Question 1015 Marks
Evaluate the integral in Exercise:
$\int\limits^{1}_{0}\sin^{-1}\bigg(\frac{2\text{x}}{1+\text{x}^{2}}\bigg)\text{dx}$
Answer$\text{Let}\text{I}=\int\limits_{0}^{1}\sin^{-1}\bigg(\frac{2\text{x}}{1+\text{x}^{2}}\bigg)\text{dx}$
$\text{put}\ \text{x}=\tan\theta\ \text{so that}\ \text{dx}=\sec^{2}\theta\ \text{d}\theta\ \text{when}\ \text{x}=0,\tan\theta=0\Rightarrow\theta=0$
$\text{when}\ \text{x}=1,\tan\theta=1\ \Rightarrow\theta=\frac{\pi}{4}$
$\therefore |=\int^{\frac{\pi}{4}}_{0}\sin^{-1}\bigg(\frac{2\tan\theta}{1+\tan^{2}\theta}\bigg).\sec^{2}\theta\ \text{d}\theta$
$ =\int^{\frac{\pi}{4}}_{0}\sin^{-1}(\sin2\theta).\sec^{2}\theta\ \text{d}\theta=\int^{\frac{\pi}{4}}_{0}2\theta.\sec^{2}\theta\ \text{d}\theta=2\int^{\frac{\pi}{4}}_{0}\theta.\sec^{2}\theta\ \text{d}\theta$
$=2\left\{[\theta\tan\theta]^{\frac{\pi}{4}}_{1}-\int^{\frac{\pi}{4}}_{0}1.\tan\theta\ \text{d}\theta\right\}=2\left\{[\theta\tan\theta]^{\frac{\pi}{4}}_{0}+[\log\cos\theta]^{\frac{\pi}{4}}_{0}\right\}$
$=2\left\{\frac{\pi}{4}\tan\frac{\pi}{4}-0+\log\cos\frac{\pi}{4}-\log\cos0\right\}=2\bigg[\frac{\pi}{4}\times1-0+\log\frac{1}{\sqrt{2}}-\log1\bigg]$
$=2\bigg[\frac{\pi}{4}+\log1-\log\sqrt{2}\bigg]=\frac{\pi}{2}-\log2$
View full question & answer→Question 1025 Marks
$\int\text{x}(1-\text{x})^{23}\text{dx}$
AnswerLet $\text{I}=\int\text{x}(1-\text{x})^{23}\text{dx}$
Substituting 1 - x = t and dx = -dt, we get
$\text{I}=\int(1-\text{t})^{23}\text{dt}$
$=-\int(\text{t}^{23}-\text{t}^{24})\text{dt}$
$=-\int\Big(\frac{\text{t}^{24}}{24}-\frac{\text{t}^{25}}{25}\Big)+\text{C}$
$=\frac{\text{t}^{25}}{24}-\frac{\text{t}^{24}}{25}+\text{C}$
$=\frac{(1-\text{x})^{25}}{25}-\frac{(1-\text{x})^{24}}{24}+\text{C}$
$\therefore\ \text{I}=\frac{(1-\text{x})^{25}}{25}-\frac{(1-\text{x})^{24}}{24}+\text{C}$
$=\frac{1}{600}(1-\text{x})^{24}\big[24(1-\text{x})-25\big]+\text{C}$
$=\frac{1}{600}(1-\text{x})^{24}\big[24-24\text{x}-25\big]+\text{C}$
$=\frac{1}{600}(1-\text{x})^{24}\big[-1-24\text{x}\big]+\text{C}$
$=\frac{1}{600}(1-\text{x})^{24}\times-\big[1+24\text{x}\big]+\text{C}$
$=\frac{1}{600}(1-\text{x})^{24}(1+24\text{x})+\text{C}$
View full question & answer→Question 1035 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\sin\text{x}|\sin\text{x}|\text{dx}$
Answer$\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\sin\text{x}|\sin\text{x}|\text{dx}$
$=\int\limits^{0}_{-\frac{\pi}{4}}\sin\text{x}|\sin\text{x}|\text{dx}+\int\limits_{0}^{\frac{\pi}{2}}\sin\text{x}|\sin\text{x}|\text{dx}$
$=\int\limits^{0}_{-\frac{\pi}{4}}\sin\text{x}(-\sin\text{x})\text{dx}+\int\limits_{0}^{\frac{\pi}{2}}\sin\text{x}(\sin\text{x})\text{dx}$ $\begin{pmatrix}|\sin\text{x}|=\begin{cases}\sin\text{x},&0\leq\text{x}\leq\frac{\pi}{2}\\-\sin\text{x},&-\frac{\pi}{4}\leq\text{x}\leq0\end{cases} \end{pmatrix} $
$=\int\limits^{0}_{-\frac{\pi}{4}}\sin^2\text{x dx}+\int\limits_{0}^{\frac{\pi}{2}}\sin^2\text{x dx}$
$=-\int\limits^{0}_{-\frac{\pi}{4}}\frac{1-\cos2\text{x}}{2}\text{ dx}+\int\limits_{0}^{\frac{\pi}{2}}\frac{1-\cos2\text{x}}{2}\text{ dx}$
$=-\frac{1}{2}\int\limits^{0}_{-\frac{\pi}{4}}\text{dx}+\frac{1}{2}\int\limits^{0}_{-\frac{\pi}{4}}\cos2\text{x dx}+\frac{1}{2}\int\limits_{0}^{\frac{\pi}{2}}\text{dx}-\frac{1}{2}\int\limits_{0}^{\frac{\pi}{2}}\cos2\text{x dx}$
$=-\frac{1}{2}\times\big[\text{x}\big]^0_\frac{\pi}{4}+\frac{1}{2}\times\Big[\frac{\sin2\text{x}}{2}\Big]^0_{\frac{\pi}{2}}+\frac{1}{2}\times\big[\text{x}\big]^{\frac{\pi}{2}}_0-\frac{1}{2}\times\Big[\frac{\sin2\text{x}}{2}\Big]^{\frac{\pi}{2}}_0$
$=-\frac{1}{2}\Big(0+\frac{\pi}{4}\Big)+\frac{1}{4}(0+1)+\frac{\pi}{4}-\frac{1}{4}(0-0)$
$=\frac{\pi}{8}+\frac{1}{4}$
View full question & answer→Question 1045 Marks
Integrate the function in Exercise: $(\text{x}^2+1)\text{log}\ \text{x}$
AnswerLet $\text{I}=\int(\text{x}^2+1)\text{log x dx}=\int\text{x}^2\text{log x dx}+\int\text{log x dx}$
Let $I = I_1 + I_2....(1)$
where, $\text{I}_1=\int\text{x}^2\ \text{log x dx} \ \text{and I}_2=\int\text{logx}\ \text{dx} $
$\text{I}_1=\int\text{x}^2\text{log x dx}$
Taking $\log x$ as first function and $x^{2 }$ as secound function and integrating by parts,
we obtain $\int\text{I}.\text{II dx}=\text{I}\int\text{II dx}-\int\Big\{\frac{\text{d}}{\text{dx}}\text{I}\int\text{II dx}\Big\}\text{dx}$
$\text{I}_1=\text{log x}\int\text{x}^2\text{dx}-\int\Bigg\{\Bigg(\frac{\text{d}}{\text{dx}}\text{log x}\Bigg)\int\text{x}^2\text{dx}\Bigg\}\text{dx}$
$=\text{log x}.\frac{\text{x}^3}{3}-\int\frac{1}{\text{x}}.\frac{\text{x}^3}{3}\text{dx}$
$=\frac{\text{x}^3}{3}\text{log x}-\frac{1}{3}\Big(\int\text{x}^2\text{dx}\Big)$
$=\frac{\text{x}^3}{3}\text{log x}-\frac{\text{x}^3}{9}+\text{C}\dots(2)$
$\text{I}_2=\int\text{log x dx}$
Taking $\log x$ as first function and $1$ as secound function and integrating by parts,
we obtain $\text{I}_2=\text{log x}\int1.\text{dx}-\int\Bigg\{\Bigg(\frac{\text{d}}{\text{dx}}\text{log x}\Bigg)\int1.\text{dx}\Bigg\}\text{dx}$
$=\text{log x}.\text{x}-\int\frac{1}{\text{x}}.\text{x dx}$
$=\text{x log x}-\int1\text{dx}$
$=\text{x log x}-\text{x}+\text{C}_2\dots(3)$
Using equations $(2)$ and $(3)$ in $(1),$ we obtain
$\text{I}=\frac{\text{x}^3}{3}\text{log x}-\frac{\text{x}^3}{9}+\text{C}_1+\text{x logx}-\text{x}+\text{C}_2$
$=\frac{\text{x}^3}{3}\text{log x}-\frac{\text{x}^3}{9}+\text{x logx}-\text{x}+(\text{C}_1+\text{C}_2)$
$=\Bigg(\frac{\text{x}^3}{3}+\text{x}\Bigg)\text{log x}-\frac{\text{x}^3}{9}-\text{x}+\text{C}$
View full question & answer→Question 1055 Marks
Evaluate the following integrals:
$\int\text{e}^{\text{x}}(\log\text{x}+\frac{1}{\text{x}^2})\text{dx}$
AnswerWe have,
$\text{I}=\int\text{e}^{\text{x}}\Big(\log\text{x}+\frac{1}{\text{x} ^2}\Big)\text{dx}$
$=\int\text{e}^{\text{x}}\Big(\log\text{x}+\frac{1}{\text{x}}-\frac{1}{\text{x}}+\frac{1}{\text{x}^2}\Big)\text{dx}$
$=\int\text{e}^{\text{x}}\Big(\log\text{x}-\frac{1}{\text{x}}\Big)\text{dx}+\int\text{e}^{\text{x}}\Big(\frac{1}{\text{x}}+\frac{1}{\text{x}^2}\Big)\text{dx}$
Integrating by parts
$=\text{e}^{\text{x}}\Big(\log\text{x}-\frac{1}{\text{x}}\Big)-\int\text{e}^{\text{x}}\frac{\text{d}}{\text{dx}}\Big(\log\text{x}-\frac{1}{\text{x}}\Big)\text{dx}+\int\text{e}^{\text{x}}\Big(\frac{1}{\text{x}}+\frac{1}{\text{x}^2}\Big)\text{dx}$
$=\text{e}^{\text{x}}\Big(\log\text{x}-\frac{1}{\text{x}}\Big)-\int\text{e}^{\text{x}}\Big(\frac{1}{\text{x}}+\frac{1}{\text{x}^2}\Big)\text{dx}+\int\text{e}^{\text{x}}\Big(\frac{1}{\text{x}}+\frac{1}{\text{x}^2}\Big)\text{dx}$
$=\text{e}^{\text{x}}\Big(\log\text{x}-\frac{1}{\text{x}}\Big)+\text{C}$
