\text{7}}{2}-\frac{16}{5}(\text{x}+2)^\frac{\text{5}}{2}+\frac{22}{3}(\text{x}+2)^\frac{\text{3}}{2}+\text{C}$
$\therefore\ \text{I}=\frac{4}{7}(\text{x}+2)^\frac{\text{7}}{2}-\frac{16}{5}(\text{x}+2)^\frac{\text{5}}{2}+\frac{22}{3}(\text{x}+2)^\frac{\text{3}}{2}+\text{C}$
View full question & answer→Question 1825 Marks
Evaluate the following integrals:
$\int\sin^3\sqrt{\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\sin^3\sqrt{\text{x}}\text{dx}$
$\sqrt{\text{x}}=\text{t}$
$\text{x = t}^2$
$\text{dx}=2\text{t dt}$
$\text{I}=2\int\text{t}\sin^3\text{t dt}$
$=2\int\text{t}\Big(\frac{3\sin\text{t}-\sin3\text{t}}{4}\Big)\text{dt}$
$=\frac{1}{2}\int\text{t}(3\sin\text{t}-\sin3\text{t})\text{dt}$
Using integration by parts,
$\text{I}=\frac{1}2{}\Big[\text{t}\Big(-3\cos\text{t}+\frac{1}{3}\cos3\text{t}\Big)-\int\Big(-3\cos\text{t}+\frac{\cos3\text{t}}{3}\Big)\text{dt}\Big]$
$=\frac{1}{2}\Big[\frac{-9\text{t}\cos\text{t}+\text{t}\cos3\text{t}}{3}-\Big\{-3\sin\text{t}+\frac{\sin3\text{t}}{9}\Big\}\Big]+\text{C}$
$=\frac{1}{2}\Big[\frac{-9\text{t}\cos\text{t}+\text{t}\cos3\text{t}}{3}+\frac{27\sin\text{t}-3\sin3\text{t}}{9}\Big]+\text{C}$
$=\frac{1}{18}\big[-27\text{t}\cos\text{t}+3\text{t}\cos3\text{t}+27\sin\text{t}-3\sin3\text{t}\big]+\text{C}$
$\text{I}=\frac{1}{18}\big[3\sqrt{\text{x}}\cos3\sqrt{\text{x}}+27\sin\sqrt{\text{x}}-27\sqrt{\text{x}}\cos\sqrt{\text{x}}-3\sin3\sqrt{\text{x}}\big]+\text{C}$
View full question & answer→Question 1835 Marks
Evalute the following integrals:
$\int\frac{1-\cot\text{x}}{1+\cot\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\frac{1-\cot\text{x}}{1+\cot\text{x}}\text{dx}$ then,
$\text{I}=\int\frac{1-\frac{\cos\text{x}}{\sin\text{x}}}{1+\frac{\cos\text{x}}{\sin\text{x}}}\text{dx}$
$=\int\frac{\frac{\sin\text{x}-\cos\text{x}}{\sin\text{x}}}{\frac{\sin\text{x}+\cos\text{x}}{\sin\text{x}}}\text{dx}$
$\Rightarrow\text{I}=\int\frac{\sin\text{x}-\cos\text{x}}{\sin\text{x}+\cos\text{x}}\text{dx}\ .....(\text{i})$
Let $\sin\text{x}+\cos\text{x}=\text{t},$ then,
$\text{d}(\sin\text{x}+\cos\text{x})=\text{dt}$
$\Rightarrow(\cos\text{x}-\sin\text{x})\text{dx}=\text{dt}$
$\Rightarrow-(\sin\text{x}-\cos\text{x})\text{dx}=\text{dt}$
$\Rightarrow\text{dx}=-\frac{\text{dt}}{\sin\text{x}-\cos\text{x}}$
Putting $\sin\text{x}+\cos\text{x}=\text{t and dx}=-\frac{\text{dt}}{\sin\text{x}-\cos\text{x}}$ in equation (i), we het
$\text{I}=\int\frac{\sin\text{x}-\cos\text{x}}{\text{t}}\times\frac{-\text{dt}}{\sin\text{x}-\cos\text{x}}$
$=\int\frac{-\text{dt}}{\text{t}}$
$=-\log|\text{t}|+\text{C}$
