Question 2513 Marks
Evaluate the following integrals:
$\int\limits^{{\pi}}_0\cos^5\text{x dx}$
AnswerLet $\text{I}=\int\limits^{{\pi}}_0\cos^5\text{x dx}$
$=\int\limits^{{\pi}}_0\cos\text{x}\big(\cos^2\text{x}\big)^2\text{dx}$
$=\int\limits^{{\pi}}_0\cos\text{x}\big(1-\sin^2\text{x}\big)^2\text{dx}$
Let $\sin\text{x}=\text{t},$ then $\cos\text{x dx}=\text{dt}$
When, $\text{x}\rightarrow0;\text{ t}\rightarrow0$ and $\text{x}\rightarrow\pi;\text{ t}\rightarrow0$
Therefore,
$\text{I}=\int\limits^0_0\big(1-\text{t}^2\big)^2\text{dt}$
$\text{I}=0$
View full question & answer→Question 2523 Marks
Evaluate the following integrals:
$\int\frac{\text{cosec }\text{x}}{\text{cosec }\text{x}-\cot\text{x}}\text{dx}$
Answer$\int\frac{\text{cosec }\text{x}}{\text{cosec }\text{x}-\cot\text{x}}\text{dx}$
$=\int\frac{\text{cosec }\text{x}}{\text{cosec }\text{x}-\cot\text{x}}\times\frac{\text{cosec x}+\cot\text{x}}{\text{cosec x}+\cot\text{x}}\times\text{dx}$
$=\int\frac{\text{cosec x}(\text{cosec x}+\cot\text{x})}{\text{cosec}^2\text{x}-\cot^2\text{x}}\text{dx}$
$=\int(\text{cosec}^2\text{x}+\text{cosec x}\cot\text{x})\text{dx}$
$=\int\text{cosec}^2\text{x dx}+\int\text{cosec x dx}$
$=-\cot\text{x}-\text{cosec x}+\text{C}$
$\therefore\ \int\frac{\text{cosec }\text{x}}{\text{cosec }\text{x}-\cot\text{x}}\text{dx}=-\cot\text{x}-\text{cosec x}+\text{C}$
View full question & answer→Question 2533 Marks
Write a value of $\int\text{x}^2\sin\text{x}^3\text{ dx}$
AnswerLet $\text{I}=\int\text{x}^2\sin\text{x}^3\text{ dx}$
Let $\text{x}^3=\text{t}$
$=3\text{x}^2\text{dx}=\text{dt}$
$=\text{x}^2\text{dx}=\frac{1}{3}\text{dt}$
$\therefore\ \text{I}=\frac{1}{3}\int\sin\text{t dt}$
$=\frac{1}{3}(-\cos\text{t})+\text{C}$
Hence, $\text{I}=\frac{-1}{3}\cos\text{x}^3+\text{C}$
View full question & answer→Question 2543 Marks
$\int(\text{e}^{\text{x}}+1)^2\text{e}^{\text{x}}\text{ dx}$
AnswerConsider $\text{I}=\int(\text{e}^{\text{x}}+1)^2\text{e}^{\text{x}}\text{dx}$
Let $(\text{e}^{\text{x}}+1)=\text{t}\rightarrow\text{e}^{\text{x}}\text{dx}=\text{dt}$
$\text{I}=\int(\text{e}^\text{x}+1)^2\text{e}^{\text{x}}\text{dx}$
$=\int(\text{t})^2\text{dt}$
$=\frac{\text{t}^3}{3}+\text{c}$
$=\frac{(\text{e}^{\text{x}}+1)^3}{3}+\text{c}$
View full question & answer→Question 2553 Marks
$\int\cos^42\text{x dx}$
Answer$\int\cos^42\text{x dx}$
$=\int(\cos^22\text{x})^2\text{dx}$
$=\int\Big(\frac{1+\cos4\text{x}}{2}\Big)^2\text{dx}$ $\Big[\therefore\cos^2\text{x}=\frac{1+\cos2\text{x}}{2}\Big]$
$=\frac{1}{4}\int(1+\cos4\text{x})^2\text{dx}$
$=\frac{1}{4}\int(1+\cos^24\text{x}+2\cos4\text{x})\text{dx}$
$=\frac{1}{4}\Big[1+\Big(\frac{1+\cos8\text{x}}{2}\Big)+2\cos4\text{x}\Big]\text{dx}$
$=\frac{1}{4}\int\Big(\frac{3}{2}+\frac{\cos8\text{x}}{2}+2\cos4\text{x}\Big]\text{dx}$
$=\frac{1}{4}\Big[\frac{3\text{x}}{2}+\frac{\sin8\text{x}}{16}+\frac{2\sin4\text{x}}{4}\Big]+\text{C}$
$=\frac{3\text{x}}{8}+\frac{\sin8\text{x}}{64}+\frac{\sin4\text{x}}{8}+\text{C}$
View full question & answer→Question 2563 Marks
Evaluate the following integrals:
$\int\limits^8_2|\text{x}-5|\text{dx}$
Answer$\int\limits^8_2|\text{x}-5|\text{dx}$
We know that,
$|\text{x}-5|=\begin{cases}-(\text{x}-5),&2\leq\text{x}\leq5\\\text{x}-5,&5<\text{x}\leq8\end{cases}$
$\therefore\ \text{I}=\int\limits^8_2|\text{x}-5|\text{dx}$
$\Rightarrow\text{I}=\int\limits^5_2-(\text{x}-5)\text{dx}+\int\limits^8_5(\text{x}-5)\text{dx}$
$\Rightarrow\text{I}=-\Big[\frac{\text{x}^2}{2}-5\text{x}\big]^5_2+\Big[\frac{}{}\frac{\text{x}^2}{2}-5\text{x}\Big]^8_5$
$\Rightarrow\text{I}=\frac{-25}{2}+25+2-10+32-40-\frac{25}{2}+25$
$\Rightarrow\text{I}=9$
View full question & answer→Question 2573 Marks
Evaluate the following intregals:
$\int\frac{1}{\cos2\text{x}+3\sin^2\text{x}}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{1}{\cos2\text{x}+3\sin^2\text{x}}\ \text{dx}$
$=\int\frac{1}{2\cos^2\text{x}-1+3\sin^2\text{x}}\ \text{dx}$
$\text{I}=\frac{1}{\sqrt{2}}\int\frac{1}{1+\text{t}^2}$
Dividing numerator and denominator by $\cos^2\text{x}$
$\text{I}=\int\frac{\sec^2\text{x}}{2-\sec^2\text{x}+3\tan^2\text{x}}\ \text{dx}$
