Mathematics

# Evaluate the given integral.$\int { { e }^{ x }\left( 1-\cot { x } +\cot ^{ 2 }{ x } \right) } dx$

$-{ e }^{ x }\cot { x } +C$

##### SOLUTION
$I=\displaystyle\int{{e}^{x}\left(1-\cot{x}+{\cot}^{2}{x}\right)dx}$

$I=\displaystyle\int{{e}^{x}\left(1+{\cot}^{2}{x}\right)dx}-\displaystyle\int{{e}^{x}\cot{x}dx}$

$I=\displaystyle\int{{e}^{x}{cosec}^{2}{x}dx}-\displaystyle\int{{e}^{x}\cot{x}dx}$

Integrating by parts, we get

$dv={cosec}^{2}{x}dx\Rightarrow\,v=-\cot{x}$

$u={e}^{x}\,\Rightarrow\,du={e}^{x}dx$

$\int u.v dx=u \int vdx-\int \left [\int vdx. \dfrac{du}{dx}.dx \right ]$......by parts formula

$I=-{e}^{x}\cot{x}-\displaystyle\int{-\cot{x}{e}^{x}dx}-\displaystyle\int{{e}^{x}\cot{x}dx}+c$

$I=-{e}^{x}\cot{x}+\displaystyle\int{\cot{x}{e}^{x}dx}-\displaystyle\int{{e}^{x}\cot{x}dx}+c$

$I=-{e}^{x}\cot{x}+c$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

#### Realted Questions

Q1 Single Correct Hard
The value of $\displaystyle \int _{ 2 }^{ 8 }{ \cfrac { \sqrt { 10-x } }{ \sqrt { x } +\sqrt { 10-x } } }dx$ is
• A. $10$
• B. $0$
• C. $8$
• D. $3$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
$\displaystyle\int \dfrac{\sin (x-\alpha)}{\sin (x+\alpha)}dx$.

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
$n\overset{Lt}{\rightarrow}\infty [\displaystyle \frac{n+1}{n^{2}+1^{2}}+\frac{n+2}{n^{2}+2^{2}}+\ldots+\frac{n+n}{n^{2}+n^{2}}]=$
• A. $\displaystyle \frac{\pi}{4}-\frac{1}{2}\log 2$
• B. $\displaystyle \frac{\pi}{2}+\frac{1}{2}\log 2$
• C. $\displaystyle \frac{\pi}{4}+\frac{1}{4}\log 2$
• D. $\displaystyle \frac{\pi}{4}+\frac{1}{2}\log 2$

1 Verified Answer | Published on 17th 09, 2020

Q4 Multiple Correct Medium
If $\displaystyle \int_{1}^{9} x^2F'(x)dx =-12$ and  $\displaystyle \int_{1}^{9} x^3F''(x)dx =40$ then the correct expression(s) is/are
• A. $9f'(3)+f'(1)-32=0$
• B. $\displaystyle \int_{1}^{3}f(x)dx=12$
• C. $9f'(3)-f'(1)+32=0$
• D. $\displaystyle \int_{1}^{3}f(x)dx=-12$

1 Verified Answer | Published on 17th 09, 2020

Q5 Passage Hard
Let us consider the integral of the following forms
$f{(x_1,\sqrt{mx^2+nx+p})}^{\tfrac{1}{2}}$
Case I If $m>0$, then put $\sqrt{mx^2+nx+C}=u\pm x\sqrt{m}$
Case II If $p>0$, then put $\sqrt{mx^2+nx+C}=u\pm \sqrt{p}$
Case III If quadratic equation $mx^2+nx+p=0$ has real roots $\alpha$ and $\beta$, then put $\sqrt{mx^2+nx+p}=(x-\alpha)u\:or\:(x-\beta)u$