Question 2513 Marks
Evaluate the following integrals:
$\int\cot^{\text{n}}\text{cosec}^2\text{x}\text{ dx},\text{ n}\neq-1$
Answer$\int\cot^{\text{n}}\text{cosec}^2\text{x}\text{ dx},\text{ n}\neq-1$
Let $\cot\text{x}=\text{t}$
$-\text{cosec}^2\text{x}\text{ dx}=\text{dt}$
$\text{cosec}^2\text{x}\text{ dx}=-\text{dt}$
Now, $\int\cot^{\text{n}}\text{x }\text{cosec}^2\text{x}\text{ dx}$
$=-\int\text{t}^{\text{n}}\text{dt}$
$=\frac{-\text{t}^{\text{n}+1}}{\text{n}+1}+\text{C}$
$=-\frac{\cot^{\text{n}+1}}{\text{n}+1}+\text{C}$
View full question & answer→Question 2523 Marks
Evaluate the following integrals:
$\int^\limits{2}_{1}|\text{x}-3|\text{dx}$
Answer$\int^\limits{2}_{1}|\text{x}-3|\text{dx}$
We know that,
$|\text{x}+1|=\begin{cases}-(\text{x}+1),&1\leq\text{x}\leq3\$\text{x}+1),&\text{x}>3\end{cases}$
$\therefore\ \text{I}=\int^\limits{2}_{1}|\text{x}-3|\text{dx}$
$\Rightarrow\text{I}=\int^\limits{2}_{1}-(\text{x}-3)\text{dx}$
$\Rightarrow\text{I}=\Big[\frac{-\text{x}^2}{2}-3\text{x}\Big]^2_1$
$\Rightarrow\text{I}=-2-6+\frac{1}{2}+3$
$\Rightarrow\text{I}=\frac{3}{2}$
View full question & answer→Question 2533 Marks
Evaluate the following integrals:$\int\frac{\cos\text{x}}{\sqrt{\sin^2\text{x}-2\sin\text{x}-3}}\text{ dx}$
Answer$\int\frac{\cos\text{x}\text{ dx}}{\sqrt{\sin^2\text{x}-2\sin\text{x}-3}}$
Let $\sin\text{x}=\text{t}$
$\cos\text{x}\text{ dx}=\text{dt}$
Now, $\int\frac{\cos\text{x}\text{ dx}}{\sqrt{\sin^2\text{x}-2\sin\text{x}-3}}$
$=\int\frac{\text{dt}}{\sqrt{\text{t}^2-2\text{t}-3}}$
$=\int\frac{\text{dt}}{\sqrt{\text{t}^2-2\text{t}+1-1-3}}$
$=\int\frac{\text{dt}}{\sqrt{(\text{t}-1)^2-2^2}}$
$=\log\Big|\text{t}-1+\sqrt{(\text{t}-1)^2-2^2}\Big|+\text{C}$
$=\log\Big|\text{t}-1+\sqrt{\text{t}^2-2\text{t}-3}\Big|+\text{C}$
$=\log\Big|\sin\text{x}-1+\sqrt{\sin^2\text{x}-2\sin\text{x}-3}\Big|+\text{C}$
View full question & answer→Question 2543 Marks
Evaluate the following definite integrals:
$\int_{0}^\limits{\frac{\pi}{2}}\text{x}^2\cos2\text{x}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\text{x}^2\cos2\text{x}\text{ dx}$ Then,
Integrating by parts
$\text{I}=\Big[\text{x}^2\frac{\sin2\text{x}}{2}\Big]^{\frac{\pi}{2}}_0-\int_{0}^\limits{\frac{\pi}{2}}2\text{x}\frac{\sin2\text{x}}{2}\text{ dx}$
$\Rightarrow\text{I}=\Big[\text{x}^2\frac{\sin2\text{x}}{2}\Big]^{\frac{\pi}{2}}_0-\Big[-\text{x}\frac{\cos2\text{x}}{2}\Big]^{\frac{\pi}{2}}_0+\int_{0}^\limits{\frac{\pi}{2}}-1\frac{\cos2\text{x}}{2}\text{ dx}$
$\Rightarrow\text{I}=\Big[\text{x}^2\frac{\sin2\text{x}}{2}\Big]^{\frac{\pi}{2}}_0-\Big[-\text{x}\frac{\cos2\text{x}}{2}\Big]^{\frac{\pi}{2}}_0-\Big[\frac{\sin2\text{x}}{4}\Big]^{\frac{\pi}{2}}_0$
$\Rightarrow\text{I}=0-\frac{\pi}{4}-0$
$\Rightarrow\text{I}=-\frac{\pi}{4}$
View full question & answer→Question 2553 Marks
Evalute the following integrals:
$\int\frac{\sec\text{x cosec x}}{\log(\tan\text{x})}\text{dx}$
AnswerNote: Here, we are considering $\log\text{x}$ as $\log_\text{e}\text{x}$.
Let $\text{I}=\int\frac{\sec\text{x cosec x}}{\log(\tan\text{x})}\text{dx}$
Putting $\log\tan\text{x}=\text{t}$
$\Rightarrow\frac{\sec^2\text{x}}{\tan\text{x}}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\sec\text{x cosec x dx}=\text{dt}$
$\therefore\text{I}=\int\frac{1}{\text{t}}\text{dt}$
$=\log|\text{t}|+\text{C}$
$=\log|\log(\tan\text{x})|+\text{C}$
View full question & answer→Question 2563 Marks
By using the properties of definite integrals, evaluate the integral in Exercise:
$\int^{\frac{\pi}{2}}_{0}\big(2\log\sin\text{x}-\log\sin2\text{x}\big)\text{dx}$
Answer$\text{Let}\ \text{I}=\int\limits_{0}^{\frac{\pi}{2}}\big(2\log\sin\text{x}-\log\sin2\text{x}\big)\text{dx}=\int\limits_{0}^{\frac{\pi}{2}}\big(\log\sin^{2}\text{x}-\log\sin2\text{x}\big)\text{dx}=\int\limits_{0}^{\frac{\pi}{2}}\log\bigg(\frac{\sin^{2}\text{x}}{\sin2\text{x}}\bigg)\text{dx}$ $=\int\limits_{0}^{\frac{\pi}{2}}\log\bigg(\frac{\sin^{2}\text{x}}{2\sin\text{x}\cos\text{x}}\bigg)\text{dx}=\int\limits_{0}^{\frac{\pi}{2}}\log\bigg(\frac{1}{2}\tan\text{x}\bigg)\text{dx}$ $\Rightarrow\ \ \text{I}=\int\limits_{0}^{\frac{\pi}{2}}\log\bigg(\frac{1}{2}\tan\bigg(\frac{\pi}{2}-\text{x}\bigg)\bigg)\text{dx}\ \bigg[\because\int\limits_{0}^{\text{a}}\text{f}\text{(x)}\text{dx}=\int\limits_{0}^{\text{a}}\text{f}\text{(a}-\text{x})\text{dx}=\bigg]$ $\Rightarrow\ \ \text{I}=\int\limits_{0}^{\frac{\pi}{2}}\log\bigg(\frac{1}{2}\cot\text{x}\bigg)\text{dx}$ Adding eq.(i) and (ii) $21=\int\limits_{0}^{\frac{\pi}{2}}\bigg[\log\bigg(\frac{1}{2}\tan\text{x}\bigg)+\log\bigg(\frac{1}{2}\cot\text{x}\bigg)\bigg]\text{dx}=\int\limits_{0}^{\frac{\pi}{2}}\bigg[\log\bigg(\frac{1}{2}\tan\text{x}\bigg)\bigg(\frac{1}{2}\cot\text{x}\bigg)\bigg]\text{dx}$ $\Rightarrow\ \ 21=\int\limits_{0}^{\frac{\pi}{2}}\bigg[\log\frac{1}{4}\bigg]\text{dx}=\log\frac{1}{4}\text{(x)}^{\frac{\pi}{2}}_{0}=\big(\log1-\log4\big)\frac{\pi}{2}=-\frac{\pi}{2}\log4$$\Rightarrow\ \ \text{I}=-\frac{\pi}{4}\log2^{2}=-\frac{2\pi}{4}\log2=-\frac{\pi}{2}\log2$
