Question 2515 Marks
Evaluate the following integrals:
$\int\frac{\text{e}^{\text{x}}}{\text{x}}\Big\{\text{x}(\log\text{x})^2+2\log\text{x}\Big\}\text{dx}$
AnswerWe have,
$\text{I}=\int\frac{\text{e}^{\text{x}}}{\text{x}}\Big\{\text{x}(\log\text{x})^2+2\log\text{x}\Big\}\text{dx}$
$=\int\text{e}^{\text{x}}\Big\{(\log\text{x})^2+\frac{2}{\text{x}}\log\text{x}\Big\}\text{dx}$
$=\int\text{e}^{\text{x}}(\log\text{x})^2+2\int\frac{\text{e}^{\text{x}}}{\text{x}}\log\text{x dx}$
Integrating by parts
$=\text{e}^{\text{x}}(\log\text{x})^2-\int\text{e}^{\text{x}}\frac{\text{d}}{\text{dx}}(\log\text{x})^2\text{dx}+2\int\text{e}^{\text{x}}\frac{1}{\text{x}}\log\text{x dx}$
$=\text{e}^{\text{x}}(\log\text{x})^2-\int\text{e}^{\text{x}}\frac{2\log\text{x}}{\text{x}}\text{dx}+2\int\text{e}^\text{x}\frac{\log\text{x}}{\text{x}}\text{dx}$
$=\text{e}^{\text{x}}(\log\text{x)}^2+\text{C}$
View full question & answer→Question 2525 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\sin^2\text{x dx}$
AnswerLet $\text{I}=\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\sin^2\text{x dx}$
Here $\text{f(x)}=\sin^2\text{x}$
$\text{f}(-\text{x})=\sin^2(-\text{x})=\sin^2\text{x}=\text{f(x)}$
Hence $\sin^2\text{x}$ is an even function
Therefore,
$\text{I}=2\int\limits^{\frac{\pi}{4}}_{0}\sin^2\text{x dx}$
$=2\int\limits^{\frac{\pi}{4}}_{0}\Big(\frac{1-\cos2\text{x}}{2}\Big)\text{dx}$
$=\int\limits^{\frac{\pi}{4}}_{0}(1-\cos2\text{x})\text{dx}$
$=\Big[\text{x}-\frac{\sin^2\text{x}}{2}\Big]^{\frac{\pi}{4}}_0$
$=\frac{\pi}{4}-\frac{1}{2}$
View full question & answer→Question 2535 Marks
By using the properties of definite integral, evaluate the integral in Exercise:
$\int^{\frac{\pi}{4}}_{0}\log(1+\tan\text{x})\text{dx}$
Answer$\text{Let}\ \text{I}=\int\limits_{0}^{\frac{\pi}{4}}\log(1+\tan\text{x})\text{dx}$$\Rightarrow\ \ \text{I}=\int^{\frac{\pi}{4}}\limits_{0}\log\bigg[1+\tan\bigg(\frac{\pi}{4}-\text{x}\bigg)\bigg]\text{dx}\ \ \bigg[\because\int\limits_{0}^{\text{a}}\text{f}\text{(x)}\ \text{dx}=\int\limits_{0}^{\text{a}}\text{f}\text{(a}-\text{x})\text{dx}=\bigg]$
$\Rightarrow\ \ \text{I}=\int\limits_{0}^{\frac{\pi}{4}}\log\bigg[1+\frac{1-\tan\text{x}}{1+\tan\text{x}}\bigg]\text{dx}=\int\limits_{0}^{\frac{\pi}{4}}\log\bigg[\frac{1+\tan\text{x}+1-\tan\text{x}}{1+\tan\text{x}}\bigg]\text{dx}=\int\limits_{0}^{\frac{\pi}{4}}\log\bigg[\frac{2}{1+\tan\text{x}}\bigg]\text{dx}$
Adding eq. (i) and (ii),
$21=\int\limits_{0}^{\frac{\pi}{4}}\bigg[\log(1+\tan\text{x})+\log\bigg(\frac{2}{1+\tan\text{x}}\bigg)\bigg]\text{dx}=\int\limits_{0 }^{\frac{\pi}{4}}\bigg[\log(1+\tan\text{x)}\bigg(\frac{2}{1+\tan\text{x}}\bigg)\bigg]\text{dx}$
$\Rightarrow\ \ \ 21=\int\limits_{0}^{\frac{\pi}{4}}\big[\log2\big]\text{dx}=(\log2)\ \text{(x)}^{\frac{\pi}{4}}_{0}=\frac{\pi}{4}\log2$
$\Rightarrow\ \ \text{I}=\frac{\pi}{8}\log2$
View full question & answer→Question 2545 Marks
Evaluate the following intregals:
$\int\frac{18}{(\text{x}+2)(\text{x}^2+4)}\text{ dx}$
AnswerLet $\frac{18}{(\text{x}+2)(\text{x}^2+4)}=\frac{\text{A}}{\text{x}+2}+\frac{\text{Bx}+\text{C}}{\text{x}^2+4}$$\Rightarrow18=\text{A}(\text{x}^2+4)+(\text{Bx}+\text{C})(\text{x}+2)$
$18=(\text{A}+\text{B})\text{x}^2+(2\text{B}+\text{C})\text{x}+(4\text{A}+2\text{C})$
Equating similar terms, get,
$\text{A}+\text{B}=0,2\text{B}+\text{C}=0,4\text{A}+2\text{C}=18$
Solving, we get,
$\text{A}=\frac{9}{4},\text{B}=-\frac{9}{4},\text{C}=\frac{9}{2}$
Thus,
$\text{I}=\frac{9}{4}\int\frac{\text{dx}}{\text{x}+2}+\Big(-\frac{9}{4}\Big)\frac{\text{x}}{\text{x}^2+4}\ \text{dx}+\frac{9}{2}\int\frac{\text{dx}}{\text{x}^2+4}$
$\text{I}=\frac{9}{4}\log|\text{x}+2|-\frac{9}{8}\log|\text{x}^2+4|+\frac{9}{4}\tan^{-1}\Big(\frac{\text{x}}{2}\Big)+\text{C}$ $\Big[\because\int\frac{\text{dx}}{\text{x}^2+\text{a}^2}=\frac{1}{\text{a}}\tan^{-1}\frac{\text{x}}{9}\Big]$
View full question & answer→Question 2555 Marks
Evaluate the following intregals:
$\int\frac{2}{2+\sin^22\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{2}{2+\sin^22\text{x}}\text{ dx}$
$\text{I}=\int\frac{2}{2+2\sin\text{x}\cos\text{x}}\ \text{dx}$
Dividing numerator and denominator by $\cos^2\text{x}$
$\text{I}=\int\frac{\frac{1}{\cos^2\text{x}}}{\frac{1}{\cos^2\text{x}}+\frac{\sin\text{x}\cos\text{x}}{\cos^2\text{x}}}\ \text{dx}$
$=\int\frac{\sec^2\text{x}}{\sec^2\text{x}+\tan\text{x}}\ \text{dx}$
$\text{I}=\int\frac{\sec^2\text{x}}{1+\tan^2\text{x}+\tan\text{x}}\ \text{dx}$
Let $\tan\text{x}=\text{t}$
$\sec^2\text{x dx}=\text{dt}$
$\text{I}=\int\frac{\text{dt}}{\text{t}^2+\text{t}+1}$
$=\int\frac{\text{dt}}{\text{t}^2+2\text{t}\Big(\frac{1}{2}\Big)+\Big(\frac{1}{2}\Big)^2-\Big(\frac{1}{2}\Big)^2+1}$
$\text{I}=\int\frac{\text{dt}}{\Big(\text{t}+\frac{1}{2}\Big)^2+\Big(\frac{\sqrt{3}}{2}\Big)^2}$
$=\frac{1}{\frac{\sqrt{3}}{2}}\tan^{-1}\Bigg(\frac{\text{t}+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\Bigg)+\text{C}$
$=\frac{2}{\sqrt{3}}\tan^{-1}\Big(\frac{2\text{t}+1}{\sqrt{3}}\Big)+\text{C}$
$\text{I}=\frac{2}{\sqrt{3}}\tan^{-1}\Big(\frac{2\tan\text{x}+1}{\sqrt{3}}\Big)+\text{C}$
View full question & answer→Question 2565 Marks
Evaluate the following definite integrals:
$\int_{0}^\limits{\frac{\pi}{6}}\cos\text{x }\cos2\text{x}\text{ dx}$
AnswerWe have,
$\int_{0}^\limits{\frac{\pi}{6}}\cos\text{x }\cos2\text{x}\text{ dx}$ $\big[\because2\cos\text{C}\cos\text{D}=\cos(\text{C}+\text{D})-\cos(\text{C}-\text{D})\big]$
$=\frac{1}{2}\int_{0}^\limits{\frac{\pi}{6}}2\cos\text{x }\cos2\text{x dx}$
$=\frac{1}{2}\int_{0}^\limits{\frac{\pi}{6}}(\cos3\text{x}+\cos\text{x})\text{dx}$
$=\frac{1}{2}\int\Big[\frac{\sin3\text{x}}{3}+\sin\text{x}\Big]_0^{\frac{\pi}{6}}$
$=\frac{1}{2}\Bigg[\bigg(\frac{\sin3\frac{\pi}{6}}{3}+\sin\frac{\pi}{6}\bigg)-(\sin0-\sin0)\Bigg]$
$=\frac{1}{2}\bigg[\frac{\sin\frac{\pi}{2}}{3}+\sin\frac{\pi}{6}\bigg]$
$=\frac{1}{2}\Big(\frac{1}{3}+\frac{1}{2}\Big)$
$=\frac{1}{2}\Big(\frac{5}{6}\Big)$
$=\frac{5}{12}$
$\therefore\ \int_{0}^\limits{\frac{\pi}{6}}\cos\text{x }\cos2\text{x}\text{ dx}=\frac{5}{12}$
View full question & answer→Question 2575 Marks
Evaluate the following integrals:
$\int^\limits\frac{\pi}{2}_{0}\frac{\cos^2\text{x}}{1+3\sin^3\text{x}}\text{ dx}$