$=\int\frac{\sec^2\text{x}}{2-(1+\tan^2\text{x})^2+3\tan^2\text{x}}\ \text{dx}$
$=\int\frac{\sec^2\text{x}}{2-1-\tan^2\text{x}+3\tan^2\text{x}}\ \text{dx}$
$=\int\frac{\text{dt}}{1+2\tan^2\text{x}}$
Let $\sqrt{2}\tan\text{x}=\text{t}$
$\sqrt{2}\sec^2\text{x dx}=\text{dt}$
$=\frac{1}{\sqrt{2}}\tan^{-1}\text{t}+\text{C}$
$\text{I}=\frac{1}{\sqrt{2}}\tan^{-1}(\sqrt{2}\tan\text{x})+\text{C}$
View full question & answer→Question 2583 Marks
Evaluate the following integrals:
$\int\frac{1+\cos\text{x}}{(\text{x}+\sin\text{x})^3}\text{dx}$
Answer$\int\frac{(1+\cos\text{x})}{(\text{x}+\sin\text{x})^3}\text{dx}$
$\text{Let x}+\sin\text{x}=\text{t}$
$\Rightarrow(1+\cos\text{x})=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow(1+\cos\text{x})\text{dx}={\text{dt}}$
$\text{Now,}\int\frac{(1+\cos\text{x})}{(\text{x}+\sin\text{x})^3}\text{dx}$
$=\int\frac{\text{dt}}{\text{t}^3}$
$=\int\text{t}^{-3}\text{dt}$
$=\frac{\text{t}^{-3+1}}{-3+1}+\text{C}$
$=\frac{-1}{2\text{t}^2}+\text{C}$
$=\frac{-1}{2(\text{x}+\sin\text{x})^2}+\text{C}$
View full question & answer→Question 2593 Marks
Evaluate the following intregals:
$\int\frac{1}{\sin^2\text{x}-\sin2\text{x}}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{1}{\sin^2\text{x}-\sin(2\text{x})}\ \text{dx}$
$=\int\frac{1}{\sin^2\text{x}+2\sin\text{x}\cos\text{x}}\text{ dx}$
Dividing numerator and denominator by $\cos^2\text{x}$
$\text{I}=\int\frac{\sec^2\text{x}}{\tan\text{x}+2\tan\text{x}}\ \text{dx}$
Let $\tan\text{x}=\text{t}$
$\sec^2\text{x dx}=\text{dt}$
$\therefore\text{I}=\int\frac{\text{dt}}{\text{t}^2+2\text{t}}$
$=\int\frac{\text{dt}}{\text{t}^2+2\text{t}+1-1}$
$=\ln\frac{\text{dt}}{(\text{t}+1)^2-(-1)^2}$
$=\frac{1}{2}\ln\Big|\frac{\text{t}}{\text{t}+2}\Big|+\text{C}$
$=\frac{1}{2}\ln\Big|\frac{\tan\text{x}}{\tan\text{x}+2}\Big|+\text{C}$
View full question & answer→Question 2603 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{2}}_0\log\Big(\frac{3+5\cos\text{x}}{3+5\sin\text{x}}\Big)\text{dx}$
AnswerLet, $\text{I}=\int\limits^{\frac{\pi}{2}}_0\log\Big(\frac{3+5\cos\text{x}}{3+5\sin\text{x}}\Big)\text{dx}\ ...(\text{i})$
$=\int\limits^{\frac{\pi}{2}}_0\log\Bigg[\frac{3+5\cos\big(\frac{\pi}{2}-\text{x}\big)}{3+5\sin\big(\frac{\pi}{2}-\text{x}\big)}\Bigg]$
$=\int\limits^{\frac{\pi}{2}}_0\log\Big(\frac{3+5\sin\text{x}}{3+5\cos\text{x}}\Big)\text{dx}\ ...(\text{ii})$
Adding (i) and (ii)
$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\bigg[\log\Big(\frac{3+5\cos\text{x}}{3+5\sin\text{x}}\Big)+\log\Big(\frac{3+5\sin\text{x}}{3+5\cos\text{x}}\Big)\bigg]\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_0\log\Big(\frac{3+5\cos\text{x}}{3+5\sin\text{x}}\times\frac{3+5\sin\text{x}}{3+5\cos\text{x}}\Big)\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_0\log1\text{ dx}=0$
Hence, $\text{I}=0$
View full question & answer→Question 2613 Marks
Evaluate:
$\int\frac{\text{e}^{\log\sqrt{\text{x}}}}{\text{x}}\text{dx}$
Answer$\int\frac{\text{e}^{\log\sqrt{\text{x}}}}{\text{x}}\text{dx}=\int\frac{\sqrt{\text{x}}}{\text{x}}\text{dx}$
$=\int\text{x}^\frac{1}{2}\times\text{x}^{-1}\text{dx}$
$=\int\text{x}^{\frac{1}{2}-1}\text{dx}$
$=\int\frac{\text{x}^{\frac{-1}{2}+1}+\text{c}}{\frac{-1}{2}+1}$
$=\frac{\text{x}^\frac{1}{2}}{\frac{1}{2}}$
$=\sqrt{\text{x}}+\text{c}$
View full question & answer→Question 2623 Marks
Evaluate the following integrals:
$\int(\text{a}\tan\text{x}+\text{b}\cot \text{x})^2\text{dx}$
Answer$\int(\text{a}\tan\text{x}+\text{b}\cot \text{x})^2\text{dx}$
$=\int(\text{a}^2\tan^2\text{x}+\text{b}^2\cot^2\text{x}+2\text{ab}\tan\text{x}\cot\text{x})\text{dx}$
$=\text{a}^2\int\tan^2\text{x dx}+\text{b}^2\int\cot^2\text{x dx}+2\text{ab}\int\text{dx}$
$=\text{a}^2\int(\sec^2\text{x}-1)\text{dx}+\text{b}^2\int(\text{cosec}^2\text{x}-1)\text{dx}+2\text{ab}\int\text{dx}$
$=\text{a}^2[\tan\text{x}-\text{x}]+\text{b}^2[-\cot\text{x}-\text{x}]+2\text{ab}\text{x}+\text{C}$
$=\text{a}^2\tan\text{x}-\text{b}^2\cot\text{x}-(\text{a}^2+\text{b}^2-2\text{ab})\text{x}+\text{C}$
View full question & answer→Question 2633 Marks
Integrate the function in Exercise.