View full question & answer→Question 2573 Marks
Evaluate the following definite integrals:
$\int_{0}^\limits{2\pi}\text{e}^{\text{x}}\cos\Big(\frac{\pi}{4}+\frac{\text{x}}{2}\Big)\text{dx}$
AnswerLet $\text{I}=\int_{0}^\limits{2\pi}\text{e}^{\text{x}}\cos\Big(\frac{\pi}{4}+\frac{\text{x}}{2}\Big)\text{dx}$ Then,
Integrating by parts,
$\text{I}=\Big[2\text{e}^{\text{x}}\sin\Big(\frac{\pi}{4}+\frac{\text{x}}{2}\Big)\Big]^{2\pi}_0-\int_{0}^\limits{2\pi}2\text{e}^{\text{x}}\sin\Big(\frac{\pi}{4}+\frac{\text{x}}{2}\Big)\text{dx}$
Integrating second term by parts,
$\text{I}=\Big[2\text{e}^{\text{x}}\sin\Big(\frac{\pi}{4}+\frac{\text{x}}{2}\Big)\Big]^{2\pi}_0+\bigg\{\Big[4\text{e}^{\text{x}}\cos\Big(\frac{\pi}{4}+\frac{\text{x}}{2}\Big)\Big]^{2\pi}_0\\+\int_{0}^\limits{2\pi}-4\text{e}^{\text{x}}\cos\Big(\frac{\pi}{4}+\frac{\text{x}}{2}\Big)\text{ dx}\bigg\}$
$\Rightarrow\text{I}=\Big[2\text{e}^{\text{x}}\sin\Big(\frac{\pi}{4}+\frac{\text{x}}{2}\Big)\Big]^{2\pi}_0+\Big[4\text{e}^{\text{x}}\cos\Big(\frac{\pi}{4}+\frac{\text{x}}{2}\Big)\Big]^{2\pi}_0-4\text{I}$
$\Rightarrow5\text{I}=-2\text{e}^{2\pi}\frac{1}{\sqrt{2}}-2\frac{1}{\sqrt{2}}-4\text{e}^{2\pi}\frac{1}{\sqrt{2}}-4\frac{1}{\sqrt{2}}$
$\Rightarrow5\text{I}=-3\sqrt{2}\text{e}^{2\pi}-3\sqrt{2}$
$\Rightarrow\text{I}=-\frac{3\sqrt{2}}{5}\big(\text{e}^{2\pi}+1\big)$
View full question & answer→Question 2583 Marks
Evaluate the following integrals:
$\int\cot^5\text{x}\text{ cosec}^4\text{x}\text{ dx}$
Answer$\int\cot^5\text{x}\text{ cosec}^4\text{x}\text{ dx}$
$=\int\cot^5\text{x}\text{ cosec}^2\text{ x}.\text{ cosec}^2\text{x}\text{ dx}$
$=\int\cot^5\text{x}.(1+\cot^2\text{x}).\text{ cosec}^2\text{x}\text{ dx}$
Let $\cot\text{x}=\text{t}$
$=-\text{ cosec}^2\text{x}\text{ dx}=\text{dt}$
$=\text{ cosec}^2\text{x}\text{ dx}=-\text{dt}$
Now, $\int\cot^5\text{x}\text{ cosec}^4\text{x}\text{ dx}$
$=\int\text{t}^5(1+\text{t}^2)\text{dt}$
$=\int(\text{t}^5+\text{t}^7)\text{dt}$
$=-\Big[\frac{\text{t}^6}{6}+\frac{\text{t}^8}{8}\Big]+\text{C}$
$=-\Big[\frac{\cot^6\text{x}}{6}+\frac{\cot^8\text{x}}{8}\Big]+\text{C}$
View full question & answer→Question 2593 Marks
Find the integrals of the functions in Exercises:
$\cos2\text{x}\cos4\text{x}\cos6\text{x}$
AnswerIt is known that, $\cos\text{A}\cos\text{B}=\frac{1}{2}\big\{\cos(\text{A}+\text{B})+\cos(\text{A}-\text{B}) \big\}$
$\therefore\int \cos2\text{x}(\cos4\text{x}\cos6\text{x) dx}$
$=\int\cos2\text{x}\bigg[\frac{1}{2}\{\cos(\text{4x+6x})+\cos(\text{4x}-\text{6x}) \big\}\bigg]\text{ dx}$
$=\frac{1}{2}\int\big\{\cos2\text{x}\cos10\text{x}+\cos2\text{x}\cos(-2\text{x})\big\}\text{ dx}$
$=\frac{1}{2}\int\big\{\cos2\text{x}\cos10\text{x}+\cos^22\text{x}\big\}\text{ dx}$
$=\frac{1}{2}\int\Bigg[\bigg\{\frac{1}{2}\cos(2\text{x}+10\text{x})+\cos(2\text{x}-10\text{x})\bigg\}+\bigg(\frac{1+\cos4\text{x}}{2}\bigg)\Bigg]\text{dx}$
$=\frac{1}{4}\int(\cos12\text{x}+\cos8\text{x}+1+\cos4\text{x})\text{dx}$
$=\frac{1}{4}\bigg[\frac{\sin12\text{x}}{12}+\frac{\sin8\text{x}}{8}+\text{x}+\frac{\sin4\text{x}}{4}\bigg]+\text{C}$
View full question & answer→Question 2603 Marks
Write a value of $\int\tan^6\text{x}\sec^2\text{x}\text{ dx}$
AnswerLet $\text{I}=\int\tan^6\text{x}\sec^2\text{x}\text{ dx}$
Let $\tan\text{x}=\text{t}$
$\sec^2\text{x dx}=\text{dt}$
$\therefore\ \text{I}=\int\text{t}^6\text{dt}$
$=\frac{\text{t}^7}{7}+\text{C}$
Thus, $\text{I}=\frac{\tan^7\text{x}}{7}+\text{C}$
View full question & answer→Question 2613 Marks
Evaluate the following integrals:$\int\frac{\text{x}+5}{3\text{x}^2+13\text{x}-10}\text{ dx}$
Answer$\text{I}=\int\frac{\text{x}+5}{3\text{x}^2+13\text{x}-10}\text{ dx}$
$=\int\frac{\text{x}+5}{3\text{x}^2+15\text{x}-2\text{x}-10}\text{ dx}$
$=\int\frac{\text{x}+5}{3\text{x}(\text{x}+5)-2(\text{x}+5)}\text{ dx}$
$=\int\frac{\text{x}+5}{(3\text{x}-2)(\text{x}+5)}\text{ dx}$
$=\int\frac{1}{3\text{x}-2}\text{ dx}$
$\therefore\ \text{I}=\frac{1}{3}\int|3\text{x}-2|+\text{C}$
View full question & answer→Question 2623 Marks
Evaluate the following integrals:
$\int\limits^{\text{a}}_0\frac{\sqrt{\text{x}}}{\sqrt{\text{x}}+\sqrt{\text{a}-\text{x}}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{\text{a}}_0\frac{\sqrt{\text{x}}}{\sqrt{\text{x}}+\sqrt{\text{a}-\text{x}}}\text{ dx}\ ...(\text{i})$
$\Rightarrow\text{I}=\int\limits^{\text{a}}_0\frac{\sqrt{\text{a}-\text{x}}}{\sqrt{\text{a}-\text{x}}+\sqrt{\text{x}}}\text{ dx}$
$\Big[\text{Using},\int\limits^{\text{a}}_0\text{f(x)}\text{dx}=\int\limits^{\text{a}}_0\text{f}(\text{a}-\text{x})\text{dx}\Big]$
$\text{I}=\int\limits^{\text{a}}_0\frac{\sqrt{\text{a}-\text{x}}}{\sqrt{\text{x}}+\sqrt{\text{a}-\text{x}}}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii)
