Mathematics

Prove that

$$\displaystyle \int {\dfrac{{{{\sin }^{ - 1}}x}}{{{{\left( {1 - {x^2}} \right)}^{\dfrac{3}{2}}}}}} dx$$


SOLUTION
$$I=\int \dfrac{\sin^{-1}x}{(1-x^{2})^{\frac{3}{2}}}dx$$

Let $$\sin^{-1}x=u\Rightarrow \dfrac{1}{\sqrt{1-x^{2}}}dx=du$$

So : $$\displaystyle\int \dfrac{\sin^{-1}x}{(1-x^{2})(1-x^{2})^{\dfrac{1}{2}}}dx=\int \frac{u}{(1-\sin^{2}u)}du$$

$$= \displaystyle\int \dfrac{u}{\cos^{2}u}dx=\int u \,cosec^{2}u\, du$$

$$\Rightarrow u\displaystyle\int cosec^{2}u\,du-\int \left(\int cosec^{2}u\,du\right)du$$

$$=-u \cot u +\int \cot u \,du$$

$$=-u\cot u+\int \cot u\, du$$

$$=-u \cot u+In \sin u+a$$

$$=-u \cot u+In \sin u+a$$

$$I=-\sin^{-1}x\cot \sin^{-1}u+ ln x+a$$
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Subjective Hard Published on 17th 09, 2020
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