Mathematics

# Evaluate: $\displaystyle\int {{sin^{ - 1}}} \left( {{{2x} \over {1 + {x^2}}}} \right)dx$

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

Simplifying the given function

$\sin^{-1}\left(\dfrac{2x}{1+x^2}\right)$

Put $x=\tan t$

$\therefore t=\tan^{-1}(x)$

$\therefore \sin^{-1}\left(\dfrac{2x}{1+x^2}\right)=\sin^{-1}\left(\dfrac{2\tan t}{1+\tan^2t}\right)$

$=\sin^{-1}(\sin 2t)$

$=2t$

$2\tan^{-1}x$

$\therefore I=\displaystyle \int{\sin^{-1}\left(\dfrac{2x}{1+x^2}\right)}=2\displaystyle \int{\tan^{-1}x dx}$

$=2\tan^{-1}x.x-2\displaystyle \int{\dfrac{x}{1+x^2}dx}$

$=2x\tan^{-1}x-\log|1+x^2|+c$

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

#### Realted Questions

Q1 Multiple Correct Hard
If $\displaystyle \int \frac{\cos\: x}{\cos(x-\alpha )}dx=Ax+B\, \log\, \left | \cos(x-\alpha ) \right |+c,$ then
• A. $A=\sin\, \alpha$
• B. $B=\cos\, \alpha$
• C. $A=\cos \alpha$
• D. $B=\sin\, \alpha$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Integrate the function    $(\sin^{-1}x)^2$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
Let $I=\displaystyle\int _{ 0 }^{ 1 }{ \sqrt { \dfrac { 1+\sqrt { x } }{ 1-\sqrt { x } } } dx }$ and $J=\displaystyle\int _{ 0 }^{ 1 }{ \sqrt { \dfrac { 1-\sqrt { x } }{ 1+\sqrt { x } } } dx }$, then the correct statement is
• A. $I-J=\pi$
• B. $I=\dfrac{2+\pi}{2}$
• C. $J=\dfrac{4-\pi}{2}$
• D. $I+J=2$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
$\displaystyle \int_{-\pi }^{\pi}\left ( \cos ax-\sin bx \right )^{2}dx$ where a and b are integer is equal to
• A. $\displaystyle -\pi$
• B. $0$
• C. $\displaystyle \pi$
• D. $\displaystyle 2\pi$

Evaluate $\int { \dfrac { dx }{ 9{ x }^{ 2 }+6x+5 } }$