$2\text{I}=\int\limits^{\text{a}}_0\frac{\sqrt{\text{x}}+\sqrt{\text{a}-\text{x}}}{\sqrt{\text{x}}+\sqrt{\text{a}-\text{x}}}\text{ dx}$
$=\int\limits^{\text{a}}_0\text{dx}=\big[\text{x}\big]^{\text{a}}_0=\text{a}$
Hence, $\text{I}=\frac{\text{a}}{2}$
View full question & answer→Question 2633 Marks
Evaluate the definite integral in Exercise:
$\int\limits^{\frac{\pi}{4}}_0\frac{\sin\text{x}\cos\text{x}}{\cos^{4}\text{x}+\sin^{4}\text{x}}\text{dx}$
Answer$\text{Let I}=\int^{\frac{\pi}{4}}\limits_{0}\frac{\sin\text{x}\cos\text{x}}{\cos^{4}\text{x}+\sin^{4}\text{x}}\text{dx}$
$\Rightarrow\text{ I}=\int^{\frac{\pi}{4}}\limits_{0}\frac{\frac{(\sin\text{x}\cos\text{x)}}{\cos^{4}\text{x}}}{\frac{\cos^{4}\text{x}+\sin^{4}\text{x}}{\cos^{4}\text{x}}}\text{dx}$
$\Rightarrow\text{ I}=\int^{\frac{\pi}{4}}\limits_{0}\frac{\tan\text{x}\sec^{2}\text{x}}{1+\tan^{4}\text{x}}\text{dx}$
$\text{Let}\ \tan^{2}\text{x}=\text{t}\ \Rightarrow2\tan\text{x}\text{dx}=\text{dt}$
when $\text{x}=0,\text{t}=0$ and when $\text{x}=\frac{\pi}{4},\text{t}=1$
$\therefore\ \text{I}=\frac{1}{2}\int^{1}\limits_{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}-\tan^{-1}0\Big]$
$=\frac{1}{2}\Big[\frac{\pi}{4}\Big]$
$=\frac{\pi}{8}$
View full question & answer→Question 2643 Marks
Evaluate the following integrals:
$\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\big(2\sin|\text{x}|+\cos|\text{x}|\big)\text{dx}$
Answer$\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\big(2\sin|\text{x}|+\cos|\text{x}|\big)\text{dx}$
$=\int^{0}_{-\frac{\pi}{4}}\big(-2\sin\text{x}+\cos\text{x}\big)\text{dx}+\int_{0}^{\frac{\pi}{2}}\big(2\sin\text{x}+\cos\text{x}\big)\text{dx}$
$=\big[2\cos\text{x}+\sin\text{x}\big]^0_{-\frac{\pi}{4}}+\big[-2\cos\text{x}+\sin\text{x}\big]_0^{\frac{\pi}{2}}$
$=2+0-0+1+0+1+2-0$
$=6$
View full question & answer→Question 2653 Marks
Evaluate the following integrals:
$\int^\limits6_{-6}\big|\text{x}+2\big|\text{dx}$
Answer$\int^\limits6_{-6}\big|\text{x}+2\big|\text{dx}$
We know that,
$|\text{x}+2|=\begin{cases}-(\text{x}+2),&-6\leq\text{x}\leq-2\\\text{x}+2,&-2<\text{x}\leq6\end{cases}$
$\therefore\ \text{I}=\int^\limits6_{-6}\big|\text{x}+2\big|\text{dx}$
$\Rightarrow\text{I}=\int^\limits{-2}_{-6}\big(\text{x}+2\big)\text{dx}+\int^\limits6_{-2}\big(\text{x}+2\big)\text{dx}$
$\Rightarrow\text{I}=\Big[\frac{-\text{x}^2}{2}-2\text{x}\Big]^{-2}_{-6}+\Big[\frac{\text{x}^2}{2}-2\text{x}\Big]^6_{-2}$
$\Rightarrow\text{I}=-2+4+18-12+18+12-2+4$
$\Rightarrow\text{I}=40$
View full question & answer→Question 2663 Marks
Evaluate the following integrals:
$\int\limits^{{\pi}}_{-\frac{\pi}{2}}\sin^{-1}(\sin\text{x})\text{dx}$
Answer$\int\limits^{{\pi}}_{-\frac{\pi}{2}}\sin^{-1}\text{x}(\sin\text{x})\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\sin^{-1}(\sin\text{x})\text{dx}\int\limits^{{\pi}}_{\frac{\pi}{2}}\sin^{-1}(\sin\text{x})\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\text{x dx}+\int\limits^{{\pi}}_{\frac{\pi}{2}}(\pi-\text{x})\text{dx}$ $\Big[\frac{\pi}{2}\leq\text{x}\leq\pi\Rightarrow-\pi\leq-\text{x}\leq-\frac{\pi}{2}\Rightarrow0\leq\pi-\text{x}\leq\frac{\pi}{2}\Big]$
$=\Big[\frac{\text{x}^2}{2}\Big]^{\frac{\pi}{2}}_{-\frac{\pi}{2}}+\bigg[\frac{(\pi-\text{x})}{2\times(-1)}\bigg]^{\pi}_{\frac{\pi}{2}}$
$=\frac{1}{2}\Big(\frac{\pi^2}{4}-\frac{\pi^2}{4}\Big)-\frac{1}{2}\Big(0-\frac{\pi^2}{4}\Big)$
$=0+\frac{\pi^2}{8}$
$=\frac{\pi^2}{8}$
View full question & answer→Question 2673 Marks
Integrate the function in Exercise:$\sqrt{\frac{1-\sqrt{\text{x}}}{1+\sqrt{\text{x}}}}$
Answer$\text{I}=\sqrt{\frac{1-\sqrt{\text{x}}}{1+\sqrt{\text{x}}}}\text{dx}$
$\text{Let }\ \text{x}=\cos^{2}\theta\Rightarrow\text{dx}=-2\sin\theta\cos\theta\ \text{d}\theta$
$\text{I}=\int\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}(-2\sin\theta\cos\theta)\text{d}\theta$
$=-\int\sqrt{\frac{2\sin^{2}\frac{\theta}{2}}{2\cos^{2}\frac{\theta}{2}}}\sin2\theta\ \text{d}\theta$
$=-\int\tan\frac{\theta}{2}.\sin\theta\cos\theta\ \text{d}\theta$
$=-2\int\frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}}\bigg(2\sin\frac{\theta}{2}\cos\frac{\theta}{2}\bigg)\cos\theta\ \text{d}\theta$
$=-4\int\sin^{2}\frac{\theta}{2}\cos\theta\ \text{d}\theta$
$=-4\int\sin^{2}\frac{\theta}{2}.\bigg(2\cos^{2}\frac{\theta}{2}-1\bigg) \text{d}\theta$
$=-4\int\bigg(2\sin^{2}\frac{\theta}{2}\cos^{2}\frac{\theta}{2}-\sin^{2}\frac{\theta}{2}\bigg) \text{d}\theta$
$=-8\int\sin^{2}\frac{\theta}{2}.\cos^{2}\frac{\theta}{2}\text{d}\theta+4\int\sin^{2}\frac{\theta}{2}\text{d}\theta$
$=-2\int\sin^{2}\theta\text{d}\theta+4\int\sin^{2}\frac{\theta}{2}\text{d}\theta$
$=-2\int\bigg(\frac{1-\cos2\theta}{2}\bigg)\text{d}\theta+4\int\frac{1-\cos\theta}{2}\text{d}\theta$
$=-2\bigg[\frac{\theta}{2}-\frac{\sin2\theta}{4}\bigg]+4\bigg[\frac{\theta}{2}-\frac{\sin\theta}{2}\bigg]+\text{C}$