View full question & answer→Question 1065 Marks
Evaluvate the following intregals:
$\int\frac{8\cot\text{x}+1}{3\cot\text{x}+2}\ \text{dx}$
AnswerLet $\text{I}=\int\Big(\frac{8\cot\text{x}+1}{3\cot\text{x}+2}\Big)\ \text{dx}$$=\int\bigg(\frac{8\frac{\cos\text{x}}{\sin\text{x}}+1}{\frac{3\cos\text{x}}{\sin\text{x}}+2}\bigg)\text{dx}$
$=\int\Big(\frac{8\cos\text{x}+\sin\text{x}}{3\cos\text{x}+2\sin\text{x}}\Big)\text{dx}$
Now, Let $8\cos\text{x}+\sin\text{x}=\text{A}(3\cos\text{x}+2\sin\text{x})+\text{B}(-3\sin\text{x}+2\cos\text{x})\ \dots(1)$
$\Rightarrow8\cos\text{x}+\sin\text{x}+\sin\text{x}=3\text{A}\cos\text{x}+2\text{A}\sin\text{x}-3\text{B}\sin\text{x}+2\text{B}\cos\text{x}$
$\Rightarrow8\cos\text{x}+\sin\text{x}-(3\text{A}+2\text{B})\cos\text{x}+(2\text{A}-3\text{B})\sin\text{x}$
Equating the coefficient of like terms we get,
$2\text{A}-3\text{B}=1\ \dots(2)$
$3\text{A}+2\text{B}=8\ \dots(3)$
Solving eq (2) and eq (3) we get,
$\text{A}=2,\text{B}=1$
Thus, by substracting the value of A and B in eq (1) we get,
$\text{I}=\int\Big[\frac{2(3\cos\text{x}+2\sin\text{x})+1(-3\sin\text{x}+2\cos\text{x})}{(3\cos\text{x}+2\sin\text{x})}\Big]\text{dx}$
$=2\int\Big(\frac{3\cos\text{x}+2\sin\text{x}}{3\cos\text{x}+2\sin\text{x}}\Big)\text{dx}+\int\Big(\frac{-3\sin\text{x}+2\cos\text{x}}{3\cos\text{x}+2\sin\text{x}}\Big)\text{dx}$
$=2\int\text{dx}+\int\Big(\frac{-3\sin\text{x}+2\cos\text{x}}{3\cos\text{x}+2\sin\text{x}}\Big)\text{dx}$
Putting $3\cos\text{x}+2\sin\text{x}=\text{t}$
$\Rightarrow(-3\sin\text{x}+2\cos\text{x})\text{dx}=\text{dt}$
$\therefore\text{I}=2\int\text{dx}+\int\frac{1}{\text{t}}\text{dt}$
$=2\text{x}+\ln|\text{t}|+\text{C}$
$=2\text{x}+\ln|3\cos\text{x}+2\sin\text{x}|+\text{C}$
View full question & answer→Question 1075 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 1085 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 1095 Marks
Evaluate the following integrals:
$\int_{0}^\limits{1}\text{x}\log(1+2\text{x})\text{dx}$
AnswerLet $\text{I}=\int_{0}^\limits{1}\text{x}\log(1+2\text{x})\text{dx}$Apply integral by part
$\text{I}=\Big[\log(1+2\text{x})\frac{\text{x}^2}{2}\Big]^{1}_0-\int_{0}^\limits{1}\Big(\frac{2}{1+2\text{x}}\Big)\times\frac{\text{x}^2}{2}\text{ dx}$
$=\frac{1}{2}\big(\log3-0\big)-\int_{0}^\limits{1}\frac{\text{x}^2}{1+2\text{x}}\text{ dx}$
$=\frac{1}{2}\log3-\frac{1}{4}\int_{0}^\limits{1}\frac{4\text{x}^2-1+1}{1+2\text{x}}\text{ dx}$
$=\frac{1}{2}\log3-\frac{1}{4}\int_{0}^\limits{1}\frac{(2\text{x}+1)(2\text{x}-1)}{1+2\text{x}}\text{ dx}-\frac{1}{4}\int_{0}^\limits{1}\frac{2}{1+2\text{x}}\text{ dx}$
$=\frac{1}{2}\log3-\frac{1}{4}\int_{0}^\limits{1}(2\text{x}-1)\text{dx}-\frac{1}{4}\int_{0}^\limits{1}\frac{1}{1+2\text{x}}\text{ dx}$
$=\frac{1}{2}\log3-\bigg[\frac{1}{4}\times\frac{(2\text{x}-1)^2}{2\times2}\bigg]^1_0-\bigg[\frac{1}{4}\times\frac{\log(1+2\text{x})}{2}\bigg]^1_0$
$=\frac{1}{2}\log3-\frac{1}{16}(1-1)-\frac{1}{8}\big(\log3-\log1\big)$
$=\frac{1}{2}\log3-0-\frac{1}{8}\log3$ $(\log1=0)$
$=\frac{3}{8}\log3$
View full question & answer→Question 1105 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 1115 Marks
Evaluate the following integrals:$\int(\text{x}+1)\text{e}^{\text{x}}\log(\text{xe}^{\text{x}})\text{dx}$
Answer$\int(\text{x}+1)\text{e}^{\text{x}}.\log(\text{xe}^{\text{x}})\text{dx}$ Let $\text{x e}^{\text{x}}=\text{t}$ $\Rightarrow\big(\text{x.e}^{\text{x}}+1.\text{e}^{\text{x}}\big)\text{dx=dt}$ $\therefore\int(\text{x}+1)\text{e}^{\text{x}}.\log(\text{x e}^\text{x})\text{dx}=\int1.\log (\text{t})\text{dt}$ $=\log\text{t}\int1\text{dt}-\int\big\{\frac{\text{d}}{\text{dt}}(\log\text{t)}-\int1\text{dt}\Big\}\text{dt}$ $=\log(\text{t})\times\text{t}-\int\frac{1}{\text{t}}\times\text{t dt}$ $=\text{t}\log(\text{t})-\text{t}+\text{C} \dots(1)$Substituting the value of t in eq (1)
$\Rightarrow\int(\text{x}+1)\text{e}^{\text{x}}.\log(\text{x e}^{\text{x}})\text{dx} = (\text{x e}^{\text{x}}).\log(\text{x e}^{\text{x}})-\text{x e}^{\text{x}}+\text{C}$ $=\text{x e}^{\text{x}}\Big\{\log(\text{x e}^{\text{x}})-1\Big\}+\text{C}$
View full question & answer→Question 1125 Marks
Evaluate the following integrals:
$\int\frac{\sin^3\text{x}}{\sqrt{\cos\text{x}}}\text{dx}$
AnswerLet I $=\int\frac{\sin^3\text{x}}{\sqrt{\cos\text{x}}}\text{dx}$$\therefore\text{I}=\int\frac{\sin^2\text{x}\sin\text{x}}{\sqrt{\cos\text{x}}}\text{dx}$
$\Rightarrow\text{I}=\int\frac{(1-\cos^2\text{x})}{\sqrt{\cos\text{x}}}\sin\text{x dx}\ ...(1)$
Let $\cos\text{x}=\text{t}$ then, $\text{d}(\cos\text{x})=\text{dt}$
$\Rightarrow-\sin\text{x dx}=\text{dt}$
$\Rightarrow\sin\text{x dx}=-\text{dt}$
Putting $\cos\text{x}=\text{t}$ and $\sin\text{x dx}=-\text{dt}$ in equation (1), we get
$\text{I}=\int\frac{(1-\text{t}^2)}{\sqrt{\text{t}}}\times-\text{dt}$
$=\int\frac{\text{t}^2-1}{\sqrt{\text{t}}}\text{dt}$
$=\int\Bigg(\frac{\text{t}^2}{\text{t}^\frac{1}{2}}-\frac{1}{\text{t}^\frac{1}{2}}\Bigg)\text{dt}$
$=\int\Big(\text{t}^{2-\frac{1}{2}}-\text{t}^\frac{-1}{2}\Big)\text{dt}$
$=\int\Big(\text{t}^\frac{3}{2}-\text{t}^{\frac{-1}{2}}\Big)\text{dt}$
$=\frac{2}{5}\text{t}^\frac{5}{2}-2\text{t}^\frac{1}{2}+\text{C}$
$\therefore\text{I}=\frac{2}{5}\cos^\frac{5}{2}\text{x}-2\cos^\frac{1}{2}\text{x}+\text{C}$
$\therefore\text{I}=\frac{2}{5}\cos^\frac{5}{2}\text{x}-2\sqrt{\cos\text{x}}+\text{C}$
View full question & answer→Question 1135 Marks
Evaluate the following integrals:$\int\frac{\text{x}^2}{\text{x}^2+7\text{x}+10}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\text{x}^2}{\text{x}^2+7\text{x}+10}\text{ dx}$ $=\int\Big\{1-\frac{7\text{x}+10}{\text{x}^2+7\text{x}+10}\Big\}\text{dx}$ $=\text{x}-\int\frac{7\text{x}+10}{\text{x}^2+7\text{x}+10}\text{ dx}+\text{C}_1\ ....(1)$ $\text{I}_1=\int\frac{7\text{x}+10}{\text{x}^2+7\text{x}+10}\text{ dx}$ 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,
$7=2\lambda\Rightarrow\lambda=\frac{7}{2}$ $7\lambda+\mu=10\Rightarrow7\Big(\frac{7}{2}\Big)+\mu=10$ $\mu=-\frac{29}{2}$ So, $\text{I}_1=\int\frac{\frac{7}{2}(2\text{x}+7)-\frac{29}{2}}{\text{x}^2+7\text{x}+10}\text{ dx}$ $\text{I}_1=\frac{7}{2}\int\frac{(2\text{x}+7)}{\text{x}^2+7\text{x}+10}\text{ dx}-\frac{29}{2}\int\frac{1}{\text{x}^2-2\text{x}\big(\frac{7}{2}\big)+\big(\frac{7}{2}\big)^2-\big(\frac{7}{2}\big)^2+10}\text{ dx}$ $\text{I}_1=\frac{7}{2}\int\frac{2\text{x}+7}{\text{x}^2+7\text{x}+10}\text{ dx}-\frac{29}{2}\int\frac{1}{\big(\text{x}+\frac{7}{2}\big)^2-\big(\frac{3}{2}\big)^2}\text{ dx}$ $\text{I}_1=\frac{2}{7}\log\big|\text{x}^2+7\text{x}+10\big|-\frac{29}{2}\times\frac{1}{2\big(\frac{3}{2}\big)}\log\bigg|\frac{\text{x}+\frac{7}{2}-\frac{3}{2}}{\text{x}+\frac{7}{2}+\frac{3}{2}}\bigg|+\text{C}_2$ $\Big[\text{since},\int\frac{1}{\text{x}^2-\text{a}^2}\text{ dx}=\frac{1}{2\text{a}}\log\Big|\frac{\text{x}-\text{a}}{\text{x}+\text{a}}\Big|+\text{C}\Big]$ $\text{I}_1=\frac{7}{2}\log\big|\text{x}^2+7\text{x}+10\big|-\frac{29}{6}\log\Big|\frac{\text{x}+2}{\text{x}+5}\Big|+\text{C}_2\ ....(2)$ Using equation (1) and (2) $\text{I}=\text{x}-\frac{7}{2}\log\big|\text{x}^2+7\text{x}+10\big|+\frac{29}{6}\log\Big|\frac{\text{x}+2}{\text{x}+5}\Big|+\text{C}$