$\sqrt{1+\frac{\text{x}^2}{9}}$
Answer$\int\sqrt{1+\frac{\text{x}^2}{9}}\text{dx}=\int\sqrt{\frac{9+\text{x}^2}{9}}\text{dx}=\int\frac{\sqrt{\text{x}^2+3^2}}{3}\text{dx}=\frac{1}{3}\int\sqrt{\text{x}^2+3^2}\text{dx}$
$=\frac{1}{3}\Bigg[\frac{\text{x}}{2}\sqrt{\text{x}^2+3^2}+\frac{3}{2}\text{log}\Big|\text{x}+\sqrt{\text{x}^2+3^2}\Big|\Bigg]+\text{c}\Bigg[\therefore\ \int\sqrt{\text{x}^2+\text{a}^2}\text{dx}=\frac{\text{x}}{2}\sqrt{\text{x}^2+\text{a}^2}+\frac{\text{a}^2}{2}\text{log}\Big|\text{x}+\sqrt{\text{x}^2+\text{a}^2}\Big|\Bigg]+\text{c}$
$=\frac{\text{x}}{6}\sqrt{\text{x}^2+9}+\frac{3}{2}\text{log}\Big|\text{x}+\sqrt{\text{x}^2+9}\Big|+\text{c}$
View full question & answer→Question 2643 Marks
Evaluate the following integrals:
$\int\sin^{-1}\Big(\frac{2\tan\text{x}}{1+\tan^2\text{x}}\Big)\text{dx}$
Answer$\int\sin^{-1}\Big(\frac{2\tan\text{x}}{1+\tan^2\text{x}}\Big)\text{dx}$
$=\int\sin^{-1}(\sin2\text{x} )\text{dx}$ $\Big[\because\ \sin2\text{x}=\frac{2\tan\text{x}}{1+\tan^2\text{x}}\Big]$
$=\int2\text{ x dx}$
$=2\int\text{x dx}$
$=\frac{2\text{x}^2}{2}+\text{C}$
$=\text{x}^2+\text{C}$
$\therefore\ \int\sin^{-1}\Big(\frac{2\tan\text{x}}{1+\tan^2\text{x}}\Big)\text{dx}=\text{x}^2+\text{C}$
View full question & answer→Question 2653 Marks
Evaluate the following integrals:
$\int\frac{1-\cos2\text{x}}{1+\cos2\text{x}}\text{dx}$
Answer$\int\frac{1-\cos2\text{x}}{1+\cos2\text{x}}\text{dx}$
$=\int\frac{2\sin^2\text{x}}{2\cos^2\text{x}}\text{dx}$
$=\int\tan^2\text{x dx}$
$=\int(\sec^2\text{x}-1)\text{dx}$
$=\int\sec^2\text{x dx}-\int\text{dx}$
$=\tan\text{x}-\text{x}+\text{C}$
$\therefore\ \int\frac{1-\cos2\text{x}}{1+\cos2\text{x}}\text{dx}=\tan\text{x}-\text{x}+\text{C}$
View full question & answer→Question 2663 Marks
Integrate the rational function in exercise:
$\frac{1}{\text{x}(\text{x}^\text{n}+1)}$
[Hint: multiply numerator and denominator by $x^{n – 1}$ and put $x^n = t]$
Answer$\text{I}=\int\frac{1}{\text{x}(\text{x}^\text{n}+1)}\text{dx}$
Multiplying both numerator and denominator by $nx^{n-1},$
$\Big[\therefore \ \frac{\text{d}}{\text{dx}}(\text{x}^\text{n}+1)=\text{nx}^{\text{n}-1}\Big]$
$\text{I}=\int\frac{\text{nx}^{\text{n}-1}}{\text{nx}^{\text{n}-1}.\text{x}(\text{x}^\text{n}+1)}\text{dx}=\frac{1}{\text{n}}\int\frac{\text{nx}^{\text{n}-1}}{\text{x}^{\text{n}}(\text{x}^\text{n}+1)}\text{dx}$
Putting $x^n = t$
$\Rightarrow nx^{n-1} = \frac{\text{dx}}{\text{dt}}$
$\Rightarrow nx^{n-1} dx = dt$
$\therefore$ From eq. $(i), $
$\text{I}=\frac{1}{\text{n}}\int\frac{\text{dt}}{\text{t}(\text{t}+1)}=\frac{1}{\text{n}}\int\frac{1}{\text{t}(\text{t}+1)}\text{dt}=\frac{1}{\text{n}}\int\frac{\text{t}+1-\text{t}}{\text{t}(\text{t}+1)}\text{dt}$
$=\frac{1}{\text{n}}\int\frac{\text{t}+1}{\text{t}(\text{t}+1)}-\int\frac{1}{\text{t}(\text{t}+1)}\text{dt}=\frac{1}{\text{n}}\Bigg[\int\frac{1}{\text{t}}\text{dt}+\int\frac{1}{\text{t}+1}\text{dt}\Bigg]$
$=\frac{1}{\text{n}}\Big[\text{log}|\text{t}|-\text{log}|\text{t}+1|\Big]+\text{c}=\frac{1}{\text{n}}\Big[\text{log}|\text{x}^\text{n}|-\text{log}|\text{x}^\text{n}+1|\Big]+\text{c}$
$=\frac{1}{\text{n}}\text{log}\Bigg|\frac{\text{x}^\text{n}}{\text{x}^\text{n}+1}\Bigg|+\text{c}$
View full question & answer→Question 2673 Marks
Evaluate the following integrals:
$\int_{0}^\limits{1}\text{xe}^{\text{x}^2}\text{ dx}$
AnswerLet $\text{x}^2=\text{t}$
Differentiating w.r.t. x, we get
$2\text{xdx}=\text{dt}$
Now, $\text{x}=0\Rightarrow\text{t}=0$
$\text{x}=1\Rightarrow\text{t}=1$
$\therefore\ \int_{0}^\limits{1}\text{xe}^{\text{x}^2}\text{ dx}=\int_{0}^\limits{1}\frac{\text{e}^{\text{t}}\text{ dt}}{2}$
$=\frac{1}{2}\int_{0}^\limits{1}\text{e}^{\text{t}}\text{ dt}$
$=\frac{1}{2}\big[\text{e}^{\text{t}}\big]^1_0$
$=\frac{1}{2}\big[\text{e}^1-\text{e}_0\big]$ $\big[\because\text{e}^0=1\big]$
$=\frac{1}{2}\big(\text{e}-1\big)$
View full question & answer→Question 2683 Marks
Evalute the following integrals:
$\int\frac{\sec\text{x}}{\sec2\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\frac{\sec\text{x}}{\sec2\text{x}}\text{dx},$ then
$\text{I}=\int\frac{\frac{1}{\cos\text{x}}}{\frac{1}{\cos2\text{x}}}\text{dx}$
$=\int\frac{\cos2\text{x}}{\cos\text{x}}\text{dx}$
$=\int\frac{2\cos^2\text{x}-1}{\cos\text{x}}\text{dx}$
$=\int2\cos\text{ x dx}-\int\frac{1}{\cos\text{x}}\text{dx}$