$=-\theta+\frac{\sin2\theta}{2}+2\theta-2\sin\theta+\text{C}$
$=\theta+\frac{\sin2\theta}{2}-2\sin\theta+\text{C}$
$=\theta+\frac{2\sin\theta\cos\theta}{2}-2\sin\theta+\text{C}$
$=\theta+\sqrt{1-\cos^{2}\theta}.\cos\theta-2\sqrt{1-\cos^{2}\theta}+\text{C}$
$=\cos^{-1}\sqrt{\text{x}}+\sqrt{1-\text{x}}.\sqrt{\text{x}}-2\sqrt{1-\text{x}}+\text{C}$
$=-2\sqrt{1-\text{x}}+\cos^{-1}\sqrt{\text{x}}+\sqrt{\text{x}(1-\text{x)}}+\text{C}$
$=-2\sqrt{1-\text{x}}+\cos^{-1}\sqrt{\text{x}}+\sqrt{\text{x}-\text{x}^{2}}+\text{C}$
View full question & answer→Question 2683 Marks
Write a value of $\int\frac{1+\log\text{x}}{3+\text{x}\log\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{1+\log\text{x}}{3+\text{x}\log\text{x}}\text{ dx}$
Let $3+\text{x}\log\text{x}=\text{t}$
$\Big(\log\text{x}+\text{x}\cdot\frac{1}{\text{x}}\Big)\text{dx}=\text{at}$
$(1+\log\text{x})\text{dx}=\text{dt}$
$\therefore\ \text{I}=\int\frac{\text{dt}}{\text{t}}$
$=\log\text{t}+\text{C}$
$\text{I}=\log(3+\text{x}\log\text{x})+\text{C}$
View full question & answer→Question 2693 Marks
$\int\limits^{\frac{\pi}{2}}_{\frac{\pi}{3}}\frac{\sqrt{1+\cos\text{x}}}{(1-\cos\text{x})^{\frac{5}{2}}}$
AnswerLet $\text{I}=\int\limits^{\frac{\pi}{2}}_{\frac{\pi}{3}}\frac{\sqrt{1+\cos\text{x}}}{(1-\cos\text{x})^{\frac{5}{2}}}$
$\int\limits^{\frac{\pi}{2}}_{\frac{\pi}{3}}\frac{\sqrt{1+\cos\text{x}}}{(1-\cos\text{x})^2\sqrt{1-\cos\text{x}}}\text{dx}$
$\int\limits^{\frac{\pi}{2}}_{\frac{\pi}{3}}\frac{1}{(1-\cos\text{x})^2}\text{dx}=\int\limits^{\frac{\pi}{2}}_{\frac{\pi}{3}}\frac{1}{(\sin^2\text{x})}\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_{\frac{\pi}{3}}\cos\text{ec}^2\text{x dx}=\big[-\cot\text{x}\big]^{\frac{\pi}{2}}_{\frac{\pi}{3}}$
$=-\Big[\cot\frac{\pi}{2}-\cot\frac{\pi}{2}\Big]$ $=-\Big[0-\frac{1}{\sqrt{3}}\Big]=+\frac{1}{\sqrt{3}}$
View full question & answer→Question 2703 Marks
Evaluate the following integrals:
$\int\sin^3\text{x}\cos^5\text{x}\text{ dx}$
Answer$\int\sin^3\text{x}\cos^5\text{x}\text{ dx}$
$=\int\sin^2\text{x}\cdot\cos^5\text{x}\cdot\sin\text{x}\text{ dx}$
$=\int(1-\cos^2\text{x})\cdot\cos^5\text{x}\sin\text{x}\text{ dx}$
Let $\cos\text{x}=\text{t}$
$-\sin\text{x}\text{ dx}=\text{dt}$
$\sin\text{x}\text{ dx}=-\text{dt}$
Now, $\int(1-\cos^2\text{x})\cdot\cos^5\text{x}\sin\text{x}\text{ dx}$
$=-\int(1-\text{t}^2)\text{t}^5\text{dt}$
$=-(\text{t}^5-\text{t}^7)\text{dt}$
$=-\int(\text{t}^7-\text{t}^5)\text{dt}$
$=\frac{\text{t}^8}{8}-\frac{\text{t}^6}{6}+\text{C}$
$=\frac{\cos^8\text{x}}{8}-\frac{\cos^6\text{6}\text{x}}{6}+\text{C}$
View full question & answer→Question 2713 Marks
Evaluate the following integrals:
$\int\text{x}^2\cos\text{x dx}$
Answer$\int\text{x}^2\cos\text{x dx}$
Taking $x^2$ as the first function and cos x as the second function.
$=\text{x}^2\int\cos\text{x dx}-\int\big(\frac{\text{d}}{\text{dx}}\text{x}^2\int\cos\text{x dx}\big)\text{dx}$
$=\text{x}^2\sin\text{x}-\int2\text{x}\sin\text{x dx}$
$=\text{x}^2\sin\text{x}-2\big[\text{x}\int\sin\text{x}-\int\big\{\frac{\text{d}}{\text{dx}}(\text{x})\int\sin\text{x dx}\big\}\text{dx}\big]$
$=\text{x}^2\sin\text{x}-2[-\text{x}\cos\text{x}+\int\cos\text{x dx}]$
$=\text{x}^2\sin\text{x}+2\text{x}\cos\text{x}-2\sin\text{x+C}$
View full question & answer→Question 2723 Marks
Evaluate the following intregals:
$\int\frac{1}{\cos\text{x}(\sin\text{x}+2\cos\text{x})}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{1}{\cos\text{x}(\sin\text{x}+2\cos\text{x})}\ \text{dx}$
$=\int\frac{1}{\sin\text{x}\cos\text{x}+2\cos^2\text{x}}$
Dividing numerator and denominator by $\cos^2\text{x}$
$\text{I}=\int\frac{\sec^2\text{x}}{\tan\text{x}+2}\ \text{dx}$
Let $2+\tan\text{x}=\text{t}$
$\sec^2\text{x dx}=\text{dt}$
$\text{I}=\int\frac{\text{dt}}{\text{t}}$
$=\log|\text{t}|+\text{C}$
$\text{I}=\log|2+\tan\text{x}|+\text{C}$
View full question & answer→Question 2733 Marks
Evaluate the following integrals:
$\int^\limits9_0\text{f(x)}\text{dx},$ Where $\text{f(x)}=\begin{cases}\sin\text{x},&0\leq\text{x}\leq\frac{\pi}{2}\\1,&\frac{\pi}{2}\leq\text{x}\leq3\\\text{e}^{\text{x}-3},&3\leq\text{x}\leq9\end{cases}$
AnswerWe have,
$\int^\limits9_0\text{f(x)}\text{dx},$ Where $\text{f(x)}=\begin{cases}\sin\text{x},&0\leq\text{x}\leq\frac{\pi}{2}\\1,&\frac{\pi}{2}\leq\text{x}\leq3\\\text{e}^{\text{x}-3},&3\leq\text{x}\leq9\end{cases}$
$\text{I}=\int^\limits9_0\text{f(x)}\text{dx}$
$\Rightarrow\text{I}=\int^\limits{\frac{\pi}{2}}_1\text{f(x)}\text{dx}+\int^\limits3_\frac{\pi}{2}\text{f(x)}\text{dx}+\int^\limits9_3\text{e}^{\text{x}-3}\text{ dx}$ [Additive property]
$\Rightarrow\text{I}=\int^\limits{\frac{\pi}{2}}_1\sin\text{x dx}+\int^\limits3_\frac{\pi}{2}\text{1 }\text{dx}+\int^\limits9_3\text{e}^{\text{x}-3}\text{ dx}$