Answer$\text{I}=\int^\limits\frac{\pi}{2}_{0}\frac{\cos^2\text{x}}{1+3\sin^3\text{x}}\text{ dx}$
$\text{I}=\int^\limits\frac{\pi}{2}_{0}\frac{\sec^2\text{x}}{\sec^2\text{x}\big(\sec^2\text{x}+3\tan^2\text{x}\big)}\text{ dx}$
Put $\tan\text{x}=\text{t}$
$\sec^2\text{x dx}=\text{dt}$
$\text{x}=0\Rightarrow\text{t}=0$ and $\text{x}=\frac{\pi}{2}\Rightarrow\text{t}=\infty$
$\Rightarrow\text{I}=\int^{\infty}\limits_0\frac{1}{\big(1+\text{t}^2\big)\big(1+4\text{t}^2\big)}\text{ dt}$
$\Rightarrow\text{I}=-\frac{1}{3}\int^{\infty}\limits_0\bigg[\frac{1}{(1+\text{t}^2)-{(1+4\text{t}^2)}}\bigg]\text{dt}$
$\Rightarrow\text{I}=-\frac{1}{3}\Big[\tan^{-1}\text{t}-2\tan^{-1}2\text{t}\Big]^{\infty}_0$
$\Rightarrow\text{I}=\frac{\pi}{6}$
View full question & answer→Question 2585 Marks
Evaluate the following integrals:
$\int\big\{\tan(\log\text{x})+\sec^2(\log\text{x})\big\}\text{dx}$
AnswerLet $\text{I}=\int\big\{\tan(\log\text{x})+\sec^2(\log\text{x})\big\}\text{dx}$
Let $\log\text{x}=\text{z}$
$\Rightarrow\text{x = e}^{\text{z}}$
$\Rightarrow\text{dx}=\text{e}^{\text{z}}\text{dz}$
$\therefore\text{I}=\int\big\{\tan\text{z}+\sec^2\text{z}\big\}\text{e}^{\text{z}}\text{dz}$
Here, $\text{f(z)}=\tan\text{z}$ and $\text{f}'\text{(z)}=\sec^2\text{z}$
And we know that
$\int\text{e}^{\text{ax}}(\text{af(x)}+\text{f}'(\text{x}))\text{dx}=\text{e}^{\text{ax}}\text{f(x) + C}$
$\therefore\int\text{e}^{\text{z}}\big\{\tan\text{z}+\sec^2\text{z}\big\}\text{dz}=\text{e}^{\text{z}}\tan\text{z + C}$
$\therefore\text{I}=\text{x}\tan(\log\text{x})+\text{C}$
View full question & answer→Question 2595 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{3}}_\frac{\pi}{6}\frac{1}{1+\cot^{\frac{3}{2}}\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{\frac{\pi}{3}}_\frac{\pi}{6}\frac{1}{1+\cot^{\frac{3}{2}}\text{x}}\text{ dx}\ ...(\text{i})$
Then,
$\text{I}=\int\limits^{\frac{\pi}{3}}_\frac{\pi}{6}\frac{1}{1+\cot^{\frac{3}{2}}\big(\frac{\pi}{3}+\frac{\pi}{6}-\text{x}\big)}\text{ dx}$ $\Bigg[\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\int\limits^{\text{b}}_\text{a}\text{f}(\text{a}+\text{b}-\text{x})\text{dx}\Bigg]$
$=\int\limits^{\frac{\pi}{3}}_\frac{\pi}{6}\frac{1}{1+\cot^{\frac{3}{2}}\big(\frac{\pi}{2}-\text{x}\big)}\text{ dx}$
$=\int\limits^{\frac{\pi}{3}}_\frac{\pi}{6}\frac{1}{1+\tan^{\frac{3}{2}}\text{x}}\text{ dx}$
$=\int\limits^{\frac{\pi}{3}}_\frac{\pi}{6}\frac{\cot^{\frac{3}{2}}\text{x}}{\cot^{\frac{3}{2}}\text{x}+1}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii)
$2\text{I}=\int\limits^{\frac{\pi}{3}}_\frac{\pi}{6}\frac{1+\cot^{\frac{3}{2}}\text{x}}{1+\cot^{\frac{3}{2}}\text{x}}\text{ dx}$
$\Rightarrow2\text{I}=\int\limits^{\frac{\pi}{3}}_\frac{\pi}{6}\text{dx}$
$\Rightarrow2\text{I}=\big[\text{x}\big]^{\frac{\pi}{3}}_\frac{\pi}{6}$
$\Rightarrow2\text{I}=\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6}$
$\Rightarrow\text{I}=\frac{\pi}{12}$
View full question & answer→Question 2605 Marks
Evaluate the following integrals:
$\int^\limits{\pi}_0\sin^3\text{x}(1+2\cos\text{x})(1+\cos\text{x})^2\text{ dx}$
AnswerLet $\text{I}=\int^\limits{\pi}_0\sin^3\text{x}(1+2\cos\text{x})(1+\cos\text{x})^2\text{ dx}$ Then,
$\text{I}=\int^\limits{\pi}_0\sin\text{x }\sin^2\text{x}(1+2\cos\text{x})(1+\cos\text{x})^2\text{ dx}$
$\Rightarrow\text{I}=\int^\limits{\pi}_0\sin\text{x}(1-2\cos^2\text{x})(1+\cos\text{x})(1+\cos\text{x})^2\text{ dx}$
$\Rightarrow\text{I}=\int^\limits{\pi}_0\sin\text{x}(1-\cos\text{x})(1+2\cos\text{x})(1+\cos\text{x})^3\text{ dx}$
Let $\cos\text{x}=\text{t}$ Then, $-\sin\text{x dx}=\text{dt}$
When $\text{x}=0,\text{t}=1$ and $\text{x}=\pi,\text{t}=-1$
$\therefore\ \text{I}=-\int^\limits{-1}_1(1-\text{t})(1+2\text{t})(1+\text{t})^3\text{ dt}$
$\Rightarrow\text{I}=\int^\limits{-1}_1\big(1+\text{t}-2\text{t}^2\big)\big(1+\text{t}^3+3\text{t}+3\text{t}^2\big)\text{dt}$
$\Rightarrow\text{I}=\int^\limits{-1}_1\big(1+\text{t}^3+3\text{t}+3\text{t}^3+\text{t}+\text{t}^4\\+3\text{t}^2+3\text{t}^3-2\text{t}^2-2\text{t}^5-6\text{t}^3-6\text{t}^4\big)\text{dt}$
$\Rightarrow\text{I}=\int^\limits{-1}_1\big(1+4\text{t}+4\text{t}^2-2\text{t}^3-5\text{t}^4-2\text{t}^5\big)\text{dt}$
$\Rightarrow\text{I}=\Big[\text{t}+2\text{t}^2+\frac{4\text{t}^3}{3}-\frac{\text{t}^4}{2}-\text{t}^5-\frac{\text{t}^6}{3}\Big]^1_{-1}$
$\Rightarrow\text{I}=1+2+\frac{4}{3}-\frac{1}{2}-1-\frac{1}{3}\\+1-2+\frac{4}{3}+\frac{1}{2}-1+\frac{1}{3}$
$\Rightarrow\text{I}=\frac{8}{3}$
View full question & answer→Question 2615 Marks
Evaluate the following definite integrals:
$\int_{1}^\limits{2}\Big(\frac{\text{x}-1}{\text{x}^2}\Big)\text{e}^{\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int_{1}^\limits{2}\Big(\frac{\text{x}-1}{\text{x}^2}\Big)\text{e}^{\text{x}}\text{ dx}$ Then,
$\text{I}=\int_{1}^\limits{2}\Big(\frac{\text{e}^{\text{x}}}{\text{x}}-\frac{\text{e}^{\text{x}}}{\text{x}^2}\Big)\text{dx}$
$\Rightarrow\text{I}=\int_{1}^\limits{2}\frac{\text{e}^\text{x}}{\text{x}}\text{ dx}-\int_{1}^\limits{2}\frac{\text{e}^\text{x}}{\text{x}^2}\text{ dx}$
Integrating first term by parts,
$\text{I}=\bigg\{\Big[\frac{\text{e}^\text{x}}{\text{x}}\Big]^2_1-\int_{1}^\limits{2}\frac{-1}{\text{x}^2}\text{e}^{\text{x}}\text{ dx}\bigg\}-\int_{1}^\limits{2}\frac{\text{e}^\text{x}}{\text{x}^2}\text{ dx}$
$\Rightarrow\text{I}=\Big[\frac{\text{e}^\text{x}}{\text{x}}\Big]^2_1+\int_{1}^\limits{2}\frac{-1}{\text{x}^2}\text{e}^{\text{x}}\text{ dx}-\int_{1}^\limits{2}\frac{\text{e}^\text{x}}{\text{x}^2}\text{ dx}$
$\Rightarrow\text{I}=\Big[\frac{\text{e}^\text{x}}{\text{x}}\Big]^2_1$
$\Rightarrow\text{I}=\frac{\text{e}^2}{2}-\text{e}$
View full question & answer→Question 2625 Marks
Evaluate the following integrals:
$\int\frac{\text{x}\sin^{-1}\text{x}^2}{\sqrt{1-\text{x}^4}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\text{x}\sin^{-1}\text{x}^2}{\sqrt{1-\text{x}^4}}\text{ dx}\ ....(1)$ Let $\sin^{-1}\text{x}^2=\text{t}$ then, $\text{d}\big(\sin^{-1}\text{x}^2\big)=\text{dt}$ $\Rightarrow2\text{x}\times\frac{1}{\sqrt{1-\text{x}^4}}\text{ dx}=\text{dt}$ $\Rightarrow\frac{\text{x}}{\sqrt{1-\text{x}^4}}\text{ dx}=\frac{\text{dt}}{2}$ Putting $\sin^{-1}\text{x}^2=\text{t}$ and $\frac{\text{x}}{\sqrt{1-\text{x}}^4}\text{ dx}=\frac{\text{dt}}{2}$ in equation (1),We get,
