Mathematics

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


ANSWER

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


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
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