View full question & answer→Question 1145 Marks
Evaluate the following integrals:$\int\frac{2\text{x}-3}{\text{x}^2+6\text{x}+13}\text{ dx}$
AnswerLet $\text{I}=\int\frac{2\text{x}-3}{\text{x}^2+6\text{x}+13}\text{ dx}$
Let $2\text{x}-3=\lambda\frac{\text{d}}{\text{dx}}\big(\text{x}^2+6\text{x}+13\big)+\mu$
$=\lambda(2\text{x}+6)+\mu$
$2\text{x}-3=(2\lambda)\text{x}+(6\lambda+\mu)$
Comparing the coefficients of like powers of x,
$2\lambda=2\Rightarrow\lambda=1$
$6\lambda+\mu=-3\Rightarrow6(1)+\mu=-3$
$\mu=-9$
So, $\text{I}=\int\frac{1(2\text{x}+6)-9}{\text{x}^2+6\text{x}+13}\text{ dx}$
$\text{I}=\int\frac{2\text{x}+6}{\text{x}^2+6\text{x}+13}\text{ dx}-9\int\frac{1}{\text{x}^2+2\text{x}(3)+(3)^2-(3)^2+13}\text{ dx}$
$\text{I}=\int\frac{2\text{x}+6}{\text{x}^2+6\text{x}+13}\text{ dx}-9\int\frac{1}{(\text{x}+3)^2+(2)^2}\text{ dx}$
$\text{I}=\log\big|\text{x}^2+6\text{x}+13\big|-9\times\frac{1}{2}\tan^{-1}\Big(\frac{\text{x}+3}{2}\Big)+\text{C}$ $\Big[\text{Since }\int\frac{1}{\text{x}^2+\text{a}^2}\text{ dx}=\frac{1}{\text{a}}\tan^{-1}\Big(\frac{\text{x}}{\text{a}}\Big)+\text{C}\Big]$
$\text{I}=\log\big|\text{x}^2+6\text{x}+13\big|-\frac{9}{2}\tan^{-1}\Big(\frac{\text{x}+3}{2}\Big)+\text{C}$
View full question & answer→Question 1155 Marks
Evaluate the following integrals:$\int\frac{\text{ax}^3+\text{bx}}{\text{x}^4+\text{c}^2}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\text{ax}^3+\text{bx}}{\text{x}^4+\text{c}^2}\text{ dx}$ Let $\text{ax}^3+\text{bx}=\lambda\frac{\text{d}}{\text{dx}}\big(\text{x}^4+\text{c}^2\big)+\mu$ $\text{ax}^3+\text{bx}=\lambda\big(4\text{x}^3\big)+\mu$ Comparing the coefficients of like powers of x, $4\lambda=\text{a}\Rightarrow\lambda=\frac{\text{a}}{4}$ $\mu=0\Rightarrow\mu=0$ So, $\text{I}=\int\frac{\frac{\text{a}}{4}(4\text{x}^3)+\text{bx}}{\text{x}^4+\text{c}^2}\text{ dx}$ $\text{I}=\frac{\text{a}}{4}\int\frac{4\text{x}^3}{\text{x}^4+\text{c}^2}\text{ dx}+\text{b}\int\frac{\text{x}}{(\text{x}^2)^2+\text{c}^2}\text{ dx}$ $\text{I}=\frac{\text{a}}{4}\int\frac{4\text{x}^3}{\text{x}^4+\text{c}^2}\text{ dx}+\frac{\text{b}}{2}\int\frac{2\text{x}}{(\text{x}^2)^2+\text{c}^2}\text{ dx}$ $\text{I}=\frac{\text{a}}{4}\log\big|\text{x}^4+\text{c}^2\big|+\frac{\text{b}}{2}\text{I}_1\ ....(1)$Now,
$\text{I}_1=\int\frac{2\text{x}}{(\text{x}^2)^2+\text{c}^2}\text{ dx}$ Put $\text{x}^2=\text{t}$ $\Rightarrow2\text{x}\text{ dx}=\text{dt}$ $\text{I}_1=\int\frac{1}{(\text{t})^2+\text{c}^2}\text{ dx}$ $\text{I}_1=\frac{1}{\text{c}}\tan^{-1}\Big(\frac{\text{t}}{\text{c}}\Big)+\text{C}_1\ ....(2)$ Using equation (2) in equation (1), $\text{I}=\frac{\text{a}}{4}\log\big|\text{x}^4+\text{c}^4\big|+\frac{\text{b}}{2\text{c}}\tan^{-1}\Big(\frac{\text{x}^2}{\text{c}}\Big)+\text{K}$ K = Integration constant.
View full question & answer→Question 1165 Marks
Evaluate the following intregals:
$\int\frac{\text{x}^2}{\text{x}^4-\text{x}^2-12}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{\text{x}^2}{\text{x}^4-\text{x}^2-12}\ \text{dx}$We express
$\frac{\text{x}^2}{\text{x}^4-\text{x}^2-12}=\frac{\text{x}^2}{\text{x}^4-4\text{x}^2+3\text{x}^2-12}$
$=\frac{\text{x}^2}{(\text{x}^2-4)(\text{x}^2+3)}$
$=\frac{\text{A}}{\text{x}^2-4}+\frac{\text{B}}{\text{x}^2+3}$
$\Rightarrow\text{x}^2=\text{A}(\text{x}^2+3)+\text{B}(\text{x}^2-4)$
Equating the coefficients of $x^2$ and constant, we get
$1 = A + B$ and $0 = 3A - 4B$ or
$\text{A}=\frac{4}{7}$ and $\text{B}=\frac{3}{7}$
$\therefore\text{I}=\int\bigg(\frac{\frac{4}{7}}{\text{x}^2-4}+\frac{\frac{3}{7}}{\text{x}^2+3}\bigg)\text{dx}$
$=\frac{4}{7}\int\frac{1}{\text{x}^2-4}\ \text{dx}+\frac{3}{7}\int\frac{1}{\text{x}^2+3}\text{dx}$
$=\frac{4}{7}\times\frac{1}{4}\log\Big|\frac{\text{x}-2}{\text{x}+2}\Big|+\frac{\sqrt{3}}{7}\tan^{-1}\frac{\text{x}}{\sqrt{3}}+\text{C}$
$=\frac{1}{7}\log\Big|\frac{\text{x}-2}{\text{x}+2}\Big|+\frac{\sqrt{3}}{7}\tan^{-1}\frac{\text{x}}{\sqrt{3}}+\text{C}$
Hence, $\int\frac{\text{x}^2}{\text{x}^4-\text{x}^2-12}\ \text{dx}=\frac{1}{7}\log\Big|\frac{\text{x}-2}{\text{x}+2}\Big|+\frac{\sqrt{3}}{7}\tan^{-1}\frac{\text{x}}{\sqrt{3}}+\text{C}$
View full question & answer→Question 1175 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\frac{\text{x}^{11}-3\text{x}^9+5\text{x}^7-\text{x}^5+1}{\cos^2\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\frac{\text{x}^{11}-3\text{x}^9+5\text{x}^7-\text{x}^5+1}{\cos^2\text{x}}\text{ dx}$$=\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\frac{\text{x}^{11}-3\text{x}^9+5\text{x}^7-\text{x}^5}{\cos^2\text{x}}\text{ dx}+\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\frac{1}{\cos^2\text{x}}\text{ dx}$
$=\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\frac{\text{x}^{11}-3\text{x}^9+5\text{x}^7-\text{x}^5}{\cos^2\text{x}}\text{ dx}+\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\sec^2\text{x}\text{ dx}$
$=\text{I}_1+\text{I}_2$
Now,
Consider, $\text{f(x)}=\frac{\text{x}^{11}-3\text{x}^9+5\text{x}^7-\text{x}^5}{\cos^2\text{x}}\text{ dx}$
$\therefore\ \text{f}(-\text{x})=\frac{(-\text{x})^{11}-3(-\text{x})^9+5(-\text{x})^7-(-\text{x})^5}{\cos^2(-\text{x})}$
$=\frac{-\text{x}^{11}+3\text{x}^9-5\text{x}^7+\text{x}^5}{\cos^2\text{x}}$
$=-\frac{\text{x}^{11}-3\text{x}^9+5\text{x}^7-\text{x}^5}{\cos^2\text{x}}=\text{f(x)}$
$\Rightarrow\text{I}_1=\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\frac{\text{x}^{11}-3\text{x}^9+5\text{x}^7-\text{x}^5}{\cos^2\text{x}}\text{ dx}=0$ $\begin{bmatrix}\int\limits^{\text{a}}_0\text{f(x)}\text{dx}=\begin{cases}\int\limits^{\text{a}}_0\text{f(x)}\text{dx},&\text{ if }\text{ f}(-\text{x})=\text{f(x)}\\0,&\text{ if }\text{ f}(-\text{x})=-\text{f(x)}\end{cases}\end{bmatrix}$
$\text{g(x)}=\sec^2\text{x}$
Let $\text{g}(-\text{x})=\sec^2(-\text{x})=\sec^2\text{x}=\text{g}(\text{x})$
$\Rightarrow\text{I}_1=\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\sec^2\text{x dx}$