$=2\int\cos\text{dx}-\int\sec\text{x dx}$
$=2\sin\text{x}-\log|\sec\text{x}+\tan\text{x}|+\text{C}$
$\because\text{I}=2\sin\text{x}-\log|\sec\text{x}+\tan\text{x}|+\text{C}$
View full question & answer→Question 2693 Marks
Evaluate the following definite integrals:
$\int_{0}^\limits{\pi}\Big(\sin^2\frac{\text{x}}{2}-\cos^2\frac{\text{x}}{2}\Big)\text{dx}$
AnswerLet $\text{I}=\int_{0}^\limits{\pi}\Big(\sin^2\frac{\text{x}}{2}-\cos^2\frac{\text{x}}{2}\Big)\text{dx}$
$=-\int_{0}^\limits{\pi}\Big(\cos^2\frac{\text{x}}{2}-\sin^2\frac{\text{x}}{2}\big)\text{dx}$
$=-\int_{0}^\limits{\pi}\cos\text{x dx}$
$\int\cos\text{x dx}=\sin\text{x}=\text{F(x)}$
By second fundamental theorem of calculus, we obtain
$\text{I}=\text{F}(\pi)-\text{F}(0)$
$=\sin(\pi)-\sin0$
$=0$
View full question & answer→Question 2703 Marks
Integrate the function in Exercise:$\tan^{-1}\sqrt{\frac{1-\text{x}}{1+\text{x}}}$
Answer$\text{I}=\tan^{-1}\sqrt{\frac{1-\text{x}}{1+\text{x}}}\text{dx}$
$\text{Let x}=\cos\theta\Rightarrow\text{dx}=-\sin\theta\text{d}\theta$
$\text{I}=\int\tan^{-1}\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}(-\sin\theta\text{d}\theta)$
$=-\int^{-1}\sqrt{\frac{2\sin^{2}\frac{\theta}{2}}{2\cos^{2}\frac{\theta}{2}}}\sin\theta\text{d}\theta$
$=-\int\tan^{-1}\tan\frac{\theta}{2}.\sin\theta\text{d}\theta$
$=-\frac{1}{2}\big[\theta.(-\cos\theta)-\int1.(-\cos\theta)\text{d}\theta\big]$
$=-\frac{1}{2}[-\theta\cos\theta+\sin\theta]$
$=+\frac{1}{2}\theta\cos\theta-\frac{1}{2}\sin\theta$
$=\frac{1}{2}\cos^{-1}\text{x}.\text{x}-\frac{1}{2}\sqrt{1-\text{x}^{2}}+\text{C}$
$=\frac{\text{x}}{2}\cos^{-1}\text{x}-\frac{1}{2}\sqrt{1-\text{x}^{2}}+\text{C}$
$=\frac{1}{2}\Big(\text{x}\cos^{-1}\text{x}-\sqrt{1-\text{x}^{2}}\Big)+\text{C}$
View full question & answer→Question 2713 Marks
Evaluate the following integrals:
$\int^\limits{\frac{\pi}{2}}_{0}|\cos2\text{x}|\text{dx}$
Answer$\int^\limits{\frac{\pi}{2}}_{0}|\cos2\text{x}|\text{dx}$
We know that,
$|\cos2\text{x}|=\begin{cases}-\cos2\text{x},&\frac{\pi}{4}\leq\text{x}\leq\frac{\pi}{2}\\\cos2\text{x},&0<\text{x}\leq\frac{\pi}{4}\end{cases}$
$\therefore\ \text{I}=\int^\limits{2}_{-2}|\cos2\text{x}|\text{dx}$
$\Rightarrow\text{I}=\int^\limits{\frac{\pi}{4}}_{0}\cos2\text{x }\text{dx}-\int^\limits{\frac{\pi}{2}}_{\frac{\pi}{4}}\cos2\text{x }\text{dx}$
$\Rightarrow\text{I}=\Big[\frac{\sin2\text{x}}{2}\Big]^{\frac{\pi}{4}}_0-\Big[\frac{\sin2\text{x}}{2}\Big]^{\frac{\pi}{2}}_{\frac{\pi}{4}}$
$\Rightarrow\text{I}=\frac{1}{2}-0-0+\frac{1}{2}$
$\Rightarrow\text{I}=1$
View full question & answer→Question 2723 Marks
Evaluate the following definite integrals:
$\int_{0}^\limits{\frac{\pi}{2}}\sin\text{x }\sin2\text{x}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\sin\text{x }\sin2\text{x}\text{ dx}$ Then,
$\text{I}=\int_{0}^\limits{\frac{\pi}{2}}2\sin^2\text{x }\cos\text{x}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{\frac{\pi}{2}}2(1-\cos^2\text{x})\cos\text{x}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\big(2\cos\text{x}-2\cos^3\text{x}\big)\text{dx}$
$\Rightarrow\text{I}=\bigg[2\sin\text{x}-2\Big(\sin\text{x}-\frac{\sin^3\text{x}}{3}\Big)\bigg]^{\frac{\pi}{2}}_0$
$\Rightarrow\text{I}=\Big[2-2\Big(1-\frac{1}{3}\Big)\Big]-0$
$\Rightarrow\text{I}=\frac{2}{3}$
View full question & answer→Question 2733 Marks
Evaluate the following integrals:
$\int\frac{1}{1-\cos\text{x}}\text{dx}$
Answer$\int\frac{1}{1-\cos\text{x}}\text{dx}$
$=\int\frac{1}{1-\cos\text{x}}\times\frac{1+\cos\text{x}}{1+\cos\text{x}}\times\text{dx}$
$=\int\frac{1+\cos\text{x}}{1-\cos^2\text{x}}\times\text{dx}$
$=\int\frac{1+\cos\text{x}}{\sin^2\text{x}}\times\text{dx}$
$=\int\frac{1}{\sin^2\text{x}}\text{dx}+\int\frac{\cos\text{x}}{\sin^2\text{x}}\text{dx}$
$=\int\text{cosec}^2\text{x dx}+\int\cot\text{x}\times\text{cosec x dx}$
$=-\cot\text{x}-\text{cosec x}+\text{C}$
$\therefore\ \int\frac{1}{1-\cos\text{x}}\text{dx}=-\cot\text{x}-\text{cosec x}+\text{C}$
View full question & answer→Question 2743 Marks
Evaluate the following integrals:
$\int\text{x e}^{\text{x}^2}=\text{dx}$
AnswerLet $\text{I}=\int\text{x e}^{\text{x}^2}=\text{dx}\ ....(1)$ Let $\text{x}^2=\text{t}$ then, $\text{d}(\text{x}^2)=\text{dt}$ $\Rightarrow2\text{x dx}=\text{dt}$ $\Rightarrow\text{x dx}=\frac{\text{dt}}{2}$ Putting $\text{x}^2=\text{t}$ and $\text{x dx}=\frac{\text{dt}}{2}$ in equation (1),we get,