$=-\log|\sin\text{x}+\cos\text{x}|+\text{C}$
View full question & answer→Question 1845 Marks
Evalute the following integrals:
$\int\tan2\text{x}\tan3\text{x}\tan5\text{x dx}$
AnswerLet $\text{I}=\int1+\tan\text{x}\tan(\text{x}+\theta)\text{dx}$
$=\int1+\tan\text{x}\Big(\frac{\tan\text{x}+\tan\theta}{1-\tan\text{x}\tan\theta}\Big)\text{dx}$
$=\int\frac{1+\tan^2\text{x}}{1-\tan\text{x}\tan\theta}\text{dx}$
$=\int\frac{\sec^2\text{x dx}}{1-\tan\text{x}\tan\theta}$
Putting $\tan\text{x}=\text{t}$
$\Rightarrow\sec^2\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{dx}=\frac{\text{dt}}{\sec^2\text{x}}$
$\therefore\text{I}\approx\int\frac{1}{1-\text{t}\tan\theta}\text{dt}$
$=\frac{-1}{\tan\theta}\text{ ln}|1-\text{t}\tan\theta|+\text{C}$
$\Big[\because\int\frac{1}{\text{ax}+\text{b}}\text{dx}=\frac{1}{\text{a}}\text{ ln}|\text{ax}+\text{b}|+\text{C}\Big]$
$=-\cot\theta\text{ ln}|1-\tan\text{ x }\tan\theta|+\text{C}$
$=\cot\theta\text{ ln}\Big|\frac{1}{1-\tan\text{ x }\tan\theta}\Big|+\text{C}$
$=\cot\theta\text{ ln}\Big|\frac{\cos\text{ x }\cos\theta}{\cos\text{x}\cos\theta-\sin\text{x}\sin\theta}\Big|+\text{C}$
$=\cot\theta\text{ ln}\Big|\frac{\cos\text{x}}{\cos(\text{x}+\theta)}\Big|+\text{C }\big[\text{Let C}'=\text{C}+\cot\theta\text{ ln}\cos\theta\big]$
View full question & answer→Question 1855 Marks
Evaluate the following integrals:
$\int \frac{\sin^5\text{x}}{\cos^4\text{x}}\text{ dx}$
Answer$\int \frac{\sin^5\text{x}}{\cos^4\text{x}}\text{ dx}$ $=\int\Big(\frac{\sin^4\text{x}.\sin\text{x}}{\cos^4\text{x}}\Big)\text{dx}$ $=\int\frac{(\sin^2\text{x})^2\sin\text{x}}{\cos^4\text{x}}\text{ dx}$ $=\int\frac{(1-\cos^2\text{x})^2\sin\text{x}}{\cos^4\text{x}}\text{ dx}$ $=\int\Big(\frac{1+\cos^4\text{x}-2\cos^2\text{x}}{\cos^4\text{x}}\Big)\sin\text{x dx }$ $=\int\Big(\frac{1}{\cos^4\text{x}}+1-\frac{2}{\cos^2\text{x}}\Big)\sin\text{x dx}$ Let $\cos\text{x}=\text{t}$ $\Rightarrow-\sin\text{x}=\frac{\text{dt}}{\text{dx}}$ $$$\Rightarrow\sin\text{x dx}=-\text{dt}$Now, $=\int\Big(\frac{1}{\cos^4\text{x}}+1-\frac{2}{\cos^2\text{x}}\Big)\sin\text{x dx}$
$=-\int\big(\text{t}^{-4}+1-2\text{t}^{-2}\big)\text{ dt}$ $=-\Big[-\frac{\text{t}^{-4+1}}{-4+1}+\text{t}-\frac{2\text{t}^{-2+1}}{-2+1}\Big]+\text{C}$ $=-\Big[-\frac{1}{3\text{t}^3}+\text{t}+\frac{2}{\text{t}}\Big]+\text{C}$ $=\frac{1}{3\text{t}^3}-\text{t}-\frac{2}{\text{t}}+\text{C}$ $=\frac{1}{3\cos^3\text{x}}-\cos\text{x}-\frac{2}{\cos\text{x}}+\text{C}$
View full question & answer→Question 1865 Marks
Evaluate the following integrals:
$\int\frac{1}{\sin^3\text{x}\cos\text{x}}\text{dx}$