$=2\int\limits^{\frac{\pi}{4}}_0\sec^2\text{x dx}$ $\begin{bmatrix}\int\limits^{\text{a}}_0\text{f(x)}\text{dx}=\begin{cases}\int\limits^{\text{a}}_0\text{f(x)}\text{dx},&\text{ if }\text{ f}(-\text{x})=\text{f(x)}\\0,&\text{ if }\text{ f}(-\text{x})=-\text{f(x)}\end{cases}\end{bmatrix}$
$=2\times\Big[\tan\text{x}\Big]^{\frac{\pi}{4}}_0$
$=2\Big(\tan\frac{\pi}{4}-\tan0\Big)$
$=2\times(1-0)$
$=2$
$\therefore\ \text{I}=\text{I}_1+\text{I}_2$
$\text{I}=0+2$
$\text{I}=2$
View full question & answer→Question 1185 Marks
By Using properties of definite integrals, evaluate the following integral in Exercise:
$\int^{\pi}_{0}\log(1+\cos\text{x})\text{dx}$
Answer$\text{Let}\ \text{I}\int^{\pi}\limits_{0}\log(1+\cos\text{x})\text{dx}$$\Rightarrow\ \ \text{I}=\int^{\pi}\limits_{0}\log(1+\cos(\pi-\text{x}))\text{dx}=\int^{\pi}\limits_{0}\log(1-\cos\text{x})\text{dx}$
Adding eq. (i) and (ii)$21=\int^{\pi}\limits_{0}\big[\log(1+\cos\text{x})+\log(1-\cos\text{x})\big]\text{dx}=\int^{\text{x}}\limits_{0}\big[\log(1+\cos\text{x})(1-\cos\text{x})\big]\text{dx}$
$\Rightarrow21=\int^{\pi}\limits_{0}\big[\log(1-\cos^{2}\text{x})\big]\text{dx}=\int^{\pi}\limits_{0}\big[\log\sin^{2}\text{x}\big]\text{dx}=2\int^{\pi}\limits_{0}\big[\log\sin\text{x}\big]\text{dx}$
$\Rightarrow\ \ \text{I}=2\int^{\frac{\pi}{2}}\limits_{0}\log\sin\text{x}\ \text{dx}$
$\Rightarrow\ \ \text{I}=2\int^{\frac{\pi}{2}}\limits_{0}\log\sin\bigg(\frac{\pi}{2}-\text{x}\bigg)\text{dx}=2\int^{\frac{\pi}{2}}\limits_{0}\log\cos\text{x}\ \text{dx}$
Adding eq. (i) and (ii),
$21=2\int^{\frac{\pi}{2}}\limits_{0}(\log\sin\text{x}+\log\cos\text{x})\text{dx}=2\int^{\frac{\pi}{2}}\limits_{0}(\log\sin\text{x}\cos\text{x})\text{dx}$
$\Rightarrow\ \ \text{I}=\int^{\frac{\pi}{2}}\limits_{0}\log\bigg(\frac{2\sin\text{x}\cos\text{x}}{2}\bigg)\text{dx}=\int^{\frac{\pi}{2}}\limits_{0}\log\bigg(\frac{\sin2\text{x}}{2}\bigg)\text{dx}$
$=\int^{\frac{\pi}{2}}\limits_{0}(\log\sin2\text{x}-\log2)\text{dx}$
$$$\Rightarrow\int\limits_{0}^{\frac{\pi}{2}}\log\sin2\text{x}\ \text{dx}-\int\limits_{0}^{\frac{\pi}{2}}\log2\ \text{dx}=\int\limits_{0}^{\frac{\pi}{2}}\log\sin2\text{dx}-\log2\text{(x)}^{\frac{\pi}{2}}_{0}=\int\limits_{0}^{\frac{\pi}{2}}\log\sin2\text{dx}-\frac{\pi}{2}\log2$
$\Rightarrow\ \ \text{I}=\text{I}_{1}-\frac{\pi}{2}\log2$
Where $\text{I}_{1}=\int^{\frac{\pi}{2}}\limits_{0}\log\sin2\text{x}\ \text{dx}$
$\text{putting}\ 2\text{x}=\text{t}\ \text{in}\ \text{eq}.\text{(vi)},\ \Rightarrow2=\frac{\text{dt}}{\text{dx}}\ \Rightarrow\ \ \text{dx}=\frac{\text{dt}}{2}$
Limits of integration when $\text{x}=0, \text{and}\ \text{x}=\frac{\pi}{2},\text{t}=\pi$
$\therefore\ \ \text{from eq. (vi)},\ \ \text{I} _{1}\int^{\pi}\limits_{0}\log\sin\text{t}\frac{\text{dt}}{2}=\frac{1}{2}\int^{\pi}\limits_{0}\log\sin\text{t}\ \text{dt} =\frac{1}{2}\times2\int^{\frac{\pi}{2}}\limits_{0}\log\sin\text{t}\ \text{dt}$
$\Rightarrow\ \ \text{I}_{1}=\int^{\frac{\pi}{2}}\limits_{0}\log\sin\text{t}\ \text{dt}=\int^{\frac{\pi}{2}}\limits_{0}\log\sin\text{x}\text{dx}\ \ \bigg[\because\int^{\text{b}}\limits_{\text{a}}\text{f}\text{(t)}\text{dt}=\int^{\text{b}}\limits_\text{a}\text{f}\text{(x)}\text{dx}\bigg]$
$\Rightarrow\ \ \text{I}=\frac{1}{2}\ \ \ \text{[from eq. (iii)]}$
$\text{putting this value in eq. (v)},\ \text{I}=\frac{1}{2}-\frac{\pi}{2}\log2\ \ \Rightarrow 21=\text{I}-\pi\log2$
$\Rightarrow\ \ \ 21-\text{I}=-\pi\log2\ \ \text{I}=-\pi\log2$
View full question & answer→Question 1195 Marks
Evaluate the following integrals:$\int\frac{\text{x}^2\tan^{-1}\text{x}}{1+\text{x}^2}\text{dx}$
AnswerLet $\text{I}=\int\Big(\frac{\text{x}^2\tan^{-1}\text{x}}{1+\text{x}^2}\Big)\text{dx}$
$=\int\Big(\frac{\text{x}^2+1-1}{\text{x}^2+1}\Big)\tan^{-1}\text{x dx}$
$=\int\Big(1-\frac{1}{\text{x}^2+1}\Big)\tan^{-1}\text{x dx}$
$=\int1.\tan^{-1}\text{x dx}-\int\frac{\tan^{-1}\text{x}}{\text{x}^2+1}\text{dx}$
$=\Big[\tan^{-1}\text{x}\int1\text{dx}-\int\Big\{\frac{\text{d}}{\text{dx}}\big(\tan^{-1}\text{x}\big)\int1\text{dx}\Big\}\text{dx}\Big]-\int\frac{\tan^{-1}\text{x}}{\text{x}^2+1}\text{dx}$
$=\Big[\tan^{-1}\text{x}\times\text{x}-\int\frac{\text{x}}{1+\text{x}^2}\text{dx}\Big]-\int\frac{\tan^{-1}\text{x}}{\text{x}^2+1}\text{dx}$
Putting $\text{x}^2+1=\text{t}$ in the first integral and $\tan^{-1}\text{x}=\text{p}$ in the second integral
$\Rightarrow2\text{x dx = dt} $ and $\frac{1}{1+\text{x}^2}\text{dx = dp}$
$\Rightarrow\text{x dx}=\frac{\text{dt}}{2}$ and $\frac{1}{1+\text{x}^2}\text{dx = dp}$
$\therefore \text{I}=\tan^{-1}\text{x.x}-\frac{1}{2}\int\frac{\text{dt}}{\text{t}}-\int\text{p.dp}$
$=\text{x}\tan^{-1}\text{x}-\frac{1}{2}\ln|\text{t}|-\frac{\text{P}^2}{2}+\text{C}$
$=\text{x}\tan^{-1}\text{x}-\frac{1}{2}\ln|1+\text{x}^2|-\frac{(\tan^{-1}\text{x})^2}{2}+\text{C}$ $[\therefore \text{t}=\text{x}^2+1\text{ and}\text{ p}=\tan^{-1}\text{x}]$
View full question & answer→Question 1205 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 1215 Marks
Evaluate the following integrals:$\int\frac{\text{x}^3}{\text{x}^4+\text{x}^2+1}\text{ dx}$
Answer$\text{I}=\int\frac{\text{x}^3}{\text{x}^4+\text{x}^2+1}\text{ dx}$ $=\int\frac{\text{x}^2\cdot\text{x}}{(\text{x}^2)^2+\text{x}^2+1}\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}}{\text{t}^2+\text{t}+1}\text{ dt}$ $=\frac{1}{4}\int\frac{2\text{t}}{\text{t}^2+\text{t}+1}\text{ dt}$$=\frac{1}{4}\int\frac{2\text{t}+1-1}{\text{t}^2+\text{t}+1}\text{ dt}$
$=\frac{1}{4}\int\Big[\frac{(2\text{t}+1)}{(\text{t}^2+\text{t}+1)}-\frac{1}{(\text{t}^2+\text{t}+1)}\Big]\text{dt}$
$=\frac{1}{4}\Big[\log\big|\text{t}^2+\text{t}+1\big|-\int\frac{1}{\big(\text{t}^2+\text{t}+\frac{1}{4}+\frac{3}{4}\big)}\text{ dt}\Big]$ $=\frac{1}{4}\begin{bmatrix}\log\big|\text{t}^2+\text{t}+1\big|-\int\frac{1}{\big(\text{t}+\frac{1}{2}\big)^2+\Big(\frac{\sqrt{3}}{2}\Big)^2}\text{ dt}\end{bmatrix}$ $=\frac{1}{4}\begin{bmatrix}\log\big|\text{t}^2+\text{t}+1\big|-\frac{2}{\sqrt3}\tan\frac{\big(\text{t}+\frac{1}{2}\big)}{\Big(\frac{\sqrt3}{2}\Big)}\end{bmatrix}+\text{C}$ $=\frac{1}{4}\Big[\log\big|\text{t}^2+\text{t}+1\big|-\frac{2}{\sqrt3}\tan\Big(\frac{2\text{t}+1}{\sqrt3}\Big)\Big]+\text{C}$ $=\frac{1}{4}\Big[\log\big|\text{x}^4+\text{x}^2+1\big|-\frac{2}{\sqrt3}\tan\Big(\frac{2\text{x}^2+1}{\sqrt3}\Big)\Big]+\text{C}$
View full question & answer→Question 1225 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 1235 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 1245 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{2}}_0\frac{\tan^{7}\text{x}}{\tan^{7}\text{x}+\cot^7\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\tan^{7}\text{x}}{\tan^{7}\text{x}+\cot^7\text{x}}\text{ dx}\ ...(\text{i})$