$\text{I}=\int\text{e}^\text{t}\frac{\text{dt}}{2}$
$=\frac{1}{2}\text{e}^\text{t}+\text{C}$
$=\frac{1}{2}\text{e}^{\text{x}^2}+\text{C}$
$\text{I}=\frac{1}{2}\text{e}^{\text{x}^2}+\text{C}$
View full question & answer→Question 2753 Marks
Evaluate the following definite integrals:
$\int_{0}^\limits{2}\frac{1}{\sqrt{3+2\text{x}-\text{x}^2}}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{2}\frac{1}{\sqrt{3+2\text{x}-\text{x}^2}}\text{ dx}$ Then,
$\text{I}=\int_{0}^\limits{2}\frac{1}{\sqrt{-\text{x}^2+2\text{x}-1+1+3}}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{2}\frac{1}{\sqrt{-(\text{x}-1)^2+4}}\text{ dx}$
$\Rightarrow\text{I}=\Big[\sin^{-1}\frac{(\text{x}-1)}{2}\Big]^{2}_0$
$\Rightarrow\text{I}=\sin^{-1}\frac{1}{2}-\sin^{-1}\Big(-\frac{1}{2}\Big)$
$\Rightarrow\text{I}=2\sin^{-1}\frac{1}{2}$
$\Rightarrow\text{I}=2\times\frac{\pi}{6}$
$\Rightarrow\text{I}=\frac{\pi}{3}$
View full question & answer→Question 2763 Marks
Evaluate the following integrals:
$\int\frac{(\text{x}+1)\text{e}^\text{x}}{\cos^2(\text{xe}^\text{x})}\text{ dx}$
Answer$\int\frac{(\text{x}+1)\text{e}^\text{x}}{\cos^2(\text{xe}^\text{x})}\text{ dx}$
Let $\text{x}\text{e}^\text{x}=\text{t}$
$\Rightarrow\big(1.\text{e}^\text{x}+\text{x}\text{e}^\text{x}\big)=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow(\text{x}+1)\text{e}^\text{x}\text{ dx}=\text{dt}$
Now, $\int\frac{(\text{x}+1)\text{e}^\text{x}}{\cos^2(\text{xe}^\text{x})}\text{ dx}$
$=\int\frac{\text{dt}}{\cos^2\text{t}}$
$=\sec^2\text{t}\text{ dt}$
$=\tan(\text{t})+\text{C}$
$=\tan\big(\text{xe}^\text{x}\big)+\text{C}$
View full question & answer→Question 2773 Marks
Find the integrals of the functions in Exercises:
$\frac{\cos\text{x}}{1+\cos\text{x}}$
Answer$\frac{\cos\text{x}}{1+\cos\text{x}}=\frac{\cos^2\frac{\text{x}}{2}-\sin^2\frac{\text{x}}{2}}{2\cos^2\frac{\text{x}}{2}}$ $\ \ \ \ \ \ \ \ \bigg[\cos\text{x}=\cos^2\frac{\text{x}}{2}-\sin^2\frac{\text{x}}{2}\text{ and }\cos\text{x}=2\cos^2\frac{\text{x}}{2}-1\bigg]$
$=\frac{1}{2}\bigg[1-\tan^2\frac{\text{x}}{2}\bigg]$
$\therefore\int\frac{\cos\text{x}}{1+\cos\text{x}}\text{ dx}=\frac{1}{2}\int\bigg(1-\tan^2\frac{\text{x}}{2}\bigg)\text{ dx}$
$=\frac{1}{2}\int\bigg(1-\sec^2\frac{\text{x}}{2}+1\bigg)\text{dx}$
$=\frac{1}{2}\int\bigg(2-\sec^2\frac{\text{x}}{2}\bigg)\text{dx}$
$=\frac{1}{2}\Bigg[2\text{x}-\frac{\tan\frac{\text{x}}{2}}{\frac{1}{2}}\Bigg]+\text{C}$
$=\text{x}-\tan\frac{\text{x}}{2}+\text{C}$
View full question & answer→Question 2783 Marks
Evaluate the following integrals:
$\int\cos^5\text{x}\text{ dx}$
Answer$\int\cos^5\text{x}\text{ dx}$
$=\int\cos^4\text{x}\cdot\cos\text{x}\text{ dx}$
$=\int(1-\sin^2\text{x})^2\cos\text{x}\text{ dx}$
Let $\sin\text{x}=\text{t}$
$\cos\text{x}\text{ dx}=\text{dt}$
Now, $\int(1-\sin^2\text{x})^2\cos\text{x}\text{ dx}$
$=\int(1-\text{t}^2)^2\text{ dt}$
$=\int(1+\text{t}^4-2\text{t}^2)\text{dt}$
$=\int\text{dt}+\int\text{t}^4\text{ dt}-2\int\text{t}^2\text{ dt}$
$=\text{t}+\frac{\text{t}^5}{5}-\frac{2\text{t}^3}{3}+\text{C}$
$=\sin\text{x}+\frac{\sin^5\text{x}}{5}-\frac{2}{3}\sin^3\text{x}+\text{C}$
View full question & answer→Question 2793 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 2803 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 2813 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 2823 Marks
Evaluate the following integrals:
$\int\frac{1}{\sqrt{\tan^{-1}\text{x}}.(1+\text{x}^2)}\text{dx}$
Answer$\int\frac{\text{dx}}{\sqrt{\tan^{-1}\text{x}}(1+\text{x}^2)}$
$\text{Let }\tan^{-1}\text{x}=\text{t}$
$\Rightarrow\frac{1}{1+\text{x}^2}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\frac{1}{1+\text{x}^2}\text{dx}=\text{dt}$
$\text{Now,}\int\frac{\text{dx}}{\sqrt{\tan^{-1}\text{x}}(1+\text{x}^2)}$
$=\int\frac{\text{dt}}{\sqrt{\text{t}}}$
$=\int\text{t}^{-\frac{1}{2}}\text{dt}$
$=\frac{\text{t}^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+\text{C}$
$=2\sqrt{\text{t}}+\text{C}$
$=2\sqrt{\tan^{-1}\text{x}}+\text{C}$
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:
$\int\frac{\sqrt{1+\text{x}^2}}{\text{x}^4}\text{dx}$