$\Rightarrow\text{I}=\big[-\cos\text{x}\big]^{\frac{\pi}{2}}_0+\big[\text{x}\big]^3_\frac{\pi}{2}+\text{e}^6-\text{e}^0$
$\Rightarrow\text{I}=0+1+3-\frac{\pi}{2}+\text{e}^6-\text{e}^0$
$\Rightarrow\text{I}=3-\frac{\pi}{2}+\text{e}^6$
View full question & answer→Question 2743 Marks
$\int \text{(2x} - 3)^{5} + \sqrt{3\text{x + 2}}\text{ dx}$
Answer$\int\big[(2\text{x}-3)^5+\sqrt{3\text{x}+2}\big]\text{dx}$
$=\int(2\text{x}-3)^5\text{dx}+\int{(3\text{x}+2)^{\frac{1}{2}}}\text{dx}$
$=\frac{(2\text{x}-3)^{5+1}}{2(5+1)}+\frac{(3\text{x}+2)^{\frac{1}{2}{+1}}}{3\Big(\frac{1}{2}+1\Big)}+\text{c}$
$=\frac{(2\text{x}-3)^6}{12}+\frac{2}{9}(3\text{x}+2)^{\frac{3}{2}}+\text{c}$
View full question & answer→Question 2753 Marks
Evalute the following integrals:
$\int\frac{\sin2\text{x}}{\sin5\text{x}\sin3\text{x}}\text{dx}$
Answer$\int\frac{\sin2\text{x}}{\sin5\text{x}\sin3\text{x}}\text{dx}$
$=\int\frac{\sin(5\text{x}-3\text{x})}{\sin5\text{x}\sin3\text{x}}\text{dx}$
$=\int\frac{5\text{x}\cos3\text{x}-\cos5\text{x}\sin3\text{x}}{\sin5\text{x}\sin3\text{x}}$
$=\int\frac{\sin5\text{x}\cos3\text{x}}{\sin5\text{x}\sin3\text{x}}-\frac{\cos5\text{x}\sin3\text{x}}{\sin5\text{x}\sin3\text{x}}\text{dx}$
$=\int\big[\cot3\text{x}-\cot5\text{x}\big]\text{dx}$
$=\int\cot3\text{x dx}-\int\cot5\text{x dx}$
$=\frac{1}{3}\text{ln}|\sin3\text{x}|-\frac{1}{5}\text{ln}|\sin5\text{x}|+\text{C}$
View full question & answer→Question 2763 Marks
Evalute the following integrals:
$\int\sqrt{\frac{1-\sin2\text{x}}{1+\sin2\text{x}}}\text{dx}$
AnswerLet $\text{I}=\int\sqrt{\frac{1-\sin2\text{x}}{1+\sin2\text{x}}}\text{dx}$ then,
$=\int\sqrt{\frac{1-\cos\Big(\frac{\pi}{2}-2\text{x}\Big)}{1+\cos\Big(\frac{\pi}{2}-2\text{x}\Big)}}\text{dx}$
$=\int\sqrt{\frac{2\sin^2\Big(\frac{\pi}{4}-\text{x}\Big)}{2\cos^2\Big(\frac{\pi}{4}-\text{x}\Big)}}\text{dx}$
$=\int\sqrt{\tan^2\Big(\frac{\pi}{4}-\text{x}\Big)}\text{dx}$
$=\int\tan\Big(\frac{\pi}{4}-\text{x}\Big)\text{dx}$
$=\log\Big|\cos\Big(\frac{\pi}{4}-\text{x}\Big)\Big|+\text{C}$
View full question & answer→Question 2773 Marks
$\text{Show that}\int^{\text{a}}\limits_{0}\text{f}\text{(x)}\text{g}\text{(x)}\text{dx}=2\int^{\text{a}}\limits_{0}\text{f}\text{(x)}\text{dx}\text,$if and are defined as $\text{f(x)}=\text{f(a}-\text{x)}$ and $\text{g(x)}+\text{g(a}-\text{x)}=4$
Answer$\text{Here}\ \text{f}\text{(x)}=\text{f}\text{(a}-\text{x)}\ .....(\text{i})\text{and}\ \ \text{g}\text{(x)}+\text{g}\text{(a}-\text{x)}=4 \ ......\text{(ii)}$$\text{Let}\ \text{I}=\int^{\text{a}}\limits_{0}\text{f}\text{(x)}\text{g}\text{(x)}\text{dx}\ ....\text{(iii)}$
$\therefore\ \ \text{I}=\int^{\text{a}}\limits_{0}\text{(a}-\text{x)}\text{g}\text{(a}-\text{x)}\text{dx}=\int^{\text{a}}\limits_{0}\text{f}\text{(x)}\text{g}\text{(a}-\text{x)}\text{dx}\ \ \ [\text{from eq.(i)}]\ ...\text{(iv)}$
Adding eq. (iii) and (iv)
$21=\int^{\text{a}}\limits_{0}(\text{f}\text{(x)}\text{g}\text{(x)}+\text{f}\text{(x)}\text{g}\text{(a}-\text{x)}\text{dx}=\int^{\text{a}}\limits_{0}\text{f}\text{(x)}(\text{g}\text{(x)}+\text{g}\text{(a}-\text{x)})\text{dx}$
$\Rightarrow\ \ \ \ 21\int^{\text{a}}_{0}\text{f}\text{(x)}(4)\text{dx}.......[\text{from eq. (ii)}]$
$\Rightarrow\ \ \ \ 21=4\int^{\text{a}}\limits_{0}\text{f}\text{(x)}\text{dx}\ \Rightarrow\ \text{I}=2\int^{\text{a}}\limits_{0}\text{f}\text{(x)}$ Hence proved.
View full question & answer→Question 2783 Marks
Evaluate the following integrals:$\int2\text{x}^3\text{e}^{\text{x}^{2}}\text{dx}$
Answer$\int2\text{x}^3\cdot\text{e}^{\text{x}^{2}}\text{dx}$
$=\int\text{x}^2\cdot\big(\text{e}^{\text{x}^2}\big)\cdot2\text{x dx}$
Let $\text{x}^2=\text{t}$
$\Rightarrow2\text{x dx = dt}$
$=\int\text{t}\cdot\text{e}^{\text{t}}\text{dt}$
$=\text{t}\cdot\text{e}^{\text{t}}-\int1\cdot\text{e}^{\text{t}}\text{dt}$
$=\text{t e}^{\text{t}}-\text{e}^{\text{t}}+\text{C}$
$=\text{x}^2\text{e}^{\text{x}^{2}}-\text{e}^{\text{x}^{2}}+\text{C}$
$=\text{e}^{\text{x}^2}(\text{x}^2-1)+\text{C}$
View full question & answer→Question 2793 Marks
Evaluate the following integrals:
$\int\text{x}^3\log\text{x dx}$
AnswerLet $\text{I}=\int\text{x}^3\log\text{x dx}$
Using integration by parts,
$\text{I}=\log\text{x}\int\text{x}^3\text{dx}-\int\Big(\frac{1}{\text{x}}\times\int\text{x}^3\text{dx}\Big)\text{dx+C}$
$=\frac{\text{x}^4}{4}\log\text{x}-\int\frac{\text{x}^4}{4\text{x}}\text{dx+C}$
$=\frac{\text{x}^4}{4}\log\text{x}-\frac{1}{4}\int\text{x}^3\text{dx+C}$
$=\frac{\text{x}^4}{4}\log\text{x}-\frac{1}{4}\int\frac{\text{x}^4}{4}\text{dx+C}$
$\text{I}=\frac{\text{x}^4}{4}\log\text{x}-\frac{1}{16}\text{x}^4+\text{C}$
View full question & answer→Question 2803 Marks
Evaluate the following integrals:$\int\frac{\log(\log\text{x})}{\text{x}}\text{dx}$
Answer$\int\frac{\log(\log\text{x})}{\text{x}}\text{dx}$
Taking log log x as the first function and $\frac{1}{\text{x}}$ as the second function.