Answer$\int\frac{1}{\sin^3\text{x}\cos\text{x}}\text{ dx}$
Dividing numerator & denominator by $\sin^4\text{x}$
$=\int\frac{\frac{1}{\sin^4\text{x}}\text{ dx}}{\frac{\sin^3\text{x}\cdot\cos\text{x}}{\sin^4\text{x}}}$
$=\int\frac{\text{cosec}^4\text{x}\text{ dx}}{\cot\text{x}}$
$=\int\frac{\text{cosec}^2\text{x}\cdot\text{cosec}^2\text{x}\text{ dx}}{\cot\text{x}}$
$=\int\frac{(1+\cot^2\text{x})\cdot\text{cosec}^2\text{x}\text{ dx}}{\cot\text{x}}$
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, $\int\frac{(1+\cot^2\text{x})\cdot\text{cosec}^2\text{x}}{\cot\text{x}}\text{ dx}$
$=\int\frac{(1+\text{t}^2)\cdot(-\text{dt})}{\text{t}}$
$=-\int\Big(\frac{1}{\text{t}}+\text{t}\Big)\text{dt}$
$=-\log|\text{t}|-\frac{\text{t}^2}{2}+\text{C}$
$=-\log|\cot\text{x}|-\frac{\cot^2\text{x}}{1}+\text{C}$
$=\log|\cot\text{x}|^{-1}-\frac{(\text{cosec}^2\text{x}-1)}{2}+\text{C}$
$=\log\Big|\frac{1}{\cot\text{x}}\Big|-\frac{\text{cosec}^2\text{x}}{2}+\frac{1}{2}+\text{C}$
$=\log|\tan\text{x}|-\frac{1}{2\sin^2\text{x}}+\text{C}'$ $\Big[\therefore\text{C}'=\text{C}+\frac{1}{2}\Big]$
View full question & answer→Question 1875 Marks
Evaluate the following intregals: $\int\frac{1}{1-2\sin\text{x}}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{1}{1-2\sin\text{x}}\ \text{dx}$
Put $\sin\text{x}=\frac{2\tan^\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}$
$\text{I}=\int\frac{1}{1-2\Bigg(\frac{2\tan\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}\Bigg)}\ \text{dx}$
$=\int\frac{1+\tan^2\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}-4\tan\frac{\text{x}}{2}}\ \text{dx}$
$=\int\frac{\sec^2\frac{\text{x}}{2}}{\tan^2\frac{\text{x}}{2}-4\tan\frac{\text{x}}{2}+1}\ \text{dx}$
Let $\tan\frac{\text{x}}{2}=\text{t}$
$\frac{1}{2}\sec^2\frac{\text{x}}{2}\text{dx}=\text{dt}$
$=\int\frac{2\text{dt}}{\text{t}^2-4\text{t}+1}$
$=\int\frac{2\text{dt}}{\text{t}^2-2\text{t}(2)+(2)^2-(2)^2+1}$
$\text{I}=2\int\frac{\text{dt}}{(\text{t}-2)^2+(\sqrt{3})^2}$
$=2\times\frac{1}{2\sqrt{3}}\log\Big|\frac{\text{t}-2-\sqrt{3}}{\text{t}-2+\sqrt{3}}\Big|+\text{C}$
$\text{I}=\frac{1}{\sqrt{3}}\log\Bigg|\frac{\tan\frac{\text{x}}{2}-2-\sqrt{3}}{\tan\frac{\text{x}}{2}-2+\sqrt{3}}\Bigg|+\text{C}$
View full question & answer→Question 1885 Marks
Evaluate the following integrals:
$\int\frac{\text{x}^2}{\text{x}^4+\text{x}^2-2}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{\text{x}^2}{\text{x}^4+\text{x}^2-2}\ \text{dx}$
we express
$\frac{\text{x}^2}{\text{x}^4+\text{x}^2-2}=\frac{\text{x}^2}{\text{x}^4+2\text{x}^2-\text{x}^2-2}$