AnswerLet $\text{I}=\int\frac{\sqrt{1+\text{x}^2}}{\text{x}^4}\text{dx}$ $=\int\frac{\sqrt{1+\text{x}^2}}{\text{x}}\cdot\frac{1}{\text{x}^3}\text{dx}$
$=\int\sqrt{\frac{1+\text{x}^2}{\text{x}^2}}\cdot\frac{1}{\text{x}^3}\text{dx}$ $=\int\sqrt{\frac{1}{\text{x}^2}+1}\cdot\frac{1}{\text{x}^3}\text{dx}$
Put $1+\frac{1}{\text{x}^2}=\text{t}^2\Rightarrow\frac{-2}{\text{x}^3}\text{dx}=2\text{tdt}$
$\Rightarrow-\frac{1}{\text{x}^3}=\text{tdt}$
$\therefore\ \text{I}=-\int\text{t}^2\text{dt}=-\frac{\text{t}^3}{3}+\text{C}$ $=-\frac{1}{3}\Big(1+\frac{1}{\text{x}^2}\Big)^{\frac{3}{2}}+\text{c}$
View full question & answer→Question 2853 Marks
Evaluate the following integrals:
$\int\frac{1}{\text{x}}(\log\text{x})^2\text{dx}$
AnswerLet I $=\int\frac{1}{\text{x}}(\log\text{x})^2\text{dx}\ .....(1)$
Let $\log\text{x}=\text{t}$ then,
$\text{d}(\log\text{x})=\text{dt}$
$\Rightarrow\frac{1}{\text{x}}\text{dx}=\text{dt}$
Putting $\log\text{x}=\text{t}$ and $\frac{1}{\text{x}}\text{dx}=\text{dt}$ in equation (1), we get
$\text{I}=\int\text{t}^2\text{dt}$
$=\frac{\text{t}^3}{3}+\text{C}$
$=\frac{(\log\text{x})^3}{3}+\text{C}$
$\therefore\text{I}=\frac{1}{3}(\log\text{x})^3+\text{C}$
View full question & answer→Question 2863 Marks
Evaluate the following integrals:
$\int\frac{(\text{x}+1)\text{e}^\text{x}}{\sin^2(\text{xe}^\text{x})}\text{ dx}$
Answer$\int\frac{(\text{x}+1)\text{e}^\text{x}}{\sin^2(\text{xe}^\text{x})}\text{ dx}$ Let $\text{xe}^\text{x}=\text{t}$ $\Rightarrow(1.\text{e}^\text{x}+\text{xe}^\text{x})=\frac{\text{dt}}{\text{dx}}$ $\Rightarrow(\text{x}+1)\text{e}^\text{x}\text{dx}=\text{dt}$ Now, $\int\frac{(\text{x}+1)\text{e}^\text{x}}{\sin^2({\text{xe}^\text{x}})}=\text{dx}$$=\int\frac{\text{dt}}{\sin^2\text{t}}$
$=\int\text{cosec}^2\text{t}\text{ dt}$
$=-\cot(\text{t})+\text{C}$
$=-\cot(\text{xe}^\text{x})+\text{C}$
View full question & answer→Question 2873 Marks
Integrate the function in Exercise:
$\frac{\sin\text{x}}{\sin(\text{x}-\text{a)}}$
Answer$\frac{\sin\text{x}}{\sin(\text{x}-\text{a)}}$
$\text{Let}\ \text{x}-\text{a}=\text{t}\Rightarrow\text{dx}=\text{dt}$
$\int\frac{\sin\text{x}}{\sin(\text{x}-\text{a)}}\text{dx}=\int\frac{\sin(\text{t}+\text{a})}{\sin\text{t}}\text{dt}$
$=\int\frac{\sin\text{t}\cos\text{a}+\cos\text{t}\sin\text{a}}{\sin\text{t}}\text{dt}$
$=\int(\cos\text{a}+\cot\text{t}\sin\text{a)}\text{dt}$
$=\text{t}\cos\text{a}+\sin\text{a}\log|\sin\text{t|}+\text{C}_{1}$
$=(\text{x}-\text{a)}\cos\text{a}+\sin\text{a}\log|\sin(\text{x}-\text{a)}|+\text{C}_{1}$
$=\text{x}\cos\text{a}+\sin\text{a}\log|\sin(\text{x}-\text{a)}|-\text{a}\cos\text{a}+\text{C}_{1}$
$=\sin\text{a}\log|\sin(\text{x}-\text{a)}|+\text{x}\cos\text{a}+\text{C}$
View full question & answer→Question 2883 Marks
Evaluate the following:
$\int\frac{(1+\cos\text{x})}{\text{x}+\sin\text{x}}\text{dx}$
AnswerConsider that, $\text{I}=\int\frac{(1+\cos\text{x})}{\text{x}+\sin\text{x}}\text{dx}$
Let $\text{x}+\sin\text{x}=\text{t}$
$\Rightarrow\ (1+\cos\text{x})\text{dx}=\text{dt}$
Substituting $\text{x}+\sin\text{x}=\text{t}$ and $(1+\cos\text{x})\text{dx}=\text{dt}$ in I, we get
$\text{I}=\int\frac{1}{\text{t}}\text{dt}$
$=\log|\text{t}|+\text{C}$ $\Big[\because\int\frac{1}{\text{x}}\text{dx}=\log|\text{x}|+\text{C}\Big]$
$=\log\big|(\text{x}+\sin\text{x})\big|+\text{C}$ $[\because\ \text{t}=\text{x}+\sin\text{x}]$
View full question & answer→Question 2893 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 2903 Marks
Evaluate the following integrals:
$\int\frac{1+\sin\text{x}}{\sqrt{\text{x}-\cos\text{x}}}\text{dx}$
Answer$\int\frac{1+\sin\text{x}}{\sqrt{\text{x}-\cos\text{x}}}\text{dx}$
$\text{Let},\text{x}-\cos\text{x}=\text{t}$
$\Rightarrow(1+\sin\text{x})=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow(1+\sin\text{x})\text{dx}=\text{dt}$
$\text{Now,}\int\frac{1+\sin\text{x}}{\sqrt{\text{x}-\cos\text{x}}}\text{dx}$
$=\int\frac{\text{dt}}{\sqrt{\text{t}}}$
$=\int\text{t}^{-\frac{1}{2}}\text{dt}$
$=\frac{\text{t}^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+\text{C}$
$=2\sqrt{\text{t}}+\text{C}$
$=2\sqrt{\text{x}-\cos\text{x}}+\text{C}$
View full question & answer→Question 2913 Marks
Evalute the following integrals:
$\int\frac{1}{\text{x}\log\text{x}\log(\log\text{x})}\text{dx}$
AnswerNote: Here, we are considering $\log\text{x}$ as $\log_\text{e}\text{x}$.