Then,
$\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\tan^{7\big(\frac{\pi}{2}-\text{x}\big)}}{\tan^{7}{\big(\frac{\pi}{2}-\text{x}\big)}+\cot^7{\big(\frac{\pi}{2}-\text{x}\big)}}\text{ dx}$ $\Bigg[\int\limits^{\text{a}}_0\text{f(x)}\text{dx}=\int\limits^{\text{a}}_0\text{f}(\text{a}-\text{x})\text{dx}\Bigg]$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\cot^7\text{x}}{\cot^7\text{x}+\tan^{7}\text{x}}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii) we get
$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\tan^{7}\text{x}+\cot^7\text{x}}{\tan^{7}\text{x}+\cot^7\text{x}}\text{ dx}$
$\Rightarrow2\text{I}=\int\limits^{\frac{\pi}{2}}_0\text{dx}$
$\Rightarrow2\text{I}=\Big[\text{x}\Big]^{\frac{\pi}{2}}_0$
$\Rightarrow2\text{I}=\frac{\pi}{2}-0=\frac{\pi}{2}$
$\Rightarrow\text{I}=\frac{\pi}{4}$
View full question & answer→Question 1255 Marks
Evaluate the following integrals:
$\int^\limits{(\pi)^\frac{2}{3}}_{0}\sqrt{\text{x}}\cos^2\text{x}^{\frac{3}{2}}\text{ dx}$
AnswerLet $\text{I}=\int^\limits{(\pi)^\frac{2}{3}}_{0}\sqrt{\text{x}}\cos^2\text{x}^{\frac{3}{2}}\text{ dx}$ Then,
Let $\text{x}^{\frac{3}{2}}=\text{t}$ Then, $\frac{3}{2}\sqrt{\text{x}}\text{ dx}=\text{dt}$
When, $\text{x}=0,\text{t}=0$ and $\text{x}=\big(\pi\big)^{\frac{2}{3}},\text{t}=\pi$
$\therefore\ \text{I}=\frac{2}{3}\int^\limits{\pi}_{0}\cos^2\text{t}\text{ dt}$
$\Rightarrow\text{I}=\frac{2}{3}\int^\limits{\pi}_{0}\frac{1+\cos2\text{x}}{2}\text{ dx}$
$\Rightarrow\text{I}=\frac{1}{3}\Big[\text{x}+\frac{\sin2\text{x}}{2}\Big]^{\pi}_0$
$\Rightarrow\text{I}=\frac{1}{3}\big(\pi+0\big)$
$\Rightarrow\text{I}=\frac{\pi}{3}$
View full question & answer→Question 1265 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 1275 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 1285 Marks
Evaluate the following integrals:
$\int^\limits{\text{a}}_{-\text{a}}\sqrt{\frac{\text{a}-\text{x}}{\text{a}+\text{x}}}\text{ dx}$
AnswerLet $\text{x}=\text{a}\cos2\theta$
Differentiating w.r.t. x, we get
$\text{dx}=-2\text{a}\sin2\theta$
Now, $\text{x}=-\text{a}\Rightarrow\theta=\frac{\pi}{2}$
$\text{x}=\text{a}\Rightarrow\theta=0$
$\therefore\ \int^\limits{\text{a}}_{-\text{a}}\sqrt{\frac{\text{a}-\text{x}}{\text{a}+\text{x}}}\text{ dx}=\int^\limits0_\frac{\pi}{2}\sqrt{\frac{\text{a}(1-\cos2\theta)}{\text{a}\big(1+\cos2\theta)}}(-2\sin2\theta\big)\text{d}\theta$
$=2\text{a}\int^\limits{\frac{\pi}{\text{2}}}_{0}\frac{\sin\theta}{\cos\theta}\cdot\sin2\theta\text{ d}\theta$ $\begin{bmatrix}\because1-\cos2\theta=2\sin^2\theta\\1+\cos2\theta=2\cos^2\theta\\-\int^\limits\text{b}_\text{a}\text{f(x)}\text{dx}=\int^\limits\text{a}_\text{b}\text{f}(\text{x})\text{dx} \end{bmatrix}$
$=2\text{a}\int^\limits{\frac{\pi}{\text{2}}}_{0}\frac{\sin\theta\cdot2\sin\theta\cos\theta}{\cos\theta}$
$=4\text{a}\int^\limits{\frac{\pi}{\text{2}}}_{0}\sin^{2}\theta\text{ d}\theta$
$=2\text{a}\int^\limits{\frac{\pi}{\text{2}}}_{0}\big(1-\cos2\theta\big)\text{d}\theta$
$=2\text{a}\Big[\theta-\frac{\sin2\theta}{2}\Big]^{\frac{\pi}{2}}_0$
$=2\text{a}\Big[\frac{\pi}{2}-0-0+0\Big]$
$=\pi\text{a}$
View full question & answer→Question 1295 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 1305 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 1315 Marks
Evaluate the following integrals:
$\int\text{x}\Big(\frac{\sec2\text{x}-1}{\sec2\text{x}+1}\Big)\text{dx}$
AnswerLet $\text{I}=\int\text{x}\Big(\frac{\sec2\text{x}-1}{\sec2\text{x}+1}\Big)\text{dx}$
$=\int\text{x}\Big(\frac{1-\cos2\text{x}}{1+\cos2\text{x}}\Big)\text{dx}$
$=\int\text{x}\Big(\frac{\sec^2\text{x}}{\cos^2\text{x}}\Big)\text{dx}$
$=\int\text{x}\tan^2\text{x dx}$
$=\int\text{x}(\sec^2\text{x}-1)\text{dx}$
$=\int\text{x}\sec^2\text{x dx}-\int\text{dx}$
$=\big[\text{x}\int\sec^2\text{x dx}-\int(1\int\sec^2\text{x dx})\text{dx}\big]-\frac{\text{x}^2}{2}$
$=\text{x}\tan\text{x}-\int\tan\text{x dx}-\frac{\text{x}^2}{2}$
$\text{I}=\text{x}\tan\text{x}-\log\sec\text{x}-\frac{\text{x}^2}{2}+\text{C}$
View full question & answer→Question 1325 Marks
Evaluate the following integrals:$\int(\log\text{x})^2\cdot\text{x dx}$
AnswerLet $\text{I}=\int(\log\text{x})^2\text{x dx}$
Using integration by parts,
$=(\log\text{x})^2\int\text{x dx }-\int\Big(2(\log\text{x})\Big(\frac{1}{\text{x}}\Big)\int\text{x dx}\Big)\text{dx}$
$=\frac{\text{x}^2}{2}(\log\text{x})^2-2\int(\log\text{x})\Big(\frac{1}{\text{x}}\Big)\Big(\frac{\text{x}^2}{2}\Big)\text{dx}$
$=\frac{\text{x}^2}{2}(\log\text{x})^2-\int\text{x}(\log\text{x})\text{dx}$
$=\frac{\text{x}^2}{2}(\log\text{x})^2-\Big[\log\text{x}\int\text{x dx}-\int\Big(\frac{1}{\text{x}}\int\text{x dx}\Big)\text{dx}\Big]$
$=\frac{\text{x}^2}{2}(\log\text{x})^2-\Big[\frac{\text{x}^2}{2}\log\text{x}-\int\Big(\frac{1}{\text{x}}\times\frac{\text{x}^2}{2}\Big)\text{dx}\Big]$
$=\frac{\text{x}^2}{2}(\log\text{x})^2-\frac{\text{x}^2}{2}\log\text{x}+\frac{1}{2}\int\text{x dx}$
$=\frac{\text{x}^2}{2}(\log\text{x})^2-\frac{\text{x}^2}{2}\log\text{x}+\frac{1}{4}\text{x}^2+\text{C}$
$\text{I}=\frac{\text{x}^2}{2}\Big[(\log\text{x})^2-\log\text{x}+\frac{1}{2}\Big]+\text{C}$
View full question & answer→Question 1335 Marks
Evaluate the following integrals:$\int\text{e}^{\text{x}}.\frac{\sqrt{1-\text{x}^2}\sin^{-1}\text{x}+1}{\sqrt{1-\text{x}^2}}\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{x}}\bigg[\frac{\sqrt{1-\text{x}^2}\sin^{-1}\text{x}+1}{\sqrt{1-\text{x}^2}}\bigg]\text{dx}$
$=\int\text{e}^{\text{x}}\Big[\sin^{-1}\text{x}+\frac{1}{\sqrt{1-\text{x}^2}}\Big]\text{dx}$
Here, $\text{f(x)}=\sin^{-1}\text{x}$
$\Rightarrow\text{f}'\text{(x)}=\frac{1}{\sqrt{1-\text{x}^2}}$
Put $\text{e}^{\text{x}}\text{f(x)}=\text{t}$
$\Rightarrow\text{e}^{\text{x}}\sin^{-1}\text{x}=\text{t}$
Diff both sides w.r.t x
$\Big(\text{e}^{\text{x}}\sin^{-1}\text{x}+\text{e}^{\text{x}}\times\frac{1}{\sqrt{1-\text{x}^2}}\Big)\text{dx = dt}$
$\because\text{I}=\int\text{dt}$
$=\text{t + C}$
$=\text{e}^{\text{x}}\sin^{-1}\text{x + C}$
View full question & answer→Question 1345 Marks
Evaluate the following integrals:$\int\sin\text{x}\log(\cos\text{x})\text{dx}$
AnswerLet $\text{I}=\int\sin\text{x}\cdot\log(\cos\text{x})\text{dx}$
Let $\cos\text{x = t}$
$\Rightarrow-\sin\text{x dx =}\text{ dt}$
$\Rightarrow\sin\text{x dx =}-\text{dt}$
$\therefore\text{I}=-\int\log\text{t dt}$
$=-\int1\cdot\log\text{t dt}$
Taking log t as the first function and 1 as the second function.