$=\frac{\text{x}^2}{(\text{x}^2+2)(\text{x}^2-1)}$
$=\frac{\text{A}}{\text{x}^2+2}+\frac{\text{B}}{\text{x}^2-1}$
$\Rightarrow\text{x}^2=\text{A}(\text{x}^2-1)+\text{B}(\text{x}^2+2)$
Equating the coefficient of x and constants, we get
1 = A + B and 0 = -A + 2B or
$\text{A}=\frac{2}{3}$ and $\text{B}=\frac{1}{3}$
$\therefore\text{I}=\int\Big(\frac{\frac{2}{3}}{\text{x}^2+2}+\frac{\frac{1}{3}}{\text{x}^2-1}\Big)\text{dx}$
$=\frac{2}{3}\int\frac{1}{\text{x}^2+2}\ \text{dx}+\frac{1}{3}\int\frac{1}{\text{x}^2-1}\ \text{dx}$
$=\frac{\sqrt{2}}{3}\tan^{-1}\frac{\text{x}}{\sqrt{2}}+\frac{1}{6}\log\Big|\frac{\text{x}-1}{\text{x}+1}\Big|+\text{C}$
Hence, $\int\frac{\text{x}^2}{\text{x}^4+\text{x}^2-2}\ \text{dx}=\frac{\sqrt{2}}{3}\tan^{-1}\frac{\text{x}}{\sqrt{2}}+\frac{1}{6}\log\Big|\frac{\text{x}-1}{\text{x}+1}\Big|+\text{C}$
View full question & answer→Question 1895 Marks
Evaluate the following integrals:
$\int\frac{3\text{x}+5}{\text{x}^3-\text{x}^2-\text{x}+1}\ \text{dx}$
AnswerTo evaluate the integrals follow the steps:
$\int\frac{3\text{x}+5}{\text{x}^3-\text{x}^2-\text{x}+1}\ \text{dx}$
Let $\frac{3\text{x}-5}{(\text{x}-1)^2(\text{x}+1)}=\frac{\text{A}}{\text{x}-1}+\frac{\text{B}}{(\text{x}-1)^2}+\frac{\text{C}}{\text{x}+1}$
$3\text{x}+5=\text{A}(\text{x}-1)(\text{x}+1)+\text{B}(\text{x}+1)+\text{C}(\text{x}-1)^2$
For x = 1 B = 4
$\text{For x} = -1\ \text{C}=\frac{1}{2}$
$\text{For x} = 0\ \text{A}=-\frac{1}{2}$
Therefore
$\int\frac{3\text{x}+5}{(\text{x}-1)^2(\text{x}+1)}\ \text{dx}=-\frac{1}{2}\int\frac{\text{dx}}{\text{x}-1}+\frac{1}{2}\int\frac{\text{dx}}{\text{x}+1}$
$=-\frac{1}{2}\ln|(\text{x}-1)|-\frac{4}{(\text{x}-1)}+\frac{1}{2}\ln|(\text{x}+1)|+\text{C}$
$=\frac{1}{2}\ln\big|\frac{\text{x}+1}{\text{x}-1}\big|-\frac{4}{(\text{x}-1)}+\text{C}$
View full question & answer→Question 1905 Marks
Evaluate the following intregals:
$\int\frac{\text{x}^2+6\text{x}-8}{\text{x}^3-4\text{x}}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{\text{x}^2+6\text{x}-8}{\text{x}^3-4\text{x}}\ \text{dx}$ $\Rightarrow\text{I}=\int\frac{\text{x}^2+6\text{x}-8}{\text{x}(\text{x}+2)(\text{x}-2)}\text{ dx}$Now,
Let $\frac{\text{x}^2+6\text{x}-8}{\text{x}(\text{x}+2)(\text{x}-2)}=\frac{\text{A}}{\text{x}}+\frac{\text{B}}{\text{x}+2}+\frac{\text{C}}{\text{x}-2}$
$\Rightarrow\text{x}^2+6\text{x}-8=\text{A}(\text{x}^2-4)+\text{B}(\text{x}-2)\text{x}+\text{C}(\text{x}+2)\text{x}$ Put x = 0 ⇒ -8 = -4A ⇒ A = 2 Put x = -2 ⇒ -16 = 8B ⇒ B = -2 Put x = 2 ⇒ 8 = 8C ⇒ C = 1 Thus,$\text{I}=\int\frac{2\text{dx}}{\text{x}}-\int\frac{2\text{dx}}{\text{x}+2}+\int\frac{\text{dx}}{\text{x}-2}$