Let $\text{I}=\int\frac{1}{\text{x}\log\text{x}\log(\log\text{x})}\text{dx}$
Putting $\log(\log\text{x})=\text{t}$
$\Rightarrow\frac{1}{\text{x}\log\text{x}}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\frac{1}{\text{x}\log\text{x}}\text{dx}=\text{dt}$
$\therefore\text{I}=\int\frac{\text{dt}}{\text{t}}$
$=\log|\text{t}|+\text{C}$
$=\log|\log(\log\text{x})|+\text{C}\ \big[\because\text{t}=\log(\log\text{x})\big]$
View full question & answer→Question 2923 Marks
Integrate the function in Exercise:$\frac{1}{\sqrt{\text{x}+\text{a}}+\sqrt{\text{x}+\text{b}}}$
Answer$\frac{1}{\sqrt{\text{x}+\text{a}}+\sqrt{\text{x}+\text{b}}}=\frac{1}{\sqrt{\text{x}+\text{a}}+\sqrt{\text{x}+\text{b}}}\times\frac{\sqrt{\text{x}+\text{a}}-\sqrt{\text{x}+\text{b}}}{\sqrt{\text{x}+\text{a}}-\sqrt{\text{x}+\text{b}}}$
$=\frac{\sqrt{\text{x}+\text{a}}-\sqrt{\text{x}+\text{b}}}{(\text{x}+\text{a)}-(\text{x}+\text{b)}}$
$=\frac{\big(\sqrt{\text{x}+\text{a}}-\sqrt{\text{x}+\text{b}}\big)}{\text{(a}-\text{b)}}$
$\Rightarrow\int\frac{1}{\sqrt{\text{x}+\text{a}}-\sqrt{\text{x}+\text{b}}}\text{dx}=\frac{1}{\text{a}-\text{b}}\int\Big(\sqrt{\text{x}+\text{a}}-\sqrt{\text{x}+\text{b}}\Big)\text{dx}$
$=\frac{1}{\text{(a}-\text{b)}}\left[\frac{\text{(x}+\text{a)}^{\frac{3}{2}}}{\frac{3}{2}}-\frac{\text{(x}+\text{b)}^{\frac{3}{2}}}{\frac{3}{2}}\right]$
$=\frac{2}{3\text{(a}-\text{b)}}\left[\text{(x}+\text{a)}^{\frac{3}{2}}-\text{(x}+\text{b)}^{\frac{3}{2}}\right]+\text{C}$
View full question & answer→Question 2933 Marks
Evaluate the following intregals:
$\int\frac{\sin2\text{x}}{\sin^4\text{x}+\cos^4\text{x}}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{\sin2\text{x}}{\sin^4\text{x}+\cos^4\text{x}}\ \text{dx}$
Dividing numerator and denominator by $\cos^2\text{x}$
$\text{I}=\int\frac{2\tan\text{x}\sec^2\text{x}}{\tan^4\text{x}+1}\ \text{dx}$
Let $\tan^2\text{x}=\text{t}$
$2\tan\text{x}\sec^2\text{x}\text{ dx}=\text{dt}$
$\text{I}=\int\frac{\text{dt}}{\text{t}^2+1}$
$=\tan^{-1}\text{t}+\text{C}$
$\text{I}=\tan^{-1}(\tan^2\text{x})+\text{C}$
View full question & answer→Question 2943 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 2953 Marks
Evaluate the following integrals:
$\int\frac{1}{\sqrt{1-\text{x}^2}(\sin^{-1}\text{x})^2}\text{dx}$
Answer$\int\frac{\text{dx}}{\sqrt{1-\text{x}^2}(\sin^{-1}\text{x})^2}$
$\text{Let,}\sin^{-1}\text{x}=\text{t}$
$\Rightarrow\frac{1}{\sqrt{1-\text{x}^2}}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\frac{1}{\sqrt{1-\text{x}^2}}\text{dx}=\text{dt}$
$\text{Now,}\int\frac{\text{dx}}{\sqrt{1-\text{x}^2}(\sin^{-1}\text{x})^2}$
$=\int\frac{\text{dt}}{\text{t}^2}$
$=\int\text{t}^{-2}\text{dt}$
$=\frac{\text{t}^{-2+1}}{-2+1}+\text{C}$
$=\frac{-1}{\text{t}}+\text{C}$
$=-\frac{1}{\sin^{-1}\text{x}}+\text{C}$
View full question & answer→Question 2963 Marks
$\int\frac{1}{\cos^2\text{x}(1-\tan\text{x})^2}\text{dx}$
AnswerLet $\text{I}=\int\frac{1}{\cos^2\text{x}(1-\tan\text{x})^2}\text{dx}$
$=\int\frac{\sec^2\text{x}}{(1-\tan\text{x})^2}\text{dx}$
$=\int\frac{\sec^2\text{x dx}}{(1-\tan\text{x})^2}$
$-\sec^2\text{x dx}=\text{dt}$
$\Rightarrow\sec^2\text{x dx}=-\text{dt}$
$\therefore\text{I}=\int\frac{-\text{dt}}{\text{t}^2}$
$=-\int\text{t}^{-2}\text{dt}$
$=-\bigg[\frac{\text{t}^{-2+1}}{-2+1}\bigg]+\text{c}$
$=\frac{1}{\text{t}}+\text{c}$
$=\frac{1}{1-\tan\text{x}}+\text{c}$
View full question & answer→Question 2973 Marks
Evaluate the following integrals:
$\int^\limits{9}_4\frac{\sqrt{\text{x}}}{\big(30-\text{x}^{\frac{3}{2}}\big)^2}\text{ dx}$
AnswerLet $\text{I}=\int^\limits{9}_4\frac{\sqrt{\text{x}}}{\big(30-\text{x}^{\frac{3}{2}}\big)^2}\text{ dx}$ Then,
Let $\Big(30-\text{x}^{\frac{3}{2}}\Big)=\text{t}$ Then, $-\frac{3}{2}\sqrt{\text{x}}\text{ dx}=\text{dt}$
When $\text{x}=4,\text{t}=22$ and $\text{x}=9,\text{t}=3$
$\therefore\ \text{I}=\int^\limits{3}_{22}-\frac{2}{3}\frac{1}{\text{t}^2}\text{ dt}$
$\Rightarrow\text{I}=\frac{2}{3}\Big[\frac{1}{\text{t}}\Big]^3_{22}$
$\Rightarrow\text{I}=\frac{2}{3}\Big(\frac{1}{3}-\frac{1}{22}\Big)$
$\Rightarrow\text{I}=\frac{19}{99}$
View full question & answer→Question 2983 Marks
Evaluate the following definite integrals:
$\int_{0}^\limits{1}\Big(\text{xe}^{2\text{x}}+\sin\frac{\pi\text{x}}{2}\Big)\text{dx}$
AnswerLet $\text{I}=\int_{0}^\limits{1}\Big(\text{xe}^{2\text{x}}+\sin\frac{\pi\text{x}}{2}\Big)\text{dx}$ Then,
$\text{I}=\int_{0}^\limits{1}\text{xe}^{2\text{x}}\text{ dx}+\int_{0}^\limits{1}\sin\frac{\pi\text{x}}{2}\text{ dx}$
Integrating first term by parts,
$\text{I}=\Big[\text{x }\frac{\text{e}^{2\text{x}}}{2}\Big]^1_0-\int_{0}^\limits{1}1\frac{\text{e}^{2\text{x}}}{2}\text{ dx}+\bigg[-\frac{\cos\frac{\pi\text{x}}{2}}{\frac{\pi}{2}}\bigg]_0^1$
$\Rightarrow\text{I}=\Big[\text{x }\frac{\text{e}^{2\text{x}}}{2}\Big]^1_0-\Big[\frac{\text{e}^{2\text{x}}}{4}\Big]^1_0-\frac{2}{\pi}\Big[\cos\frac{\pi\text{x}}{2}\Big]^1_0$
$\Rightarrow\text{I}=\frac{\text{e}^{2}}{2}-\frac{\text{e}^{2}}{4}+\frac{1}{4}+\frac{2}{\pi}$
$\Rightarrow\text{I}=\frac{\text{e}^{2}}{4}+\frac{1}{4}+\frac{2}{\pi}$
View full question & answer→Question 2993 Marks
Integrate the function in Exercise.
$\sqrt{\text{x}^2+3\text{x}}$
Answer$\int\sqrt{\text{x}^2+3\text{x}}\text{dx}=\int\sqrt{\text{x}^2+3\text{x}+\Bigg(\frac{3}{2}\Bigg)^2-\Bigg(\frac{3}{2}\Bigg)^2}\text{dx}=\int\sqrt{\Bigg(\text{x}+\frac{3}{2}\Bigg)^2-\Bigg(\frac{3}{2}\Bigg)^2}\text{dx}$
$=\frac{\text{x}+\frac{3}{2}}{2}\sqrt{\Bigg(\text{x}+\frac{3}{2}\Bigg)^2-\Bigg(\frac{3}{2}\Bigg)^2}-\frac{\Bigg(\frac{3}{2}\Bigg)^2}{2}\text{log}\Bigg|\text{x}+\frac{3}{2}+\sqrt{\Bigg(\text{x}+\frac{3}{2}\Bigg)^2-\Bigg(\frac{3}{2}\Bigg)^2}\Bigg|+\text{c}$
$\Bigg[\therefore\int\sqrt{\text{x}^2-\text{a}^2}\text{dx}=\frac{\text{x}}{2}\sqrt{\text{x}^2-\text{a}^2}-\frac{\text{a}^2}{2}\text{log}\Bigg|\text{x}+\sqrt{\text{x}^2-\text{a}^2}\Bigg|\Bigg]$
$=\frac{2\text{x}+3}{4}\sqrt{\text{x}^2+3\text{x}}-\frac{9}{8}\text{log}\Bigg|\text{x}+\frac{3}{2}+\sqrt{\text{x}^2+3\text{x}}\Bigg|+\text{c}$
View full question & answer→Question 3003 Marks
Evaluate the following integrals:
$\int\frac{5\cos^3\text{x}+6\sin^3\text{x}}{2\sin^2\text{x}\cos^2\text{x}}\text{dx}$
Answer$\int\bigg(\frac{5\cos^3\text{x}+6\sin^3\text{x}}{2\sin^2\text{x}\cos^2\text{x}}\bigg)\text{dx}$
$=\int\bigg(\frac{5\cos^3\text{x}}{2\sin^2\text{x}\cos^2\text{x}}+\frac{6\sin^3\text{x}}{2\sin^2\text{x}\cos^2\text{x}}\bigg)\text{dx}$
$=\int\Big(\frac{5}{2}\frac{\cos\text{x}}{\sin^2\text{x}}+3\frac{\sin\text{x}}{\cos^2\text{x}}\Big)\text{dx}$
$=\frac{5}{2}\int\Big(\frac{\cos\text{x}}{\sin\text{x}}\times\frac{1}{\sin\text{x}}\Big)\text{dx}+3\int\frac{\sin\text{x}}{\cos\text{x}}\times\frac{1}{\cos\text{x}}\text{dx}$
$=\frac{5}{2}\int(\text{cosec x}\cot\text{x})\text{dx}+3\int\sec\text{x}\tan\text{x dx}$
$=\frac{5}{2}(-\text{cosec x})+3\sec\text{x}+\text{C}$
$=-\frac{5}{2}\text{cosec x}+3\sec\text{x}+\text{C}$
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