$\text{I}=\int\text{t}\frac{\text{dt}}{2}$ $=\frac{1}{2}\times\frac{\text{t}^2}{2}+\text{C}$ $=\frac{1}{4}\big(\sin^{-1}\text{x}^2\big)^2+\text{C}$ $\text{I}=\frac{1}{4}\big(\sin^{-1}\text{x}^2\big)^2+\text{C}$
View full question & answer→Question 2635 Marks
Evaluate the following integrals:
$\int\cos^{-1}(4\text{x}^3-3\text{x})\text{dx}$
AnswerLet $\text{I}=\int\cos^{-1}(4\text{x}^3-3\text{x})\text{dx}$
Let $\text{x}=\cos\theta$
$\text{dx }=-\sin\theta\text{d}\theta$
$\text{I}=-\int\cos^{-1}(4\cos^3\theta-3\cos\theta)\sin\theta\text{d}\theta$
$=-\int\cos^{-1}(\cos3\theta)\sin\theta\text{d}\theta$
$=-\int3\theta\sin\theta\text{d}\theta$
$=-3[\theta\int\sin\theta\text{d}\theta-\int(1\int\sin\theta\text{d}\theta)\text{d}\theta]$
$=-3[-\theta\cos\theta+\int\cos\theta\text{d}\theta]$
$=3\theta\cos\theta-3\sin\theta+\text{C}$
$\text{I}=3\text{x}\cos^{-1}\text{x}-3\sqrt{1-\text{x}^2}+\text{C}$
View full question & answer→Question 2645 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{\sqrt{\tan\text{x}}}{\sqrt{\tan\text{x}}+\sqrt{\cot\text{x}}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{\sqrt{\tan\text{x}}}{\sqrt{\tan\text{x}}+\sqrt{\cot\text{x}}}\text{ dx}\ ...(\text{i})$
Then,
$\text{I}=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{\sqrt{\sin\big(\frac{\pi}{3}+\frac{\pi}{6}-\text{x}\big)}}{\sqrt{\sin\big(\frac{\pi}{3}+\frac{\pi}{6}-\text{x}\big)}+\sqrt{\cos\big(\frac{\pi}{3}+\frac{\pi}{6}-\text{x}\big)}}\text{ dx}$ $\Bigg[\int\limits^{\text{b}}_{\text{a}}\text{f(x)}\text{dx}=\int\limits^{\text{b}}_{\text{a}}\text{f}(\text{a}+\text{b}-\text{x})\text{dx}\Bigg]$
$=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{\sqrt{\sin\big(\frac{\pi}{2}-\text{x}\big)}}{\sqrt{\sin\big(\frac{\pi}{2}-\text{x}\big)}+\sqrt{\cos\big(\frac{\pi}{2}-\text{x}\big)}}\text{ dx}$
$=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{\sqrt{\cos\text{x}}}{\sqrt{\cos\text{x}}+\sqrt{\sin\text{x}}}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii) we get
$2\text{I}=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{\sqrt{\sin\text{x}}+\sqrt{\cos\text{x}}}{\sqrt{\sin\text{x}}+\sqrt{\cos\text{x}}}\text{ dx}$
$\Rightarrow2\text{I}=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\text{dx}$
$\Rightarrow2\text{I}=\Big[\text{x}\Big]^{\frac{\pi}{3}}_{\frac{\pi}{6}}$
$\Rightarrow2\text{I}=\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6}$
$\Rightarrow\text{I}=\frac{\pi}{12}$
View full question & answer→Question 2655 Marks
Write a value of $\int\text{e}^{\text{ax}}\cos\text{bx}\text{ dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{ax}}\cos\text{bx}\text{ dx}$
$=\cos\text{bx}\int\text{e}^{\text{ax}}\text{ dx}-\Big\{\frac{\text{d}}{\text{dx}}(\cos\text{bx})\int\text{e}^{\text{ax}}\text{ dx}\Big\}\text{dx}$
$=\cos\text{bx}\cdot\frac{\text{e}^{\text{ax}}}{\text{a}}-\int-\sin\text{bx}\cdot\text{b}\cdot\frac{\text{e}^{\text{ax}}}{\text{a}}$
$=\cos\text{bx}\cdot\frac{\text{e}^{\text{ax}}}{\text{a}}+\frac{\text{b}}{\text{a}}\int\text{e}^{\text{ax}}\cdot\sin\text{bx}\text{ dx}$
$=\cos\text{bx}\cdot\frac{\text{e}^{\text{ax}}}{\text{a}}+\frac{\text{b}}{\text{a}}\text{I}_1\ ....(\text{i})$
$\therefore\ \text{I}_1=\int\text{e}^{\text{ax}}\cdot\sin\text{bx}\text{ dx}$
$=\sin\text{bx}\int\text{e}^{\text{ax}}\text{dx}-\int\Big\{\frac{\text{d}}{\text{dx}}(\sin\text{bx})\int\text{e}^{\text{ax}}\text{dx}\Big\}\text{dx}$
$=\sin\text{bx}\cdot\frac{\text{e}^{\text{ax}}}{\text{a}}-\int\text{b}\cos\text{bx}\cdot\frac{\text{e}^{\text{ax}}}{\text{a}}\text{ dx}$
$=\sin\text{bx}\cdot\frac{\text{e}^{\text{ax}}}{\text{a}}-\frac{\text{b}}{\text{a}}\text{I}\ ....(\text{ii})$
From (i) & (ii)
$\therefore\ \text{I}=\cos\text{bx}\cdot\frac{\text{e}^{\text{ax}}}{\text{a}}+\frac{\text{b}}{\text{a}}\Big\{\sin\text{bx}\cdot\frac{\text{e}^{\text{ax}}}{\text{a}}-\frac{\text{b}}{\text{a}}\text{I}\Big\}$
$\text{I}=\cos\text{bx}\cdot\frac{\text{e}^{\text{ax}}}{\text{a}}+\frac{\text{b}}{\text{a}^2}\sin\text{bx}\text{e}^{\text{ax}}-\frac{\text{b}^2}{\text{a}^2}\text{I}$
$\text{I}+\frac{\text{b}^2}{\text{a}^2}\text{I}=\cos\text{bx}\cdot\frac{\text{e}^{\text{ax}}}{\text{a}}+\frac{\text{b}\sin\text{bx}\text{e}^{\text{ax}}}{\text{a}^2}$
$\big(\text{a}^2+\text{b}^2\big)\text{I}=(\text{a}\cos\text{bx}+\text{b}\sin\text{bx})\text{e}^{\text{ax}}$
$\text{I}=\frac{(\text{a}\cos\text{bx}+\text{b}\sin\text{bx})\text{e}^{\text{ax}}}{\text{a}^2+\text{b}^2}+\text{C}$
View full question & answer→Question 2665 Marks
Evaluate the following integrals:
$\int_{0}^\limits{\frac{\pi}{4}}\frac{\tan^{3}\text{x}}{1+\cos2\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{\frac{\pi}{4}}\frac{\tan^{3}\text{x}}{1+\cos2\text{x}}\text{ dx}$ Then,
$\text{I}=\int_{0}^\limits{\frac{\pi}{4}}\frac{\tan^{3}\text{x}}{1+\cos2\text{x}}\text{ dx}$
$\Rightarrow\text{I}=\frac{1}{2}\int_{0}^\limits{\frac{\pi}{4}}\tan^{3}\text{x}\sec^2\text{x}\text{ dx}$
Let $\tan\text{x}=\text{t}$ Then, $\sec^2\text{x dx}=\text{dt}$
When $\text{x}=0,\text{t}=0$ and $\text{x}=\frac{\pi}{4},\text{t}=1$
$\therefore\ \text{I}=\frac{1}{2}\int_{0}^\limits{1}\text{t}^3\text{ d}t$
$\Rightarrow\text{I}=\frac{1}{2}\Big[\frac{\text{t}^4}{4}\Big]^1_0$
$\Rightarrow\text{I}=\frac{1}{2}\Big(\frac{1}{4}-0\Big)$
$\Rightarrow\text{I}=\frac{1}{8}$
View full question & answer→Question 2675 Marks
Write a value of $\int\text{e}^{\log\sin\text{x}}\cos\text{x}\text{ dx}$
Answer$\int\text{e}^{\log\sin\text{x}}\cos\text{x}\text{ dx}$
Let $\text{t}=\sin\text{x}\rightarrow\text{dt}=\cos\text{x dx}$
$\int\text{e}^{\log\sin\text{x}}\cos\text{x}\text{ dx}=\int\text{e}^{\log\text{t}}\text{dt}=\text{I}$
$\text{e}^{\log\text{t}}\int1\text{dt}-\Big(\int\frac{\text{de}^{\log\text{t}}}{\text{dt}}\big(\int1\text{dt}\big)\text{dt}\Big)$
$=\text{e}^{\log\text{t}}\text{t}-\Big(\int\text{e}^{\log\text{t}}\frac{1}{\text{t}}\text{t dt}\Big)$
$=\text{e}^{\log\text{t}}\text{t}-\big(\int\text{e}^{\log\text{t}}\text{dt}\big)=\text{I}$
$\rightarrow\text{e}^{\log\text{t}}\text{t}-\text{I}=\text{I}\rightarrow2\text{I}=\text{e}^{\log\text{t}}+\text{C}$
$\text{I}=\frac{1}{2}\Big[\text{te}^{\log\text{t}}\Big]+\text{C}$
Substitute back $\text{t}=\sin\text{x}$ in above expression