$=2\log|\text{x}|-2\log|\text{x}+2|+\log|\text{x}-2|+\text{C}$
$\therefore\text{I}=\log\Big|\frac{\text{x}^2(\text{x}-2)}{(\text{x}+2)^2}\Big|+\text{C}$
View full question & answer→Question 1915 Marks
Evaluate the following intregals:
$\int\frac{\text{x}+1}{\sqrt{\text{x}^2+1}}\text{dx}$
AnswerLet $\text{I}=\int\frac{\text{x}+1}{\sqrt{\text{x}^2+1}}\text{dx}$
let $\text{x}+1=\lambda\frac{\text{d}}{\text{dx}}(\text{x}^2+1)+\mu$
$\text{x}+1=\lambda(2\text{x})+\mu$
Compairing the coefficient of like powewrs of x,
$2\lambda=1\ \Rightarrow\lambda=\frac{1}{2}$
$\Rightarrow\mu=1$
So, $\text{I}=\int\frac{\frac{1}{2}(2\text{x})+1}{\sqrt{\text{x}^2+1}}\text{dx}$
$=\frac{1}{2}\int\frac{(2\text{x})}{\sqrt{\text{x}^2+1}}\text{dx}+\int\frac{1}{\sqrt{\text{x}^2+1}}\text{dx}$
$\text{I}=\frac{1}{2}\times2\sqrt{\text{x}^2+1}+\log\big|\text{x}+\sqrt{\text{x}^2+1}\big|+\text{C}$ $\big[\text{since}, \int\frac{1}{\sqrt{\text{x}}}\text{dx}=2\sqrt{\text{x}}+\text{c},\int\frac{1}{\sqrt{\text{x}^2+1}}\text{dx}=\log\big|\text{x}+\sqrt{\text{x}^2-\text{x}^2}\big|+\text{C}\big]$
$\text{I}=\sqrt{\text{x}^2+1}+\log\big|\text{x}+\sqrt{\text{x}^2+1}\big|+\text{C}$
View full question & answer→Question 1925 Marks
Evalute the following integrals:
$\int\frac{-\sin\text{x}+2\cos\text{x}}{2\sin\text{x}+\cos\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\frac{-\sin\text{x}+2\cos\text{x}}{2\sin\text{x}+\cos\text{x}}\text{dx}\ .....\text{(i)}$
Let $2\sin\text{x}+\cos\text{x}=\text{t}$ then,
$\text{d}(2\sin\text{x}+\cos\text{x})=\text{dt}$
$\Rightarrow(2\cos\text{x}-\sin\text{x})\text{dx}=\text{dt}$
$\Rightarrow\text{dx}=\frac{\text{dt}}{-\sin\text{x}+2\cos\text{x}}$
Putting $2\sin\text{x}+\cos\text{x}=\text{t and dx}=\frac{\text{dt}}{-\sin\text{x}+2\cos\text{x}}$ in equation (i), we get,
$\text{I}=\int\frac{-\sin\text{x}+2\cos\text{x}}{\text{t}}\times\frac{\text{dt}}{-\sin\text{x}+2\cos\text{x}}$
$=\int\frac{\text{dt}}{\text{t}}$
$=\log|\text{t}|+\text{C}$
$=\log|2\sin\text{x}+\cos\text{x}|+\text{C}$
$\therefore\text{I}=\log|2\sin\text{x}+\cos\text{x}|+\text{C}$
View full question & answer→Question 1935 Marks
Evaluate the following integrals:
$\int\sec^4\text{x}\tan\text{x}\text{ dx}$
AnswerLet $\text{I}=\int\sec^4\text{x}\tan\text{x}\text{ dx}\ ....(1)$
Let $\tan\text{x}=\text{t}$ then,
$\Rightarrow\text{d}(\tan\text{x})=\text{dt}$
$\Rightarrow\sec^2\text{x}\text{ dx}=\text{dt}$
$\Rightarrow\text{dx}=\frac{\text{dt}}{\sec^2\text{x}}$
Putting $\tan\text{x}=\text{t}$ and $\text{dx}=\frac{\text{dt}}{\sec^2\text{x}}$ in equation (1), we get,