$=\log \log\text{x}\int\frac{1}{\text{x}}\text{dx}-\int\Big\{\frac{\text{d}}{\text{dx}}\log(\log\text{x})\int\frac{1}{\text{x}}\text{dx}\Big\}\text{dx}$
$=\log\text{x}.\log(\log\text{x})-\int\frac{1}{\text{x}\log\text{x}}(\log\text{x})\text{dx}$
$=\log\text{x}.\log(\log\text{x})-\int\frac{1}{\text{x}}\text{dx}$
$=\log\text{x}.\log(\log\text{x})-\log\text{x}+\text{C}$
$=\log\text{x}[\log(\log\text{x})-1]+\text{C}$
View full question & answer→Question 2813 Marks
Write a value of $\int\text{e}^{3\log\text{x}}\text{x}^{4}\text{ dx}.$
AnswerLet $\text{I}=\int\text{e}^{3\log\text{x}}\text{x}^{4}\text{ dx}$
$=\int\text{e}^{\log\text{x}^3}\cdot\text{x}^{4}\text{ dx}$
$=\int\text{x}^3\cdot\text{x}^4\text{ dx}$ $\big[\because\text{e}^{\log\text{x}}=\text{x}\big]$
$=\int\text{x}^{7}\text{ dx}$
$\therefore\ \text{I}=\frac{\text{x}^{8}}{8}+\text{C}$
View full question & answer→Question 2823 Marks
Evaluate the following integrals:
$\int\tan^{-1}\Big(\frac{\sin2\text{x}}{1+\cos2\text{x}}\Big)\text{dx}$
Answer$\int\tan^{-1}\Big[\frac{\sin2\text{x}}{1+\cos2\text{x}}\Big]\text{dx}$
$=\int\tan^{-1}\Big[\frac{2\sin\text{x}\cos\text{x}}{2\cos^2\text{x}}\Big]\text{dx}$
$=\int\tan^{-1}\Big[\frac{\sin\text{x}}{\cos\text{x}}\Big]\text{dx}$
$=\int\tan^{-1}(\tan\text{x})\text{dx}$
$=\int\text{x dx}$
$=\frac{\text{x}^2}{2}+\text{C}$
$\therefore\ \int\tan^{-1}\Big[\frac{\sin2\text{x}}{1+\cos2\text{x}}\Big]\text{dx}=\frac{\text{x}^2}{2}+\text{C}$
View full question & answer→Question 2833 Marks
Evaluate the following integrals:
$\int\text{e}^{2\text{x}}\sin\text{x }\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{2\text{x}}\sin\text{x }\text{dx}$
Considering sin x as first function and $e^{2x}$ as second function
$\text{I}=\sin\text{x}\frac{\text{e}^{2\text{x}}}{2}-\int\cos\text{x}\frac{\text{e}^{2\text{x}}}{2}\text{dx}$
$\Rightarrow\text{I}=\sin\text{x}\frac{\text{e}^{2\text{x}}}{2}-\frac{1}{2}\int\cos\text{e}^{2\text{x}}\text{dx}$
$\Rightarrow\text{I}=\frac{\sin\text{x }\text{e}^{2\text{x}}}{2}-\frac{1}{2}\Big[\cos\text{x}\frac{\text{e}^{2\text{x}}}{2}-\int(-\sin\text{x})\frac{\text{e}^{2\text{x}}}{2}\text{dx}\Big]$
$\Rightarrow\text{I}=\frac{\sin\text{x }\text{e}^{2\text{x}}}{2}-\frac{\cos\text{x }\text{e}^{2\text{x}}}{4}-\frac{1}{2}\int\frac{\text{e}^{2\text{x}}\sin\text{x}}{2}\text{dx}$
$\text{I}=\frac{\text{e}^{2\text{x}}(2\sin\text{x}-\cos\text{x})}{4}-\frac{\text{I}}{4}$
$\Rightarrow5\text{I}=\text{e}^{2\text{x}}(2\sin\text{x}-\cos\text{x})$
$\Rightarrow\text{I}=\frac{\text{e}^{2\text{x}}(2\sin\text{x}-\cos\text{x})}{5}+\text{C}$
View full question & answer→Question 2843 Marks
Evaluate the following integrals:
$\int^\limits{\frac{\pi}{4}}_{\frac{-\pi}{4}}|\sin\text{x}|\text{dx}$
Answer$\int^\limits{\frac{\pi}{4}}_{\frac{-\pi}{4}}|\sin\text{x}|\text{dx}$
We know that,
$|\sin\text{x}|=\begin{cases}-\sin\text{x},&-\frac{\pi}{4}\leq\text{x}\leq0\\\sin\text{x},&0<\text{x}\leq\frac{\pi}{4}\end{cases}$
$\therefore\ \text{I}=\int^\limits{\frac{\pi}{4}}_{\frac{-\pi}{4}}|\sin\text{x}|\text{dx}$
$\Rightarrow\text{I}=\int^\limits0_{-\frac{\pi}{4}}-\sin\text{x dx}+\int\limits^{\frac{\pi}{4}}_0\sin\text{x dx}$
$\Rightarrow\text{I}=\big[\cos\text{x}\big]^0_{\frac{-\pi}{4}}-\big[\cos\text{x}\big]^{\frac{-\pi}{4}}_0$
$\Rightarrow\text{I}=1-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}+1$
$\Rightarrow\text{I}=2-\frac{2}{\sqrt{2}}$
$\Rightarrow\text{I}=2-\sqrt{2}$
View full question & answer→Question 2853 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{2}}_0\text{e}^{\text{x}}(\sin\text{x}-\cos\text{x})\text{dx}$
Answer$\int\limits^{\frac{\pi}{2}}_0\text{e}^{\text{x}}(\sin\text{x}-\cos\text{x})\text{dx}$
$=-\int\limits^{\frac{\pi}{2}}_0\text{e}^{\text{x}}\Big[\cos\text{x}+(-\sin\text{x})\Big]\text{dx}$
$=-\big[\text{e}^{\text{x}}\cos\text{x}\big]^{\frac{\pi}{2}}_0$ $\Big\{\int\text{e}^{\text{x}}\big[\text{f(x)}+\text{f}'(\text{x})\big]\text{dx}=\text{e}^{\text{x}}\text{f(x)}+\text{C}\Big\}$
$=-\Big(\text{e}^{\frac{\pi}{2}}\cos\frac{\pi}{2}-\text{e}^0\cos0\Big)$
$=-\Big(\text{e}^{\frac{\pi}{2}}\times0-1\times1\Big)$
$=-(0-1)$
$=1$
View full question & answer→Question 2863 Marks
Evaluate the following integrals:
$\int\frac{\text{x}^{\frac{-1}{3}}+\sqrt{\text{x}}+2}{\sqrt[3]{\text{x}}}\text{dx}$
Answer$\int\Bigg(\int\frac{\text{x}^{-\frac{1}{3}}+\sqrt{\text{x}}+2}{\text{x}^{\frac{1}{3}}}\Bigg)\text{dx}$
$=\int\Bigg(\frac{\text{x}^{-\frac{1}{3}}}{\text{x}^{\frac{1}{3}}}+\frac{\text{x}^{\frac{1}{2}}}{\text{x}^{\frac{1}{3}}}+\frac{2}{\text{x}^{\frac{1}{3}}}\Bigg)\text{dx}$
$=\int\Big(\text{x}^{-\frac{2}{3}}+\text{x}^{\frac{1}{6}}+2\text{x}^{-\frac{1}{3}}\Big)\text{dx}$
$=\Bigg[\frac{\text{x}^{-\frac{2}{3}+1}}{-\frac{2}{3}+1}+\frac{\text{x}^{\frac{1}{6}+1}}{\frac{1}{6}+1}+2\frac{\text{x}^{-\frac{1}{3}+1}}{-\frac{1}{3}+1}\Bigg]$
$=\Bigg[\frac{\text{x}^{\frac{1}{3}}}{\frac{1}{3}}+\frac{\text{x}^{\frac{7}{6}}}{\frac{7}{6}}+3\text{x}^{\frac{2}{3}}\Bigg]+\text{C}$
$=3\text{x}^{\frac{1}{3}}+\frac{6}{7}\text{x}^{\frac{7}{6}}+3\text{x}^{\frac{2}{3}}+\text{C}$
View full question & answer→Question 2873 Marks
$\int\frac{1}{2-3\text{x}}+\frac{1}{\sqrt{3\text{x}-2}}\text{dx}$