We get, $\text{I}=\frac{1}{2}\big[\sin{\text{x}}\text{e}^{\log\sin\text{x}}\big]+\text{C}$
$=\frac{\sin^2\text{x}}{2}+\text{C}$ $[\because\log$ with base 10 term can be changed to in (natural log) term along with a constant$]$
View full question & answer→Question 2685 Marks
Evaluate the following integrals:
$\int\limits^{\pi}_0\text{x}\sin^3\text{x}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{\pi}_0\text{x}\sin^3\text{x}\text{ dx}\ ...(\text{i})$$=\int\limits^{\pi}_0(\pi-\text{x})\sin^3(\pi-\text{x})\text{dx}$
$=\int\limits^{\pi}_0(\pi-\text{x})\sin^3\text{x}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii) we get
$2\text{I}=\int\limits^{\pi}_0(\text{x}+\pi-\text{x})\sin^3\text{x}\text{ dx}$
$=\int\limits^{\pi}_0\pi\sin^3\text{x}\text{ dx}$
$=\int\limits^{\pi}_0\pi\frac{3\sin\text{x}-\sin3\text{x}}{4}\text{ dx}$
$=\frac{\pi}{4}\int\limits^{\pi}_0\big(3\sin\text{x}-\sin3\text{x}\big)\text{dx}$
$=\frac{\pi}{4}\Big[-3\cos\text{x}+\frac{\cos3\text{x}}{3}\Big]^{\pi}_0$
$=\frac{\pi}{4}\Big[-3\cos\pi+3\cos0+\frac{\cos3\pi}{3}-\frac{\cos0}{3}\Big]$
$=\frac{\pi}{4}\Big[3+3+\frac{-1}{3}-\frac{1}{3}\Big]$
$=\frac{\pi}{4}\Big[3-\frac{1}{3}\Big]$
$=\frac{\pi}{2}\times\frac{8}{3}$
$=\frac{4\pi}{3}$
$\therefore\ \text{I}=\frac{2\pi}{3}$
View full question & answer→Question 2695 Marks
Integrate the function in Exercise:
$\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$
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)=2\theta$
$\Rightarrow\ \int\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)\text{dx}=\int2\theta.\sec^2\theta\ \text{d}\theta=2\int\theta.\sec^2\theta\ \text{d}\theta$
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}$
$2\Big[\theta.\int\sec^2\theta\ \text{d}\theta-\int\Big\{\Big(\frac{\text{d}}{\text{dx}}\theta\Big)\int\sec^2\theta\ \text{d}\theta\Big\}\text{d}\theta\Big]$
$=2[\theta\tan\theta-\int\tan\theta\ \text{d}\theta]$
$=2[\theta\tan\theta+\text{log}|\cos\theta|]+\text{C}$
$=2\Bigg[\text{x}\tan^{-1}\text{x}+\text{log}\Bigg|\frac{1}{\sqrt{1+\text{x}^2}}\Bigg|\Bigg]+\text{C}$
$=2\text{x}\tan^{-1}\text{x}+2\text{log}(1+\text{x}^2)^\frac{1}{2}+\text{C}$
$=2\text{x}\tan^{-1}\text{x}+2\Big[-\frac{1}{2}\text{log}(1+\text{x}^2)\Big]+\text{C}$
$=2\text{x}\tan^{-1}\text{x}-\text{log}(1+\text{x}^2)+\text{C}$
View full question & answer→Question 2705 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 2715 Marks
Evaluate the following integrals:
$\int_{0}^\limits{\text{a}}\frac{\text{x}}{\sqrt{\text{a}^2+\text{x}^2}}\text{ dx}$
AnswerLet $\text{a}^2+\text{x}^2=\text{t}^2$
Differentiating w.r.t. x, we get
$2\text{xdx}=2\text{tdt}$
$\text{xdx}=\text{tdt}$
Now, $\text{x}=0\Rightarrow\text{t}=0$
$\text{x}=\text{a}\Rightarrow\text{t}=\sqrt{2}\text{a}$
$\therefore\ \int_{0}^\limits{\text{a}}\frac{\text{x}}{\sqrt{\text{a}^2+\text{x}^2}}\text{ dx}=\int_{\text{a}}^\limits{\sqrt{2\text{a}}}\frac{\text{t dt}}{\text{t}}$
$=\int_{\text{a}}^\limits{\sqrt{2\text{a}}}\text{dt}$
$=\big[\text{t}\big]^{\sqrt{2}\text{a}}_\text{a}$
$=\big[\sqrt{2}\text{a}-\text{a}\big]$
$=\text{a}\big(\sqrt{2}-1\big)$
View full question & answer→Question 2725 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 2735 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 2745 Marks
Evaluate the following intregals:
$\int\frac{1}{\text{x}\log\text{x}(2+\log\text{x})}\text{ dx}$
AnswerLet $\int\frac{1}{\text{x}\log\text{x}(2+\log\text{x})}=\frac{\text{A}}{\text{x}\log\text{x}}+\frac{\text{B}}{\text{x}(2+\log\text{x})}$
$\Rightarrow1=\text{A}(2+\log\text{x})+\text{B}\log\text{x}$
Put $x = 1$
$\Rightarrow1=2\text{A}\Rightarrow\text{A}=\frac{1}{2}$
Put $x = 10^{-2}$
$\Rightarrow1=-2\text{B}\Rightarrow\text{B}=-\frac{1}{2}$
Thus,
$\text{I}=\frac{1}{2}\int\frac{\text{dx}}{\text{x}\log\text{x}}+\Big(-\frac{1}{2}\Big)\int\frac{\text{dx}}{\text{x}(2+\log\text{x})}$
$=\frac{1}{2}\log|\log\text{x}|-\frac{1}{2}\log|2+\log\text{x}|+\text{C}$
$\text{I}=\frac{1}{2}\log\Big|\frac{\log\text{x}}{2+\log\text{x}}\Big|+\text{C}$
View full question & answer→Question 2755 Marks
Evaluate the following integrals:
$\int^\limits{\frac{1}{2}}_{0}\frac{1}{(1+\text{x}^2)\sqrt{1-\text{x}^2}}\text{ dx}$
Answer$\text{I}=\int^\limits{\frac{1}{2}}_{0}\frac{1}{(1+\text{x}^2)\sqrt{1-\text{x}^2}}\text{ dx}$
Let $\text{x}=\sin\text{u}$
$\text{dx}=\cos\text{u du}$
$\text{I}=\int^\limits{\frac{\pi}{6}}_{0}\frac{1}{(1+\sin^2\text{u})}\text{ du}$
$\text{I}=\int^\limits{\frac{\pi}{6}}_{0}\frac{\sec^2\text{u}}{(1+2\tan^2\text{u})}\text{ du}$
Let $\tan\text{u}=\text{v}$
$\text{dv}=\sec^2\text{u du}$
$\text{I}=\int^\limits{\frac{1}{\sqrt{3}}}_{0}\frac{1}{(1+2\text{v}^2)}\text{ dv}$
$\text{I}=\frac{1}{\sqrt{2}}\Big[\tan^{-1}\big(\sqrt{2}\text{v}\big)\Big]^{\frac{1}{\sqrt{3}}}_0$
$\text{I}=\frac{1}{\sqrt{2}}\Big[\tan^{-1}\Big(\sqrt{\frac{2}{3}}\Big)\Big]$
View full question & answer→Question 2765 Marks
Evaluate the following definite integrals:
$\int_{0}^\limits{1}\Big(\text{xe}^{\text{x}}+\cos\frac{\pi\text{x}}{4}\Big)\text{dx}$
AnswerWe have,
$\int_{0}^\limits{1}\Big(\text{xe}^{\text{x}}+\cos\frac{\pi\text{x}}{4}\Big)\text{dx}$
$=\int_{0}^\limits{1}\text{xe}^{\text{x}}\text{ dx}+\int_{0}^\limits{1}\cos\frac{\pi\text{x}}{4}\text{ dx}$
Applying by parts in $1^{st}$ integral we get,
$=\text{x}\int_{0}^\limits{1}\text{e}^{\text{x}}\text{ dx}-\int_{0}^\limits{1}\big(\int\text{e}^{\text{x}}\text{ dx}\big)+\int_{0}^\limits{1}\cos\frac{\pi\text{x}}{4}\text{ dx}$
$=\big[\text{xe}^{\text{x}}\big]^1_0-\int_{0}^\limits{1}\text{e}^{\text{x}}\text{ dx}+\bigg[\frac{\sin\frac{\pi\text{x}}{4}}{\frac{\pi}{4}}\bigg]^1_0$
$=\big[\text{xe}^{\text{x}}-\text{e}^{\text{x}}\big]^1_0+\frac{4}{\pi}\Big[\frac{1}{\sqrt{2}}-0\Big]$
$=\big[\text{e}^{\text{x}}(\text{x}-1)\big]^1_0+\frac{4}{\pi}\Big[\frac{1}{\sqrt{2}}\Big]$
$=0+1+\frac{4}{\pi\sqrt{2}}$
$=1+\frac{2\sqrt{2}}{\pi}$
View full question & answer→Question 2775 Marks
Evaluate the following intregals:
$\int\frac{\cos\text{x}}{\cos3\text{x}}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{\cos\text{x}}{\cos3\text{x}}\ \text{dx}$
$=\int\frac{\cos\text{x}}{(4\cos^3\text{x}-3\cos\text{x})}\ \text{dx}$ $\big[\cos3\text{A}=4\cos^3\text{A}-3\cos\text{A}\big]$
$=\int\frac{1}{4\cos^2\text{x}-3}\ \text{dx}$