$\text{I}=\int\sec^4\text{x}\tan\text{x}\frac{\text{dt}}{\sec^2\text{x}}$
$=\int\sec^2\text{x}\text{ t dt}$
$=\int\big(1+\tan^2\text{x}\big)\text{t dt}$
$=\int\big(1+\text{t}^2\big)\text{t dt}$
$=\int\big(\text{t}+\text{t}^3\big)\text{dt}$
$=\frac{\text{t}^2}{2}+\frac{\text{t}^4}{4}+\text{C}$
$=\frac{\tan^2\text{x}}{2}+\frac{\tan^4\text{x}}{4}+\text{C}$
$\text{I}=\frac{1}{2}\tan^2\text{x}+\frac{1}{4}\tan^4\text{x}+\text{C}$
View full question & answer→Question 1945 Marks
Evalute the following integrals:
$\int\frac{1+\cot\text{x}}{\text{x}+\log\sin\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\frac{1+\cot\text{x}}{\text{x}+\log\sin\text{x}}\text{dx}\ .....(\text{i})$
Let $\text{x}+\log\sin\text{x}=\text{t}$ then,
$\text{d}(\text{x}+\log\sin\text{x})=\text{dt}$
$\Rightarrow(1+\cot\text{x})\text{dx}=\text{dt}\Big[\because\frac{\text{d}}{\text{dx}}(\log\sin\text{x})=\cot\text{x}\Big]$
$\Rightarrow\text{dx}=\frac{\text{dt}}{1+\cot\text{x}}$
Putting $\text{x}+\log\sin\text{x}=\text{t}$ and $\text{dx}=\frac{\text{dt}}{1+\cot\text{x}}$ In equation (i), we get,
$\text{I}=\int\frac{1+\cot\text{x}}{\text{t}}\times\frac{\text{dt}}{1+\cot\text{x}}$
$=\int\frac{\text{dt}}{\text{t}}$
$=\log|\text{t}|+\text{C}$
$=\log|\text{x}+\log\sin\text{x}|+\text{C}$
$\therefore\text{I}=\log|\text{x}+\log\sin\text{x}|+\text{C}$
View full question & answer→Question 1955 Marks
Evaluate the following integrals:
$\int\frac{\text{x}^2}{1-\text{x}^4}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{\text{x}^2}{1-\text{x}^4}\ \text{dx}$
we express
$\frac{\text{x}^2}{1-\text{x}^4}=\frac{\text{x}^2}{(1-\text{x})^2(1+\text{x})^2}$
$=\frac{\text{A}}{1-\text{x}^2}+\frac{\text{B}}{1+\text{x}^2}$
$\Rightarrow\text{x}^2=\text{A}(1+\text{x}^2)+\text{B}(1-\text{x})^2$
Equating the coefficient of x and constants, we get
1 = A - B and 0 = A + B or
$\text{A}=\frac{1}{2}$ and $\text{B}=-\frac{1}{2}$
$\therefore\text{I}=\int\Big(\frac{\frac{1}{2}}{1-\text{x}^2}+\frac{-\frac{1}{2}}{1+\text{x}^2}\Big)\text{dx}$
$=\frac{1}{2}\int\frac{1}{1-\text{x}^2}\ \text{dx}-\frac{1}{2}\int\frac{1}{1+\text{x}^2}\ \text{dx}$
$=\frac{1}{2}\times\frac{1}{2}\log\Big|\frac{1+\text{x}}{1-\text{x}}\Big|-\frac{1}{2}\tan^{-1}\text{x}+\text{C}$
$=\frac{1}{4}\log\Big|\frac{1+\text{x}}{1-\text{x}}\Big|-\frac{1}{2}\tan^{-1}\text{t}+\text{C}$
Hence, $\int\frac{\text{x}^2}{1-\text{x}^4}\ \text{dx}=\frac{1}{4}\log\Big|\frac{1+\text{x}}{1-\text{x}}\Big|-\frac{1}{2}\tan^{-1}\text{x}+\text{C}$
View full question & answer→Question 1965 Marks
Integrate the following integrals:
$\int\sin\text{x}\cos2\text{x}\sin3\text{x dx}$
Answer$\int\sin\text{x}\cos2\text{x}\sin3\text{x dx}$
$=\frac{1}{2}\int(2\sin\text{x}\cos2\text{x})\sin3\text{x dx}$
$=\frac{1}{2}\int\big[\sin(\text{x}+2\text{x})+\sin(\text{x}-2\text{x})\big]\sin(3\text{x) dx}$
$=\frac{1}{2}\int\big[\sin(3\text{x})-\sin(\text{x})\big]\sin(3\text{x) dx}$
$=\frac{1}{2}\big[\int\sin^2(3\text{x})\text{dx}-\int\sin(\text{x})\sin(3\text{x})\text{dx}\big]$
$=\frac{1}{4}\big[\int2\sin^2(3\text{x})\text{dx}-\int2\sin(\text{x})\sin(3\text{x})\text{dx}\big]$
$=\frac{1}{4}\Big\{\int\big[1-\cos(6\text{x})\big]\text{dx}-\int\big[\cos(\text{x}-3\text{x})-\cos(\text{x}+3\text{x})\big]\text{dx}\Big\}$
$=\frac{1}{4}\big[\int1\text{dx}-\int\cos(6\text{x})\text{dx}-\int\cos(2\text{x})\text{dx}+\int\cos(4\text{x})\text{dx}\big]$
$=\frac{1}{4}\Big[\text{x}-\frac{\sin(6\text{x})}{6}-\frac{\sin(2\text{x})}{2}+\frac{\sin(4\text{x})}{4}\Big]+\text{C}$
$=\frac{\text{x}}{4}-\frac{\sin(6\text{x})}{24}-\frac{\sin(2\text{x})}{8}+\frac{\sin(4\text{x})}{16}+\text{C}$
Hence, $\int\sin\text{x}\cos2\text{x}\sin3\text{x}\text{ dx}$ $=\frac{\text{x}}{4}-\frac{\sin(6\text{x})}{24}-\frac{\sin(2\text{x})}{8}+\frac{\sin(4\text{x})}{16}+\text{C}$
View full question & answer→Question 1975 Marks
Evaluate the following integrals:
$\int(4\text{x}+2)\sqrt{\text{x}^2+\text{x}+1}\text{ dx}$
Answer$\int(4\text{x}+2)\sqrt{\text{x}^2+\text{x}+1}\text{ dx}$
$=2\int(2\text{x}+1)\sqrt{\text{x}^2+\text{x}+1}\text{ dx}$
$\text{Let }\text{x}^2+\text{x}+1=\text{t}$
$\Rightarrow(2\text{x}+1)=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow(2\text{x}+1)\text{dx}=\text{dt}$
$\text{Now, }2\int(2\text{x}+1)\sqrt{\text{x}^2+\text{x}+1}\text{ dx}$
$=2\int\sqrt{\text{t}}\text{ dt}$
$=2\int\text{t}^\frac{1}{2}\text{dt}$
$=2\bigg[\frac{\text{t}^{\frac{1}{2}+1}}{\frac{1}{2}+1}\bigg]+\text{C}$
$=2\times\frac{2}{3}\text{t}^\frac{3}{2}+\text{C}$
$=\frac{4}{3}\text{t}^\frac{3}{2}+\text{C}$
$=\frac{4}{3}(\text{x}^2+\text{x}+1)^\frac{3}{2}+\text{C}$
View full question & answer→Question 1985 Marks
Evaluate the following integrals:
$\int\text{e}^{\text{x}}\sin^2\text{x }\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{x}}\sin^2\text{x }\text{dx}$
$=\frac{1}{2}\int\text{e}^\text{x}\ 2\sin^2\text{x dx}$
$=\frac{1}{2}\int\text{e}^\text{x}(1-\cos2\text{x})\text{dx}$
$=\frac{1}{2}\int\text{e}^\text{x}\text{dx}-\frac{1}{2}\int\text{e}^\text{x}\cos2\text{ x dx}$
$\because\ \int\text{e}^{2\text{x}}\cos\text{bx dx}=\frac{\text{e}^{2\text{x}}}{\text{a}^2+\text{b}^2}\{\text{a}\cos\text{bx}-\text{b}\sin\text{bx}\}+\text{C}$