$=\log\text{t}\int1\text{dt}-\int\big\{\frac{\text{d}}{\text{dt}}(\log\text{t})\int1\text{dt}\big\}\text{dt}$
$=-[\log\text{t}\cdot\text{t}-\int\frac{1}{\text{t}}\times\text{t dt}]$
$=-[\log\text{t}\cdot\text{t}-\text{t}]+\text{C}$
$=-\text{t}(\log\text{t}-1)+\text{C} \dots(1)$
Substituting the value of t in eq (1)
$=-\cos\text{x}\{\log(\cos\text{x})-1\}+\text{C}$
$=\cos\text{x}\{1-\log(\cos\text{x})\}+\text{C}$
View full question & answer→Question 1355 Marks
Evaluate the following integrals:$\int\frac{1}{\sqrt{(\text{x}-\alpha)(\beta-\text{x})}}\text{ dx},(\beta>\alpha)$
AnswerLet $\text{I}=\int\frac{1}{\sqrt{(\text{x}-\alpha)(\beta-\text{x})}}\text{ dx},(\beta>\alpha)$
$=\int\frac{1}{-\text{x}^2-\text{x}(\alpha+\beta)-\alpha\beta}\text{ dx}$
$=\int\frac{1}{\sqrt{-\Big[\text{x}^2-2\text{x}\big(\frac{\alpha+\beta}{2}\big)+\big(\frac{\alpha+\beta}{2}\big)^2-\big(\frac{\alpha+\beta}{2}\big)^2+\alpha\beta\Big]}}\text{ dx}$
$=\int\frac{1}{\sqrt{-\Big[\big(\text{x}-\frac{\alpha+\beta}{2}\big)^2-\big(\frac{\alpha+\beta}{2}\big)^2\Big]}}\text{ dx}$
$=\int\frac{1}{\sqrt{\Big[\big(\frac{\beta-\alpha}{2}\big)^2-\big(\text{x}-\frac{\alpha+\beta}{2}\big)^2\Big]}}\text{ dx}$ $[\therefore\ \beta>\alpha]$
Let $\Big(\text{x}-\frac{\alpha+\beta}{2}\Big)=\text{t}$
$\Rightarrow\text{dx}=\text{dt}$
$\text{I}=\int\frac{1}{\sqrt{\big(\frac{\beta-\alpha}{2}\big)^2-\text{t}^2}}\text{ dt}$
$\text{I}=\sin^{-1}\bigg(\frac{\text{t}}{\frac{\beta-\alpha}{2}}\bigg)+\text{C}$ $\Big[\text{Since }\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}=\sin^{-1}\Bigg(\frac{2\big(\text{x}-\frac{\alpha+\beta}{2}\big)}{\beta-\alpha}\Bigg)+\text{C}$
$\text{I}=\sin^{-1}\Big(\frac{2\text{x}-\alpha-\beta}{\beta-\alpha}\Big)+\text{C}$
View full question & answer→Question 1365 Marks
If f'(x) = a sin x + b cos x and f'(0) = 4, f(0) = 3, $\text{f}\Big(\frac{\pi}{2}\Big)=5$, find f(x).
Answer$\text{f'(x)}=\text{a}\sin\text{x}+\text{b}\cos\text{x}$
$\text{f'}(0)=4,\text{f}(0)=3$
$\text{f}\Big(\frac{\pi}{2}\Big)=5$
$\text{f'(x)}=\text{a}\sin\text{x + b}\cos\text{x}$
$\int\text{f'(x)}\text{dx}=\int(\text{a}\sin\text{x + b}\cos\text{x})\text{dx}$
$\text{f(x)}=-\text{a}\cos\text{x}+\text{b}\sin\text{x}+\text{C}\ \dots(1)$
Now putting x = 0 in eq. (1)
$\text{f}(0)=-\text{a}\cos0+\text{b}\sin0+\text{C}$
$3=-\text{a}\times1+\text{b}\times0+\text{C}$
$3=-\text{a + C}\ \dots(2)$
Now putting $\text{x}=\frac{\pi}{2}$ in eq. (1)
$\text{f}\Big(\frac{\pi}{2}\Big)=-\text{a}\cos\frac{\pi}{2}+\text{b}\sin\frac{\pi}{2}+\text{C}$
$5=-\text{a}\cos\frac{\pi}{2}+\text{b}\sin\Big(\frac{\pi}{2}\Big)+\text{C}$
$5=-\text{a}\times0+\text{b}\times1+\text{C}$
$5=\text{b + C}\ \dots(3)$
We also have f'(0) = 4
$\text{f'(x)}=\text{a}\sin\text{x + b}\cos\text{x}$
$\text{f'(0)}=\text{a}\sin0+\text{b}\cos0$
$4=\text{a}\times0+\text{b}\times1$
$4=\text{b}\ \dots(4)$
Solving (2), (3) and (4) we get,
$\text{b}=4$
$\text{C}=1$
$\text{a}=-2$
$\therefore\ \text{f(x)}=2\cos\text{x}+4\sin\text{x}+1$
View full question & answer→Question 1375 Marks
Evaluate the following integrals:$\int\text{e}^{\text{x}}\Big(\frac{\sin4\text{x}-4}{1-\cos4\text{x}}\Big)\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{x}}\Big(\frac{\sin4\text{x}-4}{2\sin^22\text{x}}\Big)\text{dx}$
$=\int\text{e}^{\text{x}}\Big\{\frac{2\sin2\text{x}\cos2\text{x}}{2\sin^22\text{x}}-\frac{4}{2\sin^{2}2\text{x}}\Big\}\text{dx}$
$=\int\text{e}^{\text{x}}\big(\cot2\text{x}-2\text{cosec}^22\text{x}\big)\text{dx}$
$=\int\text{e}^{\text{x}}\cot2\text{x dx}-2\int\text{e}^{\text{x}}\text{cosec}^22\text{x dx}$
integrating by parts
$=\text{e}^{\text{x}}\cot2\text{x}-\int\text{e}^{\text{x}}\frac{\text{d}}{\text{dx}}(\cot2\text{x})\text{dx}-2\int\text{e}^{\text{x}}\text{cosec}^22\text{x dx}$
$=\text{e}^{\text{x}}\cot2\text{x}+2\int\text{e}^{\text{x}}\text{cosec}^22\text{x}-2\int\text{e}^{\text{x}}\text{cosec}^22\text{x dx}$
$=\text{e}^{\text{x}}\cot2\text{x}+\text{C}$
View full question & answer→Question 1385 Marks
Evaluate the following integrals:$\int\frac{\sqrt{1-\sin\text{x}}}{1+\cos\text{x}}\text{e}^{\frac{-\text{x}}{2}}\text{dx}$
AnswerLet $\text{I}=\int\frac{\sqrt{1-\sin\text{x}}}{1+\cos\text{x}}\text{e}^{\frac{-\text{x}}{2}}\text{dx}$
Put $=\frac{\text{x}}{2}=\text{t}$
$\Rightarrow\text{x}=2\text{t}$
$\text{dx}=2\text{dt}$
$\therefore\int\frac{\sqrt{1-\sin\text{x}}}{1+\cos\text{x}}\text{e}^{-\frac{\text{x}}{2}}\text{dx}$
$=2\int\frac{\sqrt{1-\sin2\text{t}}}{1+\cos2\text{t}}\text{e}^{-\text{t}}\text{dt}$ $\big[\because\sin^2\text{t}+\cos^2\text{t}=1\big]$
$=2\int\frac{\sqrt{\sin^2\text{t}+\cos^2\text{t}-2\sin\text{t}\cos\text{t}}}{1+\cos2\text{t}}\text{e}^{-\text{t}}\text{dt}$
$=2\int\frac{\sqrt{(\cos\text{t}-\sin\text{t})^2}}{2\cos^2\text{t}}\text{e}^{-\text{t}}\text{dt}$
$=2\int\frac{(\cos\text{t}-\sin\text{t})}{2\cos^2\text{t}}\text{e}^{-\text{t}}\text{dt}$
$=\int(\sec\text{t}-\tan\text{t}\sec\text{t})\text{e}^{-\text{t}}\text{dt}$
$=\int\sec\text{e}^{-\text{t}}\text{dt}-\int\tan\text{t}\sec\text{e}^{-\text{t}}\text{dt}$
Integrating by parts
$=\text{e}^{-\text{t}}\sec\text{t}+\int\text{e}^{-\text{t}\frac{\text{d}}{\text{dt}}}(\sec\text{t})\text{dt}-\int\tan\text{t}\sec\text{t}\text{ e}^{-\text{t}}\text{dt}$
$=-\text{e}^{-\text{t}}\sec\text{t}+\int\text{e}^{-\text{t}}\sec\text{t}\tan\text{t dt}-\int\sec\text{t}\tan\text{t}\text{ e}^{-\text{}t}\text{dt}$
$=-\text{e}^{-\text{t}}\sec\text{t+C}$
Putting the value of t
$=\text{-e}^{-\frac{\text{x}}{2}}\sec\frac{\text{x}}{2}+\text{C}$
View full question & answer→Question 1395 Marks
If f'(x) = x + b, f'(1) = 5, f'(2) = 13, find f'(x).