Dividing numerator and denominator by $\cos^2\text{x}$
$\Rightarrow\text{I}=\int\frac{\sec^2\text{x}}{4-3\sec^2\text{x}}\ \text{dx}$
$=\int\frac{\sec^2\text{x}}{4-3(1+\tan^2\text{x})}\ \text{dx}$
$=\int\frac{\sec^2\text{x}}{1-3\tan^2\text{x}}\ \text{dx}$
$=\int\frac{\sec^2\text{x}}{1-(\sqrt{3}\tan\text{x})^2}\ \text{dx}$
Let $\sqrt{3}\tan\text{x}=\text{t}$
$\Rightarrow\sqrt{3}\sec^2\text{x dx}=\text{dt}$
$\Rightarrow\sec^2\text{x}\text{ dx}=\frac{\text{dt}}{\sqrt{3}}$
$\therefore\text{I}=\frac{1}{\sqrt{3}}\int\frac{\text{dt}}{1^2-\text{t}^2}$
$=\frac{1}{\sqrt{3}}\times\frac{1}{2}\ln\big|\frac{1+\text{t}}{1-\text{t}}\big|+\text{C}$
$=\frac{1}{2\sqrt{3}}\ln\Big|\frac{1+\sqrt{3}\tan\text{x}}{1-\sqrt{3}\tan\text{x}}\Big|+\text{C}$
View full question & answer→Question 2785 Marks
Evaluvate the following intregals:
$\int\frac{5\cos\text{x}+6}{2\cos\text{x}+\sin\text{x}+3}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{5\cos\text{x}+6}{2\cos\text{x}+\sin\text{x}+3}\ \text{dx}$
Let $(5\cos\text{x}+6)=\lambda\frac{\text{d}}{\text{dx}}(2\cos\text{x}+\sin\text{x}+3)+\mu(2\cos\text{x}+\sin\text{x}+3)+\text{v}$
$(5\cos\text{x}+6)=\lambda(-2\sin\text{x}+\cos\text{x})+\mu(2\cos\text{x}+\sin\text{x}+3)+\text{v}$
$(5\cos\text{x}+6)=(-2\lambda+\mu)\sin\text{x}(\lambda+2\mu)\cos\text{x}+(3\mu+\text{v})$
Comparing the coefficient of $\sin\text{x}\ \&\cos\text{x}$ on the both the sides,
$-2\lambda+\mu=0\dots\dots(1)$
$\lambda+2\mu=5\dots\dots(2)$
$3\mu+\text{v}=6\dots\dots(3)$
Solving equations (1), (2) and (3),
$\text{I}=\int\frac{(-2\sin\text{x}+\cos\text{x})}{(2\cos\text{x}+\sin\text{x}+3)}\text{dx}+2\int\text{dx}$
$\text{I}=\log|2\cos\text{x}+\sin\text{x}+3|+2\text{x}+\text{C}$
View full question & answer→Question 2795 Marks
Evaluate the following definite integrals:
$\int_{0}^\limits{\pi}\text{e}^{2\text{x}}\sin\Big(\frac{\pi}{4}+{\text{x}}\Big)\text{dx}$
AnswerLet $\text{I}=\int_{0}^\limits{\pi}\text{e}^{2\text{x}}\sin\Big(\frac{\pi}{4}+{\text{x}}\Big)\text{dx}$
Integrating by parts, we get
$\text{I}=\frac{1}{2}\Big[\text{e}^{2\text{x}}\sin\Big(\frac{\pi}{4}+{\text{x}}\Big)\text{dx}\Big]^\pi_0-\frac{1}{2}\int_{0}^\limits{\pi}\text{e}^{2\text{x}}\cos\Big(\frac{\pi}{4}+{\text{x}}\Big)\text{dx}$
Now, integrating the second term by parts, we get
$\Rightarrow\text{I}=\frac{1}{2}\Big[\text{e}^{2\text{x}}\sin\Big(\frac{\pi}{4}+{\text{x}}\Big)\text{dx}\Big]^\pi_0-\frac{1}{2}\bigg\{\Big[\frac{1}{2}\text{e}^{2\text{x}}\cos\Big(\frac{\pi}{4}+{\text{x}}\Big)\Big]^{\pi}_0\\+\frac{1}{2}\int_{0}^\limits{\pi}\text{e}^{2\text{x}}\sin\Big(\frac{\pi}{4}+{\text{x}}\Big)\text{dx}\bigg\}$
$\Rightarrow\text{I}=\frac{1}{2}\Big[\text{e}^{2\text{x}}\sin\Big(\frac{\pi}{4}+{\text{x}}\Big)\text{dx}\Big]^\pi_0-\frac{1}{4}\Big[\text{e}^{2\text{x}}\cos\Big(\frac{\pi}{4}+{\text{x}}\Big)\text{dx}\Big]^{\pi}_0-\frac{1}{4}\text{I}$
$\Rightarrow\frac{5}{4}\text{I}=\frac{1}{2}\Big[\text{e}^{2\pi}\sin\Big(\pi+\frac{\pi}{4}\Big)-\sin\Big(\frac{\pi}{4}\Big)\Big]\\-\frac{1}{4}\Big[\text{e}^{2\pi}\cos\Big(\pi+\frac{\pi}{4}\Big)-\cos\Big(\frac{\pi}{4}\Big)\Big]$
$\Rightarrow\frac{5}{4}\text{I}=\frac{1}{2}\Big[-\text{e}^{2\pi}\times\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\Big]-\frac{1}{4}\Big[-\text{e}^{2\pi}\times\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\Big]$
$\Rightarrow\frac{5}{4}\text{I}=-\frac{1}{2\sqrt{2}}\text{e}^{2\pi}-\frac{1}{2\sqrt{2}}+\frac{1}{4\sqrt{2}}\text{e}^{2\pi}+\frac{1}{4\sqrt{2}}$
$\Rightarrow\text{I}=-\frac{1}{5\sqrt{2}}\big(\text{e}^{2\pi}+1\big)$
View full question & answer→Question 2805 Marks
Evaluate the following intregals:
$\int\frac{\text{x}}{\sqrt{\text{x}^2+6\text{x}+10}}\ \text{dx}$
AnswerLet $\text{I}\int\frac{\text{x}}{\sqrt{\text{x}^2+6\text{x}+10}}\ \text{dx}$
Let $\text{x}=\lambda\frac{\text{d}}{\text{dx}}(\text{x}^2+6\text{x}+10)+\mu$
$=\lambda(2\text{x}+6)+\mu$
$\text{x}=(2\lambda)\text{x}+6\lambda+\mu$
Comparing the co-efficient of like powers of x.
$2\lambda=1\Rightarrow\lambda=\frac{1}{2}$
$6\lambda+\mu=0\Rightarrow6\Big(\frac{1}{2}\Big)+\mu=0$
$\mu=-3$
So, $\text{I}_1=\int\frac{\frac{1}{2}(2\text{x}+6)=3}{\sqrt{\text{x}^2+6\text{x}+10}}\ \text{dx}$
$\frac{1}{2}\int\frac{{2\text{x}+6}=3}{\sqrt{\text{x}^2+6\text{x}+10}}\ \text{dx}-3\text{ I }\frac{1}{\sqrt{\text{x}^2+2\text{x}(3)+(3)^2+10}}$
$\text{I}_1=\frac{1}{2}\int\frac{2\text{x}+6}{\sqrt{\text{x}^2+6\text{x}+10}}\ \text{dx}-3\int\frac{1}{\sqrt{(\text{x}+3)^2+(1)^2}}\text{dx}$
$\text{I}=\sqrt{\text{x}^2+6\text{x}+10}-3\log\big|\text{x}+3+\sqrt{\text{x}^2+6\text{x}+10}\big|+\text{c}$
$\text{I}_1=\frac{1}{2}(2\sqrt{\text{x}^2+6\text{x}+10})-3\log\big|\text{x}+3+\sqrt{(\text{x}+3)^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+\text{a}^2}}\text{dx}-\log\big|\text{x}+\sqrt{\text{x}}^2+\text{a}^2\big|+\text{c}\Big]$
$\text{I}=\sqrt{\text{x}^2+6\text{x}+10}-3\log\big|\text{x}+3+\sqrt{\text{x}^2+6\text{x}+10}\big|+\text{c}$
View full question & answer→Question 2815 Marks
Integrate the function in Exercise:$\frac{1}{\big(\text{x}^{2}+1\big)\big(\text{x}^{2}+4\big)}$
Answer$\therefore\frac{1}{\big(\text{x}^{2}+1\big)\big(\text{x}^{2}+4\big)}=\frac{\text{Ax}+\text{B}}{\big(\text{x}^{2}+1\big)}+\frac{\text{Cx}+\text{D}}{\big(\text{x}^{2}+4\big)}$
$\Rightarrow1=\big(\text{Ax}+\text{B}\big)\big(\text{x}^{2}+4\big)+\big(\text{Cx}+\text{D}\big)\big(\text{x}^{2}+1\big)$
$\Rightarrow1=\text{Ax}^{3}+4\text{Ax}+\text{Bx}^{2}+4\text{B}+\text{Cx}^{3}+\text{Cx}+\text{Dx}^{2}+\text{D}$
Equating the coefficients of $\text{x}^{3},\text{x}^{2}$ and constant term, we obtain
$\text{A}+\text{C}=0$
$\text{B}+\text{D}=0$
$4\text{A}+\text{C}=0$
$4\text{B}+\text{D}=1$
On solving these equations, we obtain
$\text{A}=0,\text{B}=\frac{1}{3},\text{C}=0,$and $\text{D}=-\frac{1}{3}$
From equation (1), we obtain
$\frac{1}{\big(\text{x}^{2}+1\big)\big(\text{x}^{2}+4\big)}=\frac{1}{3\big(\text{x}^{2}+1\big)}-\frac{1}{3\big(\text{x}^{2}+4\big)}$
$\int\frac{1}{\big(\text{x}^{2}+4\big)\big(\text{x}^{2}+4\big)}\text{dx}=\frac{1}{3}\int\frac{1}{\text{x}^{2}+1}\text{dx}-\frac{1}{3}\int\frac{1}{\text{x}^{2}+4}\text{dx}$
$=\frac{1}{3}\tan^{-1}\text{x}-\frac{1}{3}.\frac{1}{2}\tan^{-1}\frac{\text{x}}{2}+\text{C}$