Answer$\int\Big(\frac{1}{2-3\text{x}}+\frac{1}{\sqrt{3\text{x}-2}}\Big)\text{dx}$
$=\int\frac{\text{dx}}{2-3\text{x}}+\int(3\text{x}-2)^{-\frac{1}{2}}\text{dx}$
$=\frac{\ln(2-3\text{x})}{-3}+\Bigg[\frac{(3\text{x}-2)^{-\frac1{2}+1}}{3\big(-\frac{1}{2}+1\big)}\Bigg]+\text{c}$
$=\frac{\ln(2-3\text{x})}{-3}+\frac{2}{3}(3\text{x}-2)^{\frac{1}{2}}+\text{c}$
$=-\frac{1}{3}\ln(2-3\text{x})+\frac{2}{3}\sqrt{3\text{x}-2}+\text{c}$
View full question & answer→Question 2883 Marks
Evalute the following integrals:
$\int\frac{1}{\sqrt{1-\text{x}^2}(2+3\sin^{-1}\text{x})}\text{dx}$
AnswerLet $\text{I}=\int\frac{1}{\sqrt{1-\text{x}^2}(2+3\sin^{-1}\text{x})}\text{dx}$
Putting $\sin^{-1}\text{x}=\text{t}$
$\Rightarrow\frac{1}{\sqrt{1-\text{x}^2}}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\frac{1}{\sqrt{1-\text{x}^2}}\text{dx}=\text{dt}$
$\therefore\text{I}=\frac{1}{2+3\text{t}}\text{dt}$
$=\frac{1}{3}\text{ln}|2+3\text{t}|+\text{C}$
$=\frac{1}{3}\text{ln}|2+3\sin^{-1}\text{tx}|+\text{C }\big[\because\text{t}=\sin^{-1}\text{x}\big]$
View full question & answer→Question 2893 Marks
Evalute the following integrals:
$\int\big\{1+\tan\text{x}\tan(\text{x}+\theta)\big\}\text{dx}$
AnswerSince,
$\tan(\text{A}-\text{B})=\frac{\tan\text{A}-\tan\text{B}}{1+\tan\text{A}\tan\text{B}}$
$\therefore\tan(\text{x}+\theta-\text{x})=\frac{\tan(\text{x}+\theta)-\tan\text{x}}{1+\tan(\text{x}+\theta)\tan\text{x}}$
$\Rightarrow 1+\tan(\text{x}+\theta)\tan\text{x}=\frac{\tan(\text{x}+\theta)-\tan\text{x}}{\tan\theta}$
$\Rightarrow\int1+\tan(\text{x}+\theta)\tan\text{x dx}$
$=\frac{1}{\tan\theta}\big[\int\tan(\text{x}+\theta)\text{dx}-\int\tan\text{x dx}\big]$
$=\frac{1}{\tan\theta}\big[-\log|\cos(\text{x}+\theta)++\log|\cos\text{x}|\big]+\text{C}$
$=\frac{1}{\tan\theta}\big[\log|\cos\text{x}|-\log|\cos(\text{x}+\theta)|\big]+\text{C}$
$=\frac{1}{\tan\theta}\log\Big|\frac{\cos\text{x}}{\cos(\text{x}+\theta)}\Big|+\text{C}$
View full question & answer→Question 2903 Marks
Evaluate the following definite integrals:
$\int_{1}^\limits{\text{e}}\frac{\text{e}^{\text{x}}}{\text{x}}(1+\text{x}\log\text{x})\text{dx}$
AnswerLet $\text{I}=\int_{1}^\limits{\text{e}}\frac{\text{e}^{\text{x}}}{\text{x}}(1+\text{x}\log\text{x})\text{dx}$
$\text{I}=\int_{1}^\limits{\text{e}}\frac{\text{e}^{\text{x}}}{\text{x}}+\int_{1}^\limits{\text{e}}\text{e}^{\text{x}}\log\text{x}\text{ dx}$
$\text{I}=\big[\text{e}^{\text{x}}\log\text{x}\big]^{\text{e}}_1-\int_{1}^\limits{\text{e}}\text{e}^{\text{x}}\log\text{x}+\int_{1}^\limits{\text{e}}\text{e}^{\text{x}}\log\text{x}$
$\text{I}=\big[\text{e}^{\text{x}}\log\text{x}\big]^{\text{e}}_1$
$\text{I}=\big[\text{e}^{\text{x}}\log\text{e}-\text{e}^1\log1\big]$
$\text{I}=\big[\text{e}^{\text{e}}1-0\big]$
$\text{I}=\text{e}^{\text{e}}$
View full question & answer→Question 2913 Marks
Integrate the rational function in exercise:
$\frac{3\text{x}-1}{(\text{x}+2)^2}$
AnswerLet, $\text{I}=\int\frac{3\text{x}-1}{(\text{x}+2)^2}\text{dx}\dots(\text{i})$Putting x + 2 = t
⇒ x = t - 2
$\frac{\text{dx}}{\text{dt}}=1$
⇒ dx = dt
Putting this value in eq. (i),
$\text{I}=\int\frac{3(\text{t}-2)-1}{(\text{t})^2}\text{dt}=\int\frac{3\text{t}-6-1}{\text{t}^2}\text{dt}=\int\frac{3\text{t}-7}{\text{t}^2}\text{dt}$
$=\int\Bigg(\frac{3\text{t}}{\text{t}^2}-\frac{7}{\text{t}^2}\Bigg)\text{dt}=\int\Bigg(\frac{3}{\text{t}}-\frac{7}{\text{t}^2}\Bigg)\text{dt}=3\int\frac{1}{\text{t}}\text{dt}-7\int\text{t}^{-2}\text{dt}$
$=3\text{log}|\text{t}|-7\frac{\text{t}^{-1}}{-1}+\text{c}=3\text{log}|\text{t}|+\frac{7}{\text{t}}+\text{c}=3\text{log}|\text{x}+2|+\frac{7}{\text{x}+2}+\text{c}$
View full question & answer→Question 2923 Marks
Evaluate the following integrals:$\int\sec^{-1}\sqrt{\text{x}}\text{dx}$
Answer$\int1.\sec^{-1}\sqrt{\text{x}}\text{dx}$
$=\sec^{-1}\sqrt{\text{x}}\int1\text{dx}-\int\Big\{\frac{\text{d}}{\text{dx}}\Big(\sec^{-1}\sqrt{\text{x}}\Big)\int1\text{dx}\Big\}\text{dx}$
$=\sec^{-1}\sqrt{\text{x}}.\text{x}-\int\frac{1}{\sqrt{\text{x}}\sqrt{1-\text{x}}}\times\frac{1}{2\sqrt{\text{x}}}\times\text{x dx}$
$=\text{x}\sec^{-1}\sqrt{\text{x}}-\frac{1}{2}\int(1-\text{x})^{-\frac{1}{2}\text{dx}}$
$=\text{x}\sec^{-1}\text{x}-\frac{1}{2}\Bigg[\frac{(1-\text{x})^{-\frac{1}{2}+1}}{\big(-\frac{1}{2}+1\big)(-1)}\Bigg]+\text{C}$
$=\text{x}\sec^{-1}\text{x}+(1-\text{x})^{\frac{1}{2}}+\text{C}$
View full question & answer→Question 2933 Marks
Find the integrals of the functions in Exercises:
$\tan^32\text{x}\sec2\text{x}$
Answer$\tan^32\text{x }\sec2\text{x}=\tan^22\text{x }\tan2\text{x }\sec2\text{x}$
$=(\sec^22\text{x}-1)\tan2\text{x }\sec2\text{x}$
$=\sec^22\text{x }\tan2\text{x }\sec2\text{x}-\tan2\text{x }\sec2\text{x}$
$\therefore\int\tan^32\text{x }\sec2\text{x}\text{ dx}$
$=\int\sec^22\text{x }\tan2\text{x }\sec2\text{x}\text{ dx}-\int\tan2\text{x }\sec2\text{x}\text{ dx}$
$=\int\sec^22\text{x }\tan2\text{x }\sec2\text{x}\text{ dx}-\frac{\sec2\text{x}}{2}+\text{C}$
$\text{Let }\sec2\text{x}=\text{t}$
$\therefore2\sec2\text{x }\tan2\text{x}\text{ dx}=\text{dt}$
$\therefore\int\tan^32\text{x }\sec2\text{x}\text{ dx}\text=\frac{1}{2}\int\text{t}^2\text{dt}-\frac{\sec2\text{x}}{2}+\text{C}$