Answer$\text{f}'\text{(x)}=\text{x + b},\text{f}'(1)=5,\text{f}'(2)=13$
$\text{f}'\text{(x)}=\text{x + b}$
$\int\text{f}'\text{(x) dx}=\int(\text{x + b})\text{dx}$
$\text{f}'\text{(x)}=\frac{\text{x}^2}{2}+\text{bx}+\text{C}\ \dots(1)$
$\text{f}'(1)=5,\text{f}'(2)=13$ (Given)
putting x = 1 in (1)
$\text{f}'(1)=\frac{1^2}{2}+\text{b}_1+\text{C}$
$5=\frac{1}{2}+\text{b + C}\ \dots(2)$
Putting x = 2 in (1)
$\text{f}'(2)=\frac{2^2}{2}+\text{b}_2+\text{C}$
$13=\frac{4}{2}+2\text{b + C}$
$13=2+2\text{b + C}\ \dots(3)$
Solving (2) and (3) we get,
$\text{b}=\frac{13}{2}\text{ and }\text{C}=-2$
Thus, $\text{f}'\text{(x)}=\frac{\text{x}^2}{2}+\frac{13}{2}\text{x}-2$
View full question & answer→Question 1405 Marks
Evaluate the following definite integral as limit of sum:$\int_\limits{0}^{4}\text{(x}+\text{e}^{2\text{x}})\ \text{dx}$
Answer$\text{we}\ \text{know}\ \text{that}\ \int\limits_{\text{a}}^{\text{b}}\text{f}\text{(x)}\ \text{dx}=\lim\limits_{\text{h}\rightarrow0}\ \text{h}\big[\text{f}\ \text{(a)}+\text{f}\text{(a+h)}+\text{f}\text{(a}+2\text{h})+.....+\text{f}\text{(a}+\text{(n}-1)\text{h)}\big] $ $\text{where}\ \text{nh}=\text{b}-\text{a}$ $\text{Here},\ \text{a}=0,\text{b}=4,\text{nh}=4\ \text{and}\ \text{f}\ \text{(x)}=\text{x}+\text{e}^{2\text{x}}$ $\therefore\ \ \int\limits_{0}^{4}\text{(x+e}^{2\text{h}})\ \text{dx}=\lim\limits_{\text{h}\rightarrow0}\ \text{h}\bigg[1+\text{(h}+\text{e}^{2\text{h}})+\text{(2h}+\text{e}^{4\text{h}})+....+\big(\text{(n}-1)\text{h+e}^{2\text{(n-1)}\text{h}}\big)\bigg]$ $=\lim\limits_{\text{h}\rightarrow0}\text{h}\bigg[\text{(h}+2\text{h}+.....+\text{(n}-1)\text{h)}+\big(1+\text{e}^{2\text{h}}+\text{e}^{4\text{h}}+....+\text{e}^{2(\text{n}-1)\text{h}}\big)\bigg]$ $=\lim\limits_{\text{h}\rightarrow0}\text{h}\bigg[\text{h}(1+2+......+\text{(n}-1)+\text{a}\bigg(\frac{\text{r}^{\text{n}}-1}{\text{r}-1}\bigg)\bigg]=\lim\limits_{\text{h}\rightarrow0}\bigg[\text{h.h}\frac{\text{n}\text{(n}-1)}{2}+\frac{1\big(\text{e}^{2\text{h}^{\text{n}}}\big)-1}{\text{e}^{2\text{h}}-1}\bigg]$ $=\lim\limits_{\text{h}\rightarrow0}=\bigg[\frac{\text{nh}\text{(nh}-\text{h)}}{2}+\frac{\text{h}\big(\big(\text{e}^{2\text{nh}}\big)-1\big)}{\text{e}^{2\text{h}}-1}\bigg] =\lim\limits_{\text{h}\rightarrow0}\bigg[\frac{4(4-\text{h)}}{2}+\frac{\text{h}\big(\big(\text{e}^{2.4}\big)-1\big)}{\text{e}^{2\text{h}}-1}\bigg]$$=\frac{4(4-0)}{2}+\big(\text{e}^{8}-1\big)\lim\limits_{\text{h}\rightarrow0}\frac{\text{h}}{\text{e}^{2\text{h}}-1}=8\big(\text{e}^{8}-1\big)\frac{1}{2}\lim\limits_{\text{h}\rightarrow0}\frac{2\text{h}}{\text{e}^{2\text{h}}-1}=8+\frac{\big(\text{e}^{8}-1\big)}{2}$
$=\bigg[\because\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}}{\text{e}^{\text{x}}-1}=1\bigg]$
View full question & answer→Question 1415 Marks
Evaluate the following integrals:$\int\frac{\cos\text{x}-\sin\text{x}}{\sqrt{8-\sin2\text{x}}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\cos\text{x}-\sin\text{x}}{\sqrt{8-\sin2\text{x}}}\text{ dx}$
$=\int\frac{\cos\text{x}-\sin\text{x}}{\sqrt{9-1-\sin2\text{x}}}\text{ dx}$
$=\int\frac{\cos\text{x}-\sin\text{x}}{\sqrt{9-\sin^2\text{x}-\cos^2\text{x}-2\sin\text{x}\cos\text{x}}}\text{ dx}$
$=\int\frac{\cos\text{x}-\sin\text{x}}{\sqrt{9-(\sin\text{x}+\cos\text{x})^2}}\text{ dx}$
Let $(\sin\text{x}+\cos\text{x})=\text{t}$
On differentiating both sides, we get
$(\cos\text{x}-\sin\text{x})\text{dx}=\text{dt}$
$\therefore\ \text{I}=\int\frac{1}{(3)^2-(\text{t})^2}\text{ dt}$
$=\sin^{-1}\Big(\frac{\text{t}}{3}\Big)+\text{C}$
$=\sin^{-1}\Big(\frac{\sin\text{x}-\cos\text{x}}{3}\Big)+\text{C}$
Hence, $\int\frac{\cos\text{x}-\sin\text{x}}{\sqrt{8-\sin2\text{x}}}\text{ dx}=\sin^{-1}\Big(\frac{\cos\text{x}-\sin\text{x}}{3}\Big)+\text{C}$
View full question & answer→Question 1425 Marks
Evaluate the following integrals:$\int\cos^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)\text{dx}$
AnswerLet $\text{I}=\int\cos^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)\text{dx}$
Let $\text{x}=\tan\text{t}$
$\text{dx}=\sec^2\text{t dt}$
$\text{I}=\int\cos^{-1}\Big(\frac{1-\tan^2\text{t}}{1+\tan^2\text{t}}\Big)\sec^2\text{t dt}$
$=\int\cos^{-1}(\cos2\text{t})\sec^2\text{t dt}$
$=\int2\text{t}\sec^2\text{x dx}$
$=2\Big[\text{t}\int\sec^2\text{t dt}-\int(1\int\sec^2\text{t dt})\text{dt}\Big]$
$=2[\text{t}\tan^2\text{t}-\int\tan\text{t dt}]$
$=2[\text{t}\tan^2\text{t}-\log\sec\text{t}]+\text{C}$
$=2\Big[\text{x}\tan^{-1}\text{x}-\log\sqrt{1+\text{x}^2}\Big]+\text{C}$
$\text{I}=2\text{x}\tan^{-1}\text{x}-\log|1+\text{x}^2|+\text{C}$
View full question & answer→Question 1435 Marks
Evaluate the following integrals:$\int\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)\text{dx}$
AnswerLet $\text{x}=\tan\theta\Rightarrow\text{dx}=\sec^2\theta\text{d}\theta$
$\therefore\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)=\sin^{-1}\Big(\frac{2\tan\theta}{1+\tan^2\theta}\Big)=\sin^{-1}(\sin2\theta)=\theta$
$\Rightarrow\int\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)\text{dx}=\int2\theta\cdot\sec^2\theta\text{d}\theta=2\int\theta\cdot\sec^2\theta\text{d}\theta$
Integrating by parts, we obtain
$2\Big[\theta\cdot\int\sec^2\theta\text{d}\theta-\int\Big\{\Big(\frac{\text{d}}{\text{d}\theta}\theta\Big)\int\sec^2\theta\text{d}\theta\Big\}\text{d}\theta\Big]$
$=2\big[\theta\cdot\tan\theta-\int\tan\theta\text{d}\theta\big]$
$=2\big[\theta\tan\theta+\log|\cos\theta|\big]+\text{C}$
$=2\Big[\text{x}\tan^{-1}\text{x}+\log\Big|\frac{1}{\sqrt{1+\text{x}^2}}\Big|\Big]+\text{C}$
$=2\text{x}\tan^{-1}\text{x}+2\log(1+\text{x}^2)^{-\frac{1}{2}}+\text{C}$
$=2\text{x}\tan^{-1}\text{x}+2\Big[-\frac{1}{2}\log(1+\text{x}^2)\Big]+\text{C}$
$=2\text{x}\tan^{-1}\text{x}-\log(1+\text{x}^2)+\text{C}$
View full question & answer→Question 1445 Marks
Evaluate the following integrals:$\int\frac{(3\sin\text{x}-2)\cos\text{x}}{13-\cos^2\text{x}-7\sin\text{x}}\text{ dx}$
Answer$\text{I}=\int\frac{(3\sin\text{x}-2)\cos\text{x}}{13-\cos^2\text{x}-7\sin\text{x}}\text{ dx}$ $=\int\frac{(3\sin\text{x}-2)\cos\text{x}}{13-(1-\sin^2\text{x})-7\sin\text{x}}\text{ dx}$ $\big(\because\ \cos^2\text{x}=1-\sin^2\text{x}\big)$ $=\int\frac{(3\sin\text{x}-2)\cos\text{x}}{\sin^2\text{x}-7\sin\text{x}+12}\text{ dx}$ $=\int\frac{(3\sin\text{x}-2)\cos\text{x}}{\sin^2\text{x}-4\sin\text{x}-3\sin\text{x}+12}\text{ dx}$ $=\int\frac{(3\sin\text{x}-2)\cos\text{x}}{\sin\text{x}(\sin\text{x}-4)-3(\sin\text{x}-4)}\text{dx}$ $=\int\frac{(3\sin\text{x}-2)\cos\text{x}}{(\sin\text{x}-3)(\sin\text{x}-4)}\text{ dx}$Let $\sin\text{x}=\text{t}$
$\Rightarrow\cos\text{x}\text{ dx}=\text{dt}$
$\therefore\ \text{I}=\int\frac{(3\text{t}-2)}{(\text{t}-3)(\text{t}-4)}\text{ dt}$
Using partial fraction, we get $\frac{(3\text{t}-2)}{(\text{t}-3)(\text{t}-4)}=\frac{\text{A}}{(\text{t}-3)}+\frac{\text{B}}{(\text{t}-4)}$ $=\frac{\text{A}(\text{t}-4)+\text{B}(\text{t}-3)}{(\text{t}-3)(\text{t}-4)}$ $\Rightarrow3\text{t}-2=(\text{A}+\text{B})\text{t}-4\text{A}-3\text{B}$ Comparing coefficients, we get A = -7 and B = 10 So, $\text{I}=-7\int\frac{1}{(\text{t}-3)}\text{ dt}+10\int\frac{1}{(\text{t}-4)}\text{ dt}$$\Rightarrow\text{I}=-7\ln|\text{t}-3|+10\ln|\text{t}-4|+\text{C}$