$=\frac{1}{3}\tan^{-1}\text{x}-\frac{1}{6}\tan^{-1}\frac{\text{x}}{2}+\text{C}$
View full question & answer→Question 2825 Marks
If f is an integrable function such that f(2a - x) = f(x), then prove that:
$\int\limits^{2\text{a}}_0\text{f(x)}\text{dx}=2\int\limits^{\text{a}}_0\text{f(x)}\text{dx}$
AnswerLet $\text{I}=\int\limits^{2\text{a}}_0\text{f(x)}\text{dx}$
By Additive property
$\text{I}=\int\limits^{\text{a}}_0\text{f(x)}\text{dx}+\int\limits^{2\text{a}}_0\text{f(x)}\text{dx}$
Consider the integral $\int\limits^{2\text{a}}_0\text{f(x)}\text{dx}$
Let $\text{x}=2\text{a}-\text{t},$ then $\text{dx}=-\text{dt}$
When $\text{x}=\text{a},\text{ t}=\text{a},\text{ x}=2\text{x},\text{t}=0$
Hence $\int\limits^{2\text{a}}_0\text{f(x)}\text{dx}=-\int\limits^{0}_\text{a}\text{f(2a}-\text{t})\text{dt}$
$=\int\limits_{0}^\text{a}\text{f(2a}-\text{t})\text{dt}$
$=\int\limits_{0}^\text{a}\text{f(2a}-\text{x})\text{dx}$ (Changeing the varible)
Therefore,
$\text{I}=\int\limits_{0}^\text{a}\text{f(x})\text{dx}+\int\limits_{0}^\text{a}\text{f(2a}-\text{x})\text{dx}$
$=\int\limits_{0}^\text{a}\text{f(x})\text{dx}+\int\limits_{0}^\text{a}\text{f(x})\text{dx}$ $\Bigg[\text{Given}\int\limits_{0}^\text{a}\text{f(x})\text{dx}+\int\limits_{0}^\text{a}\text{f(2a}-\text{x})\text{dx}\Bigg]$
$=2\int\limits_{0}^\text{a}\text{f(x})\text{dx}$
View full question & answer→Question 2835 Marks
Evaluate the following integrals:
$\int_{0}^\limits{\text{a}}\sqrt{\text{a}^2-\text{x}^2}\text{ dx}$
AnswerLet $\text{x}=\text{a}\sin\theta$
Differentiating w.r.t. x, we get
$\text{dx}=\text{a}\cos\theta\text{ d}\theta$
Now, $\text{x}=0\Rightarrow\theta=0$
$\text{x}=\text{a}\Rightarrow\theta=\frac{\pi}{2}$
$\therefore\ \int_{0}^\limits{\text{a}}\sqrt{\text{a}^2-\text{x}^2}\text{ dx}$
$=\int_{0}^\limits{\frac{\pi}{2}}\sqrt{\text{a}^2(1-\sin^2\theta)}\text{a}\cos\theta\text{ d}\theta$
$=\text{a}^2\int_{0}^\limits{\frac{\pi}{2}}\cos^2\theta\text{ d}\theta$ $\Big[\because(1-\sin^2\theta)=\cos^2\theta\text{ and }\frac{1+\cos2\theta}{2}=\cos2\theta\Big]$
$=\frac{\text{a}^2}{2}\int_{0}^\limits{\frac{\pi}{2}}\big(1+\cos2\theta\big)\text{d}\theta$
$=\frac{\text{a}^2}{2}\Big[\frac{\pi}{2}+0-0-0\big]$
$=\frac{\pi\text{a}^2}{4}$
View full question & answer→Question 2845 Marks
Evaluate the following integrals:
$\int\limits^{\pi}_0\frac{\text{x}\tan\text{x}}{\sec\text{x}\text{ cosecx}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{\pi}_0\frac{\text{x}\tan\text{x}}{\sec\text{x}\text{ cosecx}}\text{ dx}\ ....(\text{i})$
$=\int\limits^{\pi}_0\frac{(\pi-\text{x})\tan\text{x}}{\sec(\pi-\text{x})\text{ cosec}(\pi-\text{x})}\text{ dx}$ $\Bigg[\text{Using}\ \int\limits^{\text{a}}_{0}\text{f(x)}\text{dx}=\int\limits^{\text{a}}_{\text{0}}\text{f}(\text{a}-\text{x})\text{dx}\Bigg]$
$=\int\limits^{\pi}_0\frac{-(\pi-\text{x})\tan\text{x}}{-\sec\text{x}\text{ cosec}\text{x}}\text{ dx}$
$=\int\limits^{\pi}_0\frac{(\pi-\text{x})\tan\text{x}}{\sec\text{x}\text{ cosecx}}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii)
$2\text{I}=\int\limits^{\pi}_0\frac{\text{x}\tan\text{x}}{\sec\text{x}\text{ cosecx}}+\frac{(\pi-\text{x})\tan\text{x}}{\sec\text{x}\text{ cosecx}}\text{ dx}$
$=\int\limits^{\pi}_0(\text{x}+\pi-\text{x})\frac{\tan\text{x}}{\sec\text{x}\text{ cosecx}}\text{ dx}$
$=\int\limits^{\pi}_0\frac{\pi\tan\text{x}}{\sec\text{x}\text{ cosecx}}\text{ dx}$
$=\int\limits^{\pi}_0\pi\sin^2\text{x dx}$
$=\pi\int\limits^{\pi}_0\big(1-\cos^2\text{x}\big)\text{dx}$
$=\pi\big[\text{x}\big]^{\pi}_0-\frac{\pi}{2}\int\limits^{\pi}_0\big(1-\cos^2\text{x}\big)\text{dx}$
$=\frac{\pi}{2}\big[\text{x}\big]^{\pi}_0-\frac{\pi}{2}\Big[\frac{\sin2\text{x}}{2}\Big]^{\pi}_0$
$=\frac{\pi^2}{4}$
Hence, $\text{I}=\frac{\pi^2}{4}$
View full question & answer→Question 2855 Marks
Evaluate the following definite integrals:
$\int_{0}^\limits{\frac{\pi}{2}}\cos^4\text{x}\text{ dx}$
AnswerWe have,
$\int_{0}^\limits{\frac{\pi}{2}}\cos^4\text{x}\text{ dx}$
$=\frac{1}{4}\int_{0}^\limits{\frac{\pi}{2}}(1+\cos2\text{x})^2\text{dx}$ $\big[\because2\cos^2\text{x}=1+\cos2\text{x}\big]$
$=\frac{1}{4}\int_{0}^\limits{\frac{\pi}{2}}\big(1+\cos^22\text{x}+2\cos2\text{x}\big)\text{dx}$
$=\frac{1}{4}\int_{0}^\limits{\frac{\pi}{2}}\Big(1+\frac{1+\cos4\text{x}}{2}+2\cos2\text{x}\Big)\text{dx}$
$=\frac{1}{4}\Big[\text{x}+\frac{1}{2}\text{x}+\frac{\sin4\text{x}}{8}+\sin2\text{x}\Big]^{\frac{\pi}{2}}_0$ $\Big[\because\int\cos4\text{x dx}=\frac{\sin4\text{x}}{4}\Big]$
$=\frac{1}{4}\Big[\frac{\pi}{2}+\frac{\pi}{4}+0+0-0-0-0-0\Big]$
$=\frac{1}{4}\times\frac{3\pi}{4}$
$=\frac{3\pi}{16}$
View full question & answer→Question 2865 Marks
Evaluate the following integrals:
$\int_{0}^\limits{1}\text{x}\tan^{-1}\text{x}\text{ dx} $
AnswerWe have,
$\int_{0}^\limits{1}\text{x}\tan^{-1}\text{x}\text{ dx}=\tan^{-1}\text{x}\int_{0}^\limits{1}\text{x dx}-\int_{0}^\limits{1}\big(\int\text{dx}\big)\frac{\text{d}}{\text{dx}}\big(\tan^{-1}\text{x}\big)\text{dx}$
$=\Big[\frac{\text{x}^2}{2}\tan^{-1}\text{x}\Big]^1_0-\frac{1}{2}\int_{0}^\limits{1}\frac{\text{x}^2}{1+\text{x}^2}\text{ dx}$
$=\Big[\frac{\text{x}^2}{2}\tan^{-1}\text{x}\Big]^1_0-\frac{1}{2}\int_{0}^\limits{1}\frac{1+\text{x}^2-1}{1+\text{x}^2}\text{ dx}$
$=\frac{1}{2}\Big(\frac{\pi}{4}\Big)-\frac{1}{2}\Bigg[\int_{0}^\limits{1}\text{dx}-\int_{0}^\limits{1}\frac{\text{dx}}{1+\text{x}^2}\Bigg]$
$=\frac{\pi}{8}-\frac{1}{2}\big[\text{x}-\tan^{-1}\text{x}\big]^{1}_0$
$=\frac{\pi}{8}-\frac{1}{2}\Big[1-\frac{\pi}{4}\Big]$
$=\frac{\pi}{8}-\frac{1}{2}+\frac{\pi}{8}$
$=\frac{\pi}{4}-\frac{1}{2}$
View full question & answer→Question 2875 Marks
Evaluate the following integrals:
$\int_{0}^\limits{\frac{\pi}{2}}\frac{\sin\text{x}\cos\text{x}}{1+\sin^4\text{x}}\text{ dx}$
AnswerLet $\sin^2\text{x}=\text{t}$
Differentiating w.r.t. x, we get
$2\sin\text{x}\cos\text{x dx}=\text{dt}$
Now, $\text{x}=0\Rightarrow\text{t}=0$
$\text{x}=\frac{\pi}{2}\Rightarrow\text{t}=1$
$\int_{0}^\limits{\frac{\pi}{2}}\frac{\sin\text{x}\cos\text{x}}{1+\sin^4\text{x}}\text{ dx}$
$=\frac{1}{2}\int\limits^1_0\frac{\text{dt}}{1+\text{t}^2}$
$=\frac{1}{2}\big[\tan^{-1}\text{t}\big]^1_0$
$=\frac{1}{2}\Big[\tan^{-1}(1)-\tan^{-1}(0)\Big]$
$=\frac{1}{2}\Big[\tan^{-1}\Big(\tan\frac{\pi}{4}\Big)-\tan^{-1}(\tan0)\Big]$