$=\frac{\text{t}^3}{6}-\frac{\sec2\text{x}}{2}+\text{C}$
$=\frac{(\sec2\text{x})^3}{6}-\frac{\sec2\text{x}}{2}+\text{C}$
View full question & answer→Question 2943 Marks
Write a value of $\int\text{e}^{2\text{x}^2+\ln\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\text{e}^{2\text{x}^2+\ln\text{x}}\text{ dx}$
$=\int\text{e}^{2\text{x}^2}\cdot\text{e}^{\ln{\text{x}}}\text{dx}$
$=\int\text{x}\cdot\text{e}^{2\text{x}^2}\text{dx}$ $\big[\because\text{e}^{\ln\text{x}}=\text{x}\big]$
$=\int\text{x}\cdot\big(\text{e}^{\text{x}^2}\big)\text{dx}$
Let $\text{e}^{\text{x}^2}=\text{t}$
$\text{e}^{\text{x}^2}\cdot2\text{x dx}=\text{dt}$
$\therefore\ \frac{1}{2}\int\text{t dt}$
$=\frac{1}{2}\frac{\text{t}^2}{2}+\text{C}$
$=\frac{1}{4}\text{e}^{2\text{x}^2}+\text{C}$
$\therefore\ \text{I}=\frac{1}{4}\text{e}^{2\text{x}^2}+\text{C}$
View full question & answer→Question 2953 Marks
Evaluate the following integrals:
$\int\text{e}^{\text{x}}(\cos\text{x}-\sin\text{x})\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{x}}(\cos\text{x}-\sin\text{x})\text{dx}$
$=\int\text{e}^{\text{x}}\cos\text{x dx}-\int\text{e}^{\text{x}}\sin\text{x dx}$
Integrating by parts
$=\text{e}^{\text{x}}\cos\text{x}-\int\text{e}^{\text{x}}\Big(\frac{\text{d}}{\text{dx}}\cos\text{x}\Big)\text{dx}-\int\text{e}^{\text{x}}\sin\text{x dx}$
$=\text{e}^{\text{x}}\cos\text{x}+\int\text{e}^{\text{x}}\sin\text{x dx}-\int\text{e}^{\text{x}}\sin\text{x dx}$
$=\text{e}^{\text{x}}\cos\text{x}+\text{C}$
$\therefore\int\text{e}^{\text{x}}(\cos\text{x}-\sin\text{x})\text{dx}=\text{e}^{\text{x}}\cos\text{x}+\text{C}$
View full question & answer→Question 2963 Marks
Evaluate the following integrals:
$\int\frac{\tan\text{x}}{\sec\text{x}+\tan\text{x}}\text{dx}$
Answer$\int\frac{\tan\text{x}}{\sec\text{x}+\tan\text{x}}\text{dx}$
$=\int\frac{\tan\text{x}}{(\sec\text{x}+\tan\text{x})}\times\Big(\frac{\sec\text{x}-\tan\text{x}}{\sec\text{x}-\tan\text{x}}\Big)\text{dx}$
$=\int\frac{\tan\text{x}(\sec\text{x}-\tan\text{x})}{(\sec^2\text{x}-\tan^2\text{x})}\text{dx}$
$=\int\Big(\frac{\sec\text{x}\tan\text{x}-\tan^2\text{x}}{1}\Big)\text{dx}$
$=\int\sec\text{x}\tan\text{x dx}-\int(\sec^2\text{x}-1)\text{dx}$
$=\sec\text{x}-\tan\text{x}+\text{x}+\text{C}$
View full question & answer→Question 2973 Marks
By using the properties of definite integral, evaluate the integral in Exercise:
$\int\limits_{2}^{8}|\text{x}-5|\ \text{dx}$
Answer$\text{Let}\ \text{I}=\int\limits_{2}^{8}|\text{x}-5|\ \text{dx}$ $\text{putting}\ \text{x}-5=0\ \ \Rightarrow\ \text{x}=5\in(2,8)$ $\therefore\ \ \text{from eq. (i)},\ \text{I}=\int\limits_{2}^{5}|\text{x}-5|\ \text{dx}+\int\limits_{5}^{8}|\text{x}-5| \text{dx}$ $=\int\limits_{2}^{5}-(\text{x}-5)\ \text{dx}+\int\limits_{5}^{8}(\text{x}-5)\text{dx}$ $=-\bigg(\frac{\text{x}^{2}}{2}-5\text{x}\bigg)^{-2}_{-5}+\bigg(\frac{\text{x}^{2}}{2}-5\text{x}\bigg)^{5}_{-2}$$=-\bigg[\bigg(\frac{25}{2}-25\bigg)-(10-2)\bigg]+\bigg[(32-40)-\bigg(\frac{25}{2}-25\bigg)\bigg]$
$=25-\frac{25}{2}-8-8-\frac{25}{2}+25=34-\frac{50}{2}=34-25=9$
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Evaluate the following integrals:$\int\text{e}^{\text{x}}(\log\text{x}+\frac{1}{2})\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{x}}(\log\text{x}+\frac{1}{2})\text{dx}$
Here, $\text{f(x)}=\log\text{x}$
$\Rightarrow\text{f}'\text{(x)}=\frac{1}{\text{x}}$
Put $\text{e}^{\text{x}}\text{f(x)}=\text{t}$
$\Rightarrow\text{e}^{\text{x}}\log\text{x}=\text{t}$
Diff. both sides w.r.t x
$\text{e}^{\text{x}}\log\text{x}+\text{e}^{\text{x}}\frac{1}{\text{x}}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{e}^{\text{x}}(\log\text{x}+\frac{1}{\text{x}})\text{dx = dt}$
$\therefore\int\text{e}^{\text{x}}\big[\log\text{x}+\frac{1}{\text{x}}\big]\text{dx}=\int\text{dt}$
$\Rightarrow\text{I}=\text{t}+\text{C}$
$=\text{e}^{\text{x}}\log\text{x}+\text{C}$
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$\int\frac{1}{\sqrt{\text{x+a}}+\sqrt{\text{x+b}}}\text{dx}$
AnswerLet $\text{l}=\int\frac{1}{\sqrt{\text{x+a}}+\sqrt{\text{x+b}}}\text{dx}. $ Then,
$\text{I}=\int\frac{1}{\sqrt{\text{x+a}}+\sqrt{\text{x+b}}}\times\frac{\sqrt{\text{x+a}}-\sqrt{\text{x+b}}}{\sqrt{\text{x+a}}-\sqrt{\text{x+b}}}\times\text{dx}$
$=\int\frac{\sqrt{\text{x+a}}-\sqrt{\text{x+b}}}{\text{x+a}-\text{x-b}}\times\text{dx}$
$=\int\frac{\sqrt{\text{x+a}}-\sqrt{\text{x+b}}}{\text{a}-\text{b}}\times\text{dx}$
$=\frac{1}{\text{a}-\text{b}}\bigg[\frac{2}{3}(\text{x+a})^{\frac{3}{2}}-\frac{2}{3}(\text{x+b})^{\frac{3}{2}}\bigg]+\text{c}$
$=\frac{2}{3(\text{a}-\text{b})}\Big[(\text{x+a})^{\frac{3}{2}}-(\text{x+b})^{\frac{3}{2}}\Big]+\text{c}$
$ \text{I}=\frac{2}{3(\text{a}-\text{b})}\Big[(\text{x+a})^{\frac{3}{2}}-(\text{x+b})^{\frac{3}{2}}\Big]+\text{c}$
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Evaluate $\int\frac{1}{\text{x}(1+\log\text{x})}\text{ dx}$
Answer$\text{I}=\int\frac{1}{\text{x}(1+\log\text{x})}\text{ dx}$
Let $(1+\log\text{x})=\text{t}$
Or, $\frac{1}{\text{x}}\text{dx}=\text{dt}$
$\text{I}=\int\frac{1}{\text{t}}\text{dt}$
$\text{I}=\log|\text{t}|+\text{C}$
$\therefore\ \text{I}=\log|1+\log\text{x}|+\text{C}$
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