Mathematics

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


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

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

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

$$=\dfrac{-\sin^{-1}x}{x}+\dfrac{\ln \left ( \left | \sqrt{1-x^2}-1 \right | \right )-\ln \left ( \sqrt{1-x^2}+1 \right )}{2}+C$$
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Subjective Medium Published on 17th 09, 2020
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