$=\frac{1}{2}\times\frac{\pi}{4}$
$=\frac{\pi}{8}$
View full question & answer→Question 2885 Marks
Evaluate the following integrals:
$\int^\limits1_{-1}|2\text{x}+1|\text{dx}$
AnswerWe know that,
$\int^\limits1_{-1}|2\text{x}+1|\text{dx}$
$=\int^\limits\frac{1}{2}_{-1}-(2\text{x}+1)\text{dx}+\int\limits_{-\frac{1}{2}}^{1}(2\text{x}+1)\text{dx}$
$=-\Big[\frac{2\text{x}^2}{2}+\text{x}\Big]^{-\frac{1}{2}}_{-1}+\Big[\frac{2\text{x}^2}{2}+\text{x}\Big]^{1}_{-\frac{1}{2}}$
$=-\bigg[\Big(\frac{2}{8}-\frac{1}{2}\Big)-\Big(\frac{2}{2}-1\Big)\bigg]+\bigg[\Big(\frac{2}{2}+1\Big)-\Big(\frac{2}{8}-\frac{1}{2}\Big)\bigg]$
$=-\bigg[\Big(\frac{1}{4}-\frac{1}{2}\Big)-(1-1)\bigg]+\bigg[(1+1)+\Big(\frac{1}{4}-\frac{1}{2}\Big)\bigg]$
$=-\Big[-\frac{1}{4}\Big]++\Big[2+\frac{1}{4}\Big]$
$=\frac{1}{4}+2+\frac{1}{4}$
$=2\frac{1}{2}$
View full question & answer→Question 2895 Marks
Evaluate the following integrals:
$\int\limits^{\text{a}}_0\frac{1}{\text{x}+\sqrt{\text{a}^2-\text{x}^2}}\text{ dx}$
AnswerWe have,
$\text{I}=\int\limits^{\text{a}}_0\frac{1}{\text{x}+\sqrt{\text{a}^2-\text{x}^2}}\text{ dx}$
Putting $\text{x}=\text{a}\sin\theta$
$\text{dx}=\text{a}\cos\theta\text{ d}\theta$
When $\text{x}\rightarrow0;\theta\rightarrow0$
And $\text{x}\rightarrow\text{a};\theta\rightarrow\frac{\pi}{2}$
$\therefore\ \text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\text{a}\cos\theta}{\text{a}\sin\theta+\sqrt{\text{a}^2-(\text{a}\sin\theta)^2}}\text{ d}\theta$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\text{a}\cos\theta}{\text{a}\sin\theta+\text{a}\cos\theta}\text{ d}\theta$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\cos\theta}{\sin\theta+\cos\theta}\text{ dx}\ ....(\text{i})$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\cos\big(\frac{\pi}{2}-\theta\big)}{\sin\big(\frac{\pi}{2}-\theta\big)+\cos\big(\frac{\pi}{2}-\theta\big)}\text{ d}\theta$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\sin\theta}{\cos\theta+\sin\theta}\text{ d}\theta$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\sin\theta}{\sin\theta+\cos\theta}\text{ d}\theta\ ....(\text{ii})$
By adding (i) and (ii) we get
$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\cos\theta+\sin\theta}{\sin\theta+\cos\theta}\text{ d}\theta$
$\Rightarrow2\text{I}=\int\limits^{\frac{\pi}{2}}_0\text{d}\theta$
$\Rightarrow2\text{I}=\Big[\theta\Big]^{\frac{\pi}{2}}_0$
$\Rightarrow2\text{I}=\frac{\pi}{2}$
$\Rightarrow\text{I}=\frac{\pi}{4}$
View full question & answer→Question 2905 Marks
Evaluate the following integrals:
$\int^\limits{\frac{\pi}{2}}_{0}\frac{\tan\text{x}}{1+\text{m}^2\tan^2\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int^\limits{\frac{\pi}{2}}_{0}\frac{\tan\text{x}}{1+\text{m}^2\tan^2\text{x}}\text{ dx}$
$=\int^\limits{\frac{\pi}{2}}_{0}\frac{\frac{\sin\text{x}}{\cos\text{x}}}{1+\text{m}^2\frac{\sin^2\text{x}}{\cos^2\text{x}}}\text{ dx}=\int^\limits{\frac{\pi}{2}}_{0}\frac{{\sin\text{x}}{\cos\text{x}}}{\cos^2\text{x}+\text{m}^2{\sin^2\text{x}}}\text{ dx}$
Put $\cos^2\text{x}+\text{m}^2\sin^2\text{x}=\text{z}$
$\therefore\ 2\cos\text{x}(-\sin\text{x})\text{dx}+\text{m}^2\times2\sin\text{x}\cos\text{x dx}=\text{ dz}$
$\Rightarrow2(\text{m}^2-1)\sin\text{x}\cos\text{x dx}=\text{dz}$
$\Rightarrow\sin\text{x}\cos\text{x dx}=\frac{\text{dz}}{2(\text{m}^2-1)}$
When $\text{x}\rightarrow0,\text{ z}\rightarrow1$ $\big(\text{z}=\cos^2\text{x}+\text{m}^2\sin^2\text{x}=1+\text{m}^2\times0=1\big)$
When $\text{x}\rightarrow\frac{\pi}{2},\text{ z}\rightarrow\text{m} ^2$ $\big(\text{z}=\cos^2\text{x}+\text{m}^2\sin^2\text{x}=0+\text{m}^2\times0=\text{m}^2\big)$
$\therefore\ \text{I}=-\frac{1}{2(\text{m}^2-1)}\int\limits^{\text{m}^2}_1\frac{\text{dz}}{\text{z}}$
$=\frac{1}{2(\text{m}^2-1)}\big[\log\text{z}\big]^{\text{m}^2}_1$
$=\frac{1}{2(\text{m}^2-1)}\big(\log\text{m}^2-\log1\big)$
$=\frac{1}{2(\text{m}^2-1)}\big(2\log|\text{m}|-0\big)$
$=\frac{\log|\text{m}|}{\text{m}^2-1}$
View full question & answer→Question 2915 Marks
Evaluate the following integrals as limit of sum:
$\int\limits^{3}_{2}\text{x}^2\text{ dx}$
Answer$\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\text{f}(\text{a}+2\text{h})\ +\\ ....\ +\text{f}(\text{a}+(\text{n}-1)\text{h})\Big]$
Where, $\text{h}=\frac{\text{b}-\text{a}}{\text{n}}$
Here, $\text{a}=2,\text{ b}=3,\text{ f(x)}=\text{x}^2,\text{ h}=\frac{3-2}{\text{n}}=\frac{1}{\text{n}}$
Therefore, $\text{I}=\int\limits^{3}_{2}\text{x}^2\text{ dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{f}(2)+\text{f}(2+\text{h})\ ....\ +\text{f}\big\{2+(\text{n}-1)\text{h}\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[2^2+(2+\text{h})^2+\ ....+\ \big\{2(\text{n}-1)\text{h}\big\}^2\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[4\text{n}+\text{h}\big\{1^2+2^2+3^2+\ ....\ +(\text{n}-1)^2+4\text{h}\big\}\\+4\text{h}\big\{1+2+\ ....+\ (\text{n}-1)\text{h}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[4\text{n}+\text{h}^2\frac{\text{n}(\text{n}-1)(\text{n}-2)}{6}+4\text{h}\frac{\text{n}(\text{n}-1)}{2}\Big]$
$=\lim\limits_{\text{h}\rightarrow\infty}\frac{1}{\text{n}}\Big[4\text{n}+\frac{(\text{n}-1)(2\text{n}-1)}{6\text{n}}+2\text{n}-2\Big]$
$=\lim\limits_{\text{h}\rightarrow\infty}5\Big[6+\frac{1}{6}\Big(1-\frac{1}{\text{n}}\Big)\Big(2-\frac{1}{\text{n}}\Big)-\frac{2}{\text{n}}\Big]$
$=6+\frac{1}{3}$
$=\frac{19}{3}$
View full question & answer→Question 2925 Marks
Evaluate the following integrals:
$\int\limits^{1}_0\frac{\log(1+\text{x})}{1+\text{x}^2}\text{ dx}$
AnswerWe have,
$\text{I}=\int\limits^{1}_0\frac{\log(1+\text{x})}{1+\text{x}^2}\text{ dx}$
Putting $\text{x}=\tan\theta$
$\text{dx}=\sec^2\theta\text{ d}\theta$
When $\text{x}\rightarrow0;\theta\rightarrow0$
And $\text{x}\rightarrow1;\theta\rightarrow\frac{\pi}{4}$
Now, integral becomes,
