Mathematics

# $\displaystyle \int \dfrac {\sqrt {1-x^{2}}}{x^{4}}dx=A(x)(\sqrt {1-x^{2}})^{m}+c$ then $(A(x))^{m}$ is

$-\dfrac {1}{27x^{9}}$

##### SOLUTION
$\int \dfrac{\sqrt{1-x^2}}{x^4}dx$

substitute $x=\sin u\rightarrow dx=\cos u du$

$\int \dfrac{\cot ^2u}{\sin ^2u}du$

substitute $v=\cot u\rightarrow dv = -\csc ^2u du$

$-\int v^2 dv$

$=-\dfrac{v^3}{3}$

$=-\dfrac{\cot ^3u}{3}$

$=-\dfrac{\cot ^3(\sin ^{-1}x)}{3}$

$=-\dfrac{\sqrt{(1-x^2)^3}}{3x^3}+C$

$\therefore m=3$

$A(x)=-\dfrac{1}{3x^3}$

$\therefore A(x)^m=\left ( -\dfrac{1}{3x^3} \right )^3=-\dfrac{1}{27x^9}$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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