Mathematics

# Solve : $\displaystyle \int \sin^{-1} x . \dfrac{1}{x^2} dx$

##### SOLUTION
$\int \sin^{-1}x \dfrac{1}{x^2}dx$

$=\sin^{-1}x\dfrac{-1}{x}-\int \dfrac{1}{\sqrt{1-x^2}}\left ( \dfrac{-1}{x} \right )$

$=\dfrac{-\sin^{-1}x}{x}+\int \dfrac{1}{x\sqrt{1-x^2}} dx$

$=\dfrac{-\sin^{-1}x}{x}+\dfrac{\ln \left ( \left | \sqrt{1-x^2}-1 \right | \right )-\ln \left ( \sqrt{1-x^2}+1 \right )}{2}+C$

Its FREE, you're just one step away

Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 105

#### Realted Questions

Q1 Single Correct Medium
Evaluate $\displaystyle \int \frac{\sec^{2}\:x}{\sqrt{\tan^{2}\:x+4}}dx.$
• A. $\displaystyle=\log\left | \tan\:x+\sqrt{4 \tan^{2}\:x+4} \right |+C$
• B. $\displaystyle=\log\left | \tan\:x-\sqrt{\tan^{2}\:x+4} \right |+C$
• C. $\displaystyle=\log\left | \tan\:x+\sqrt{\tan^{2}\:x-4} \right |+C$
• D. $\displaystyle=\log\left | \tan\:x+\sqrt{\tan^{2}\:x+4} \right |+C$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Hard
Evaluate the following integral:
$\int { \cfrac { 2x+5 }{ { x }^{ 2 }-x-2 } } dx$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
The value of $\displaystyle \int_{0}^{1}\left(\prod_{r=1}^{n}(x+r)\right)\left( \sum_{k=1}^{n}\dfrac{1}{x+k}\right)dx$ equals to
• A. $n$
• B. $n!$
• C. $(n+1)!$
• D. $n.n!$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Hard
The value(s) of $\int _{ 0 }^{ 1 }{ \cfrac { { x }^{ 4 }{ \left( 1-x \right) }^{ 4 } }{ 1+{ x }^{ 2 } } } dx$ is (are)
• A. $\cfrac{2}{105}$
• B. $0$
• C. $\cfrac{71}{15}-\cfrac{3\pi}{2}$
• D. $\cfrac{22}{7}-\pi$

Evaluate $\displaystyle \int _ { 0 } ^ { \pi / 2 } \log ( \tan x ) d x$