$\text{I}=\int\limits^{\frac{\pi}{4}}_0\frac{\log(1+\tan\theta)}{\sec^2\theta}\sec^2\theta\text{ d}\theta$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{4}}_0\big[\log(\tan\theta)\big]\text{d}\theta\ ...(\text{i})$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{4}}_0\bigg[\log\Big\{1+\tan\Big(\frac{\pi}{4}-\theta\Big)\Big\}\bigg]\text{d}\theta$ $\Bigg[\because\ \int\limits^{\text{a}}_{0}\text{f(x)}\text{dx}=\int\limits^{\text{a}}_{\text{0}}\text{f}(\text{a}-\text{x})\text{dx}\Bigg]$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{4}}_0\Bigg[\log\bigg\{1+\frac{\tan\frac{\pi}{4}-\tan\theta}{1+\tan\frac{\pi}{4}\tan\theta}\bigg\}\Bigg]\text{d}\theta$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{4}}_0\bigg[\log\Big\{1+\frac{1-\tan\theta}{1+\tan\theta}\Big\}\bigg]\text{d}\theta$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{4}}_0\bigg[\log\Big\{\frac{2}{1+\tan\theta}\Big\}\bigg]\text{d}\theta$
$\Rightarrow\text{I}=\int\limits^{\frac{\pi}{4}}_0\big[\log2-\log(1+\tan\theta)\big]\text{d}\theta\ ...(\text{ii})$
Adding (i) and (ii) we get
$2\text{I}=\int\limits^{\frac{\pi}{4}}_0\big(\log2\big)\text{d}\theta$
$\Rightarrow2\text{I}=\big(\log2\big)\Big[\theta\Big]^{\frac{\pi}{4}}_0$
$\Rightarrow2\text{I}=\frac{\pi}{4}\log2$
$\Rightarrow\text{I}=\frac{\pi}{8}\log2$
$\therefore\ \int\limits^{1}_0\frac{\log(1+\text{x})}{1+\text{x}^2}\text{ dx}=\frac{\pi}{8}\log2$
View full question & answer→Question 2935 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 2945 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 2955 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 2965 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{3}}_{-\frac{\pi}{3}}\frac{1}{1+\text{e}^{\tan\text{x}}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{\frac{\pi}{3}}_{-\frac{\pi}{3}}\frac{1}{1+\text{e}^{\tan\text{x}}}\text{ dx}\ ...(\text{i})$
Then,
$\text{I}=\int\limits^{\frac{\pi}{3}}_{-\frac{\pi}{3}}\frac{1}{1+\text{e}^{\tan\big[\frac{\pi}{3}+\big(-\frac{\pi}{3}\big)-\text{x}\big]}}\text{ dx}$ $\Bigg[\int\limits^{\text{a}}_{0}\text{f(x)}\text{dx}=\int\limits^{\text{a}}_{\text{0}}\text{f}(\text{a}-\text{x})\text{dx}\Bigg]$
$=\int\limits^{\frac{\pi}{3}}_{-\frac{\pi}{3}}\frac{1}{1+\text{e}^{\tan(-\text{x})}}\text{ dx}$
$=\int\limits^{\frac{\pi}{3}}_{-\frac{\pi}{3}}\frac{1}{1+\text{e}^{-\tan\text{x}}}\text{ dx}$
$=\int\limits^{\frac{\pi}{3}}_{-\frac{\pi}{3}}\frac{\text{e}^{\tan\text{x}}}{\text{e}^{\tan\text{x}}+1}\text{ dx}\ ...{\text{(ii)}}$
Adding (i) and (ii) we get
$2\text{I}=\int\limits^{\frac{\pi}{3}}_{-\frac{\pi}{3}}\frac{\text{e}^{1+\tan\text{x}}}{\text{e}^{1+\tan\text{x}}}\text{ dx}$
$\Rightarrow2\text{I}=\int\limits^{\frac{\pi}{3}}_{-\frac{\pi}{3}}\text{dx}$
$\Rightarrow2\text{I}=\big[\text{x}\big]^{\frac{\pi}{3}}_{-\frac{\pi}{3}}$
$\Rightarrow2\text{I}=\frac{\pi}{3}-\Big(-\frac{\pi}{3}\Big)$
$\Rightarrow2\text{I}=\frac{2\pi}{3}$
$\Rightarrow\text{I}=\frac{\pi}{3}$
View full question & answer→Question 2975 Marks
Evaluate the following integrals:
$\int^\limits{\frac{\pi}{2}}_{0}\frac{\sin\text{x}\cos\text{x}}{\cos^2\text{x}+3\cos\text{x}+2}\text{ dx}$
AnswerLet $\text{I}=\int^\limits{\frac{\pi}{2}}_{0}\frac{\sin\text{x}\cos\text{x}}{\cos^2\text{x}+3\cos\text{x}+2}\text{ dx}$ Then,
Let $\cos\text{x}=\text{t},$ Then, $-\sin\text{x dx}=\text{dt}$
When, $\text{x}=0,\text{ t}=1$ and $\text{x}=\frac{\pi}{2},\text{ t}=0$
$\therefore\ \text{I}=-\int^\limits0_1\frac{\text{t dt}}{\text{t}^2+3\text{t}+2}$
$\Rightarrow\text{I}=\int\limits^0_1\frac{-\text{t dt}}{\text{t}^3+3\text{t}+2}$
$\Rightarrow\text{I}=\int\limits^0_1\Big(\frac{1}{(\text{t}+1)}-\frac{2}{(\text{t}+2)}\Big)\text{dt}$
$\Rightarrow\text{I}=\Big[\log(\text{t}+1)-2\log(\text{t}+2)\Big]^0_1$
$\Rightarrow\text{I}=\bigg[\log\frac{(\text{t}+1)}{(\text{t}+2)^2}\bigg]^1_0$
$\Rightarrow\text{I}=\bigg[\log\Big(\frac{1}{4}\Big)-\log\Big(\frac{2}{9}\Big)\bigg]^1_0$
$\Rightarrow\text{I}=\log\frac{9}{8}$
View full question & answer→Question 2985 Marks
Evaluate the following integrals:
$\int\frac{\text{e}^{\text{3x}}}{4\text{e}^{6\text{x}}-9}\text{dx}$
AnswerLet $\text{I}=\int\frac{\text{e}^{3\text{x}}}{4\text{e}^{6\text{x}}-9}\text{dx}$
Let $\text{e}^{3\text{x}}=\text{t}$
$\Rightarrow3\text{e}^{3\text{x}}\text{ dx = dt}$
$\Rightarrow\text{e}^{3\text{x}} \text{dx}=\frac{\text{dt}}{3}$
$\text{I}=\frac{1}3{}\int\frac{\text{dt}}{4\text{t}^2-9}$
$=\frac{1}{12}\int\frac{\text{dt}}{\text{t}^2-\frac{9}{4}}$
$=\frac{1}{12}\int\frac{\text{dt}}{\text{t}^2-\big(\frac{3}{2}\big)^2}$
$=\frac{1}{12}\times\frac{1}{2\big(\frac{3}{2}\big)}\log\Bigg|\frac{\text{t}-\frac{3}{2}}{\text{t}+\frac{3}{2}}\Bigg|+\text{C}$ $\bigg[\text{Since,}\int\frac{1}{\text{x}^2+\text{a}^2}\text{dx}=\frac{1}{2\text{a}}\bigg|\frac{\text{x}-\text{a}}{\text{x+a}}\bigg|+\text{C}\bigg]$
$\text{I}=\frac{1}{36}\log\bigg|\frac{2\text{t}-3}{2\text{t}+3}\bigg|+\text{C}$
$\text{I}=\frac{1}{36}\log\bigg|\frac{2\text{e}^{3\text{x}}-3}{2\text{e}^{3\text{x}}+3}\bigg|+\text{C}$
View full question & answer→Question 2995 Marks
Evaluate the following integrals:$\int\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)\text{dx}$
AnswerLet $\text{I}=-\int\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)\text{dx}$
Let $\text{x}=\tan\theta$
$\text{dx}=\sec^2\theta\text{d}\theta$
$\text{I}=\int\tan^{-1}\Big(\frac{2\tan\theta}{1-\tan^2\theta}\Big)\sec^2\theta\text{d}\theta$
$=\int\tan^{-1}(\tan2\theta)\sec^2\theta\text{d}\theta$
$=\int2\theta\sec^2\theta\text{d}\theta$
$=2\Big[\theta\int\sec^2\theta\text{d}\theta-\int(1\int\sec^2\theta\text{d}\theta)\text{d}\theta\Big]$
$=2\big[\theta\tan\theta-\int\tan\theta\text{d}\theta\big]$
$=2\big[\theta\tan\theta-\log\sec\theta\big]+\text{c}$
$=2\Big[\text{x}\tan^{-1}\text{x}-\log\sqrt{1+\text{x}^2}\Big]+\text{C}$
$\text{I}=2\text{x}\tan^{-1}\text{x}-\log\big|1+\text{x}^2\big|+\text{C}$
View full question & answer→Question 3005 Marks
Evaluate the following integrals:
$\int_{0}^\limits{1}\frac{24\text{x}^3}{(1+\text{x}^2)^4}\text{ dx}$
AnswerLet $1+\text{x}^2=\text{t}$
Differentiating w.r.t. x, we get
$2\text{xdx}=\text{dt}$
Now, $\text{x}=0\Rightarrow\text{t}=1$
$\text{x}=1\Rightarrow\text{t}=2$
$\int_{0}^\limits{1}\frac{24\text{x}^3}{(1+\text{x}^2)^4}\text{ dx}=\int_{1}^\limits{2}\frac{12(\text{t}-1)}{\text{t}^4}\text{ dt}$
$=12\int_{1}^\limits{2}\Big(\frac{1}{\text{t}^3}-\frac{1}{\text{t}^4}\Big)\text{dt}$
$=12\Big[-\frac{1}{2\text{t}^2}-\frac{1}{3\text{t}^3}\Big]^2_1$
$=12\Big[-\frac{1}{8}+\frac{1}{24}+\frac{1}{2}-\frac{1}{3}\Big]$
$=12\Big[\frac{-3+1+12-8}{24}\Big]$
$=\frac{12\times2}{24}=1$
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