$\therefore\ \text{I}=\frac{1}{2}\big[\text{e}^\text{x}-\frac{\text{e}^\text{x}}{5}\{\cos2\text{x}+2\sin2\text{x}\}\big]+\text{C}$
$\therefore\ \text{I}=\frac{\text{e}^\text{x}}{2}-\frac{\text{e}^\text{x}}{10}\{\cos2\text{x}+2\sin2\text{x}\}+\text{C}$
View full question & answer→Question 1995 Marks
Evaluate the following intregals:
$\int\frac{5\text{x}}{(\text{x}+1)(\text{x}^2-4)}\text{dx}$
Answer$\int\frac{5\text{x}}{(\text{x}+1)(\text{x}^2-4)}=\frac{5\text{x}}{(\text{x}+1)(\text{x}+2)(\text{x}-2)}$
Let $\frac{5\text{x}}{(\text{x}+1)(\text{x}+2)(\text{x}-2)}=\frac{\text{A}}{(\text{x}+1)}+\frac{\text{B}}{(\text{x}+2)}+\frac{\text{C}}{(\text{x}-2)}$
$5\text{x}=\text{A}(\text{x}+2)(\text{x}-2)+\text{B}(\text{x}+1)(\text{x}-2)\\+\text{C}(\text{x}+1)(\text{x}+2)\ \dots(1)$
Substituting x = -1, -2 and 2 respectively in equation (1), we obtain
$\text{A}=\frac{5}{3},\text{B}=\frac{5}{2},\text{and }\text{C}=\frac{5}{6}$
$\therefore\frac{5\text{x}}{(\text{x}+1)(\text{x}+2)(\text{x}-2)}=\frac{5}{3(\text{x}+1)}-\frac{5}{2(\text{x}+2)}+\frac{5}{6(\text{x}-2)}$
$\Rightarrow\frac{5\text{x}}{(\text{x}+1)(\text{x}^2-4)}\ \text{dx}=\frac{5}{3}\int\frac{1}{(\text{x}+1)}\ \text{dx}-\frac{5}{2}\int\frac{1}{(\text{x}+2)}\\\ \text{dx}+\frac{5}{6}\int\frac{1}{(\text{x}-2)}\ \text{dx}$
$=\frac{5}{3}\log|\text{x}+1|-\frac{5}{2}\log|\text{x}+2|+\frac{5}{6}\log|\text{x}-2|+\text{C}$
View full question & answer→Question 2005 Marks
Evaluate the following integrals:$\int\frac{1}{2\text{x}^2-\text{x}-1}\text{dx}$
AnswerLet $\text{I}=\int\frac{1}{2\text{x}^2-\text{x}-1}\text{dx}$
$=\frac{1}{2}\int\frac{1}{\text{x}^2-\frac{\text{x}}{2}-\frac{1}{2}}\text{dx}$
$=\frac{1}{2}\int\frac{1}{\text{x}^2-2\text{x}\times\frac{1}{4}+\big(\frac{1}{4}\big)^2-\big(\frac{1}{4}\big)^2-\frac{1}{2}}\text{dx}$
$=\frac{1}{2}\int\frac{1}{\big(\text{x}-\frac{1}{4}\big)^2-\frac{9}{16}}\text{dx}$
Let $\text{x}-\frac{1}{4}=\text{t}$
$\Rightarrow\text{dx = dt}$
$\text{I}=\frac{1}{2}\int\frac{1}{\text{t}^2-\big(\frac{3}{4}\big)^2}\text{dt}$
$\text{I}=\frac{1}{2}\times\frac{1}{2\times\big(\frac{3}{4}\big)}\log\Bigg|\frac{\text{t}-\frac{3}{4}}{\text{t}+\frac{3}{4}}\Bigg|+\text{C} $ $\Big[\text{Since,}\int\frac{1}{\text{x}^2-\text{a}^2}\text{dx}=\frac{1}{2\text{a}}\log\bigg|\frac{\text{x}-\text{a}}{\text{x+a}}\bigg|+\text{C}\Big]$
$\text{I}=\frac{1}{3}\log\Bigg|\frac{\text{x}-\frac{1}{4}-\frac{3}{4}}{\text{x}-\frac{1}{4}+\frac{3}{4}}\Bigg|+\text{C}$
$\text{I}=\frac{1}{3}\log\bigg|\frac{\text{x}-1}{2\text{x}+1}\bigg|+\text{C}$
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