$\therefore\ \text{I}=-7\ln|\sin\text{x}-3|+10\ln|\sin\text{x}-4|+\text{C}$
View full question & answer→Question 1455 Marks
Evaluate the following integrals:$\int\sin^{-1}\sqrt{\text{x}}\text{dx}$
Answer$\int\sin^{-1}\sqrt{\text{x}}\text{dx}$
Let $\text{x} = \text{t}^{2}\ \ \ [\therefore\text{dx = 2tdt}]$
$\int\sin^{-1}\sqrt{\text{x}}\text{dx}=\int\sin^{-1}\sqrt{\text{t}^2}2\text{tdt}=\int\sin^{-1}\text{t}2\text{tdt}$
$=\sin^{-1}\text{t}\int2\text{tdt}-\Big(\int\frac{\text{d}\sin^{-1}\text{t}}{\text{dt}}\big(\int2\text{tdt}\big)\text{dt}\Big)$
$=\sin^{-1}\text{t}(\text{t}^2)-\int\frac{1}{\sqrt{1-\text{t}^2}}(\text{t}^2)\text{dt}$
Lets solve $\int\frac{1}{\sqrt{1-\text{t}^2}}(\text{t}^2)\text{dt}$
$\int\frac{1}{\sqrt{1-\text{t}^2}}(\text{t}^2)\text{dt}=\int\frac{\text{t}^2-1+1}{\sqrt{1-\text{t}^2}}\text{dt}=\int\frac{\text{t}^2-1}{\sqrt{1-\text{t}^2}}\text{dt}+\int\frac{1}{\sqrt{1-\text{t}^2}}\text{dt}$
We know that, value of $\int\frac{1}{\sqrt{1-\text{t}^2}}\text{dt}=\sin^{-1}\text{t}$
Remaining integral to evalute is $\int\frac{\text{t}^2-1}{\sqrt{1-\text{t}^2}}\text{dt}=\int-\sqrt{1-\text{t}^2}\text{dt}$
sub $\text{t}=\sin\text{u},\text{dt}=\cos\text{u du}$
$\int-\sqrt{1-\text{t}^2}\text{dt}=\int-\cos^2\text{u du}=-\int\Big[\frac{1+\cos2\text{u}}{2}\Big]\text{du}$
$=-\frac{\text{u}}{2}-\frac{\sin2\text{u}}{4}$
Substitute back $\text{u}=\sin^{-1}\text{t}$ and $\text{t}=\sqrt{\text{x}}$
$=-\frac{\sin^{-1}\sqrt{\text{x}}}{2}-\frac{\sin(2\sin^{-1}\sqrt{\text{x}})}{4}$
$\int\sin^{-1}\sqrt{\text{x}}\text{dx}=\text{x}\sin^{-1}\sqrt{\text{x}}-\frac{\sin^{-1}\sqrt{\text{x}}}{2}-\frac{\sin(2\sin^{-1}\sqrt{\text{x}})}{4}$
$\sin(2\sin^{-1}\sqrt{\text{x}})=2\sqrt{\text{x}}\sqrt{1-\text{x}}$
$\int\sin^{-1}\sqrt{\text{x}}\text{dx}=\text{x}\sin^{-1}\sqrt{\text{x}}-\frac{\sin^{-1}\sqrt{\text{x}}}{2}-\frac{\sqrt{\text{x}(1-\text{x})}}{2}$
View full question & answer→Question 1465 Marks
Evaluate the following integrals:
$\int^\limits2_1\frac{1}{\text{x}(1+\log\text{x})^2}\text{ dx}$
AnswerLet $\text{I}=\int^\limits2_1\frac{1}{\text{x}(1+\log\text{x})^2}\text{ dx}$ Then,
Let $(1+\log\text{x})=\text{t}$ Then, $\frac{1}{\text{x}}\text{ dx}=\text{dt}$
When, $\text{x}=1,\text{t}=1$ and $\text{x}=2,\text{t}=1+\log2$
$\therefore\ \text{I}=\int^\limits{(1+\log2)}_{1}\frac{1}{\text{t}}\text{ dt}$
$\Rightarrow\text{I}=\Big[\frac{-1}{\text{t}}\Big]^{(1+\log2)}_1$
$\Rightarrow\text{I}=-\frac{1}{1+\log2}+1$
$\Rightarrow\text{I}=\frac{\log2}{1+\log2}$
View full question & answer→Question 1475 Marks
Evaluate the following integrals:$\int\text{cosec}^3\text{x dx}$
AnswerLet $\text{I}=\int\text{cosec}^3\text{dx}$
$=\int\text{cosec x}-\text{cosec}^2\text{x dx}$
using integration by parts,
$=\text{cosec x}\times\int\text{cosec}^2\text{x dx}+\int(\text{cosec x}\cot\text{x}\int\text{cosec}^2\text{x dx})\text{dx}$
$=\text{cosec x}\times(-\cot\text{x})+\int\text{cosec x}\cot\text{x}(-\cot\text{x})\text{dx}$
$=-\text{cosec x}\cot\text{x}-\int\text{cosec x}\cot^2\text{x dx}$
$=-\text{cosec x}\cot \text{x}-\int\text{cosec x}(\text{cosec}^2\text{x}-1)\text{dx}$
$=-\text{cosec x}\cot\text{x}-\int\text{cosec}^3\text{x dx}+\int\text{cosec x dx}$
$\text{I}=-\text{cosec x}\cot\text{x}-\text{I}+\log\Big|\tan\frac{\text{x}}{2}\Big|+\text{C}_1$
$2\text{I}=-\text{cosec x}\cot\text{x}+\log\Big|\tan\frac{\text{x}}{2}\Big|+\text{C}_1$
$\text{I}=-\frac{1}{2}\text{cosec x}\cot\text{x}+\frac{1}{2}\log\Big|\tan\frac{\text{x}}{2}\Big|+\text{C}$
View full question & answer→Question 1485 Marks
Evaluate the following integrals:
$\int^\limits{\frac{\pi}{2}}_{0}\sqrt{\cos\text{x}-\cos^3\text{x}}(\sec^3\text{x}-1)\cos^2\text{x dx}$
AnswerLet $\text{I}=\int^\limits{\frac{\pi}{2}}_{0}\sqrt{\cos\text{x}-\cos^3\text{x}}(\sec^3\text{x}-1)\cos^2\text{x dx}$
$=\int^\limits{\frac{\pi}{2}}_{0}\sqrt{\cos\text{x}(1-\cos^2\text{x}\big)}(-\tan^2\text{x})\cos^2\text{x dx}$
$=-\int^\limits{\frac{\pi}{2}}_{0}\sqrt{\cos\text{x}(\sin^2\text{x})}\sin^2\text{x dx}$
$=-\int^\limits{\frac{\pi}{2}}_{0}\sqrt{\cos\text{x}}(\sin\text{x})\sin^2\text{x dx}$
$=-\int^\limits{\frac{\pi}{2}}_{0}\sqrt{\cos\text{x}}\big(1-\cos^2\text{x}\big)\sin\text{x dx}$ $\Big(\sin\text{x}=\sin\text{x}\text{ for }0\leq\text{x}\leq\frac{\pi}{2}\Big)$
Put $\cos\text{x}=\text{z}^2$
$\therefore\ -\sin\text{x dx}=2\text{z dz}$
When $\text{x}\rightarrow0,\text{ z}\rightarrow0$
When $\text{x}\rightarrow\frac{\pi}{2},\text{ z}\rightarrow0$
$\therefore\ \text{I}=-\int^\limits0_1\text{z}(1-\text{z}^4)2\text{z dz}$
$=-2\int^0\limits_1\text{z}^2\text{ dz}+2\int^\limits0_1\text{z}^6\text{ dz}$
$=-2\times\Big[\frac{\text{z}^3}{3}\Big]^0_1+2\times\Big[\frac{\text{z}^7}{7}\Big]^0_1$
$=-\frac{2}{3}\big(0-1\big)+\frac{2}{7}\big(0-1\big)$
$=\frac{2}{3}-\frac{2}{74}$
$=\frac{8}{21}$
View full question & answer→Question 1495 Marks
Evaluate the following definite integrals:
$\int_{0}^\limits{\frac{\pi}{2}}\sin^3\text{x}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\sin^3\text{x}\text{ dx}$ Then,
$\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\sin\text{x }\sin^2\text{x}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\sin\text{x}(1-\cos^2\text{x})\text{dx}$
Let $\text{u}=\cos\text{x},\text{ du}=-\sin\text{x dx}$
$\therefore\ \text{I}=\int-(1-\text{u}^2)\text{du}$
$\Rightarrow\text{I}=\Big[\frac{\text{u}^3}{3}-\text{u}\Big]$
$\Rightarrow\text{I}=\Big[\frac{\cos^3\text{x}}{3}-\cos\text{x}\Big]^{\frac{\pi}{2}}_0$
$\Rightarrow\text{I}=0-\frac{1}{3}+1$
$\Rightarrow\text{I}=\frac{2}{3}$
View full question & answer→Question 1505 Marks
Evaluate the following definite integrals:
$\int_{0}^\limits{1}\sqrt{\text{x}(1-\text{x})}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{1}\sqrt{\text{x}(1-\text{x})}\text{ dx}$ Then,
$\text{I}=\int_{0}^\limits{1}\sqrt{\frac{1}{4}-\big(\text{x}-\frac{1}{2}\big)^2}\text{ dx}$
$\Rightarrow\text{I}=\frac{1}{2}\int_{0}^\limits{1}\sqrt{1-\frac{\big(\text{x}-\frac{1}{2}\big)^2}{\frac{1}{4}}}\text{ dx}$
$\Rightarrow\text{I}=\frac{1}{2}\int_{0}^\limits{1}\sqrt{1-\Bigg(\frac{\text{x}-\frac{1}{2}}{\frac{1}{2}}\Bigg)^2}\text{ dx}$
Let $\Bigg(\frac{\text{x}-\frac{1}{2}}{\frac{1}{2}}\Bigg)=\sin\text{u}$
$\Rightarrow2\text{dx}=\cos\text{u du}$
$\therefore\ \text{I}=\frac{1}{4}\int_{-\frac{\pi}{2}}^\limits{\frac{\pi}{2}}\sqrt{1-\sin^2\text{u}}\cos\text{u du}$
$\Rightarrow \text{I}=\frac{1}{4}\int_{-\frac{\pi}{2}}^\limits{\frac{\pi}{2}}\cos^2\text{u du}$
$\Rightarrow \text{I}=\frac{1}{4}\int_{-\frac{\pi}{2}}^\limits{\frac{\pi}{2}}\Big(\frac{\cos2\text{u}+1}{2}\Big)\text{du}$
$\Rightarrow \text{I}=\frac{1}{8}\Big[\frac{\sin2\text{u}}{2}+\text{u}\Big]^{\frac{\pi}{2}}_{-\frac{\pi}{2}}$
$\Rightarrow \text{I}=\frac{1}{8}\Big[\frac{\pi}{2}+\frac{\pi}{2}\Big]$
$\Rightarrow\text{I}=\frac{\pi}{8}$
View full question & answer→