Evaluate: _0^ 1 2 ^ -1 x (1-x^2 )^ 3 2 d x - Vidyadip

Question

Evaluate: $\int_0^{\frac{1}{\sqrt{2}}} \frac{\sin ^{-1} x}{\left(1-x^2\right)^{\frac{3}{2}}} d x$

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Answer

Let $I =\int_0^{\frac{1}{\sqrt{2}}} \frac{\sin ^{-1} x}{\left(1-x^2\right)^{\frac{3}{2}}} d x$
Put $\sin ^{-1} x=t$
$\therefore x =\sin t$
$\therefore d x=\cos t d t$
When $x =0, t =0$ and when $x =\frac{1}{\sqrt{2}}, t =\frac{\pi}{4}$
$\therefore I =\int_0^{\frac{\pi}{4}} \frac{ t }{\left(1-\sin ^2 t \right)^{\frac{3}{2}}} \times \cos t d t$
$=\int_0^{\frac{\pi}{4}} \frac{ t }{\left(\cot ^2 t \right)^{\frac{3}{2}}} \times \cos t d t$
$=\int_0^{\frac{\pi}{4}} t \sec ^2 t dt$
$=\left[ t \int \sec ^2 t d t \right]_0^{\frac{\pi}{4}}-\int_0^{\frac{\pi}{4}}\left[\frac{ d }{ dt }( t ) \int \sec ^2 t dt \right] dt$
$=[ t \cdot \tan t ]_0^{\frac{\pi}{4}}-\int_0^{\frac{\pi}{4}} 1 \cdot \tan t d t$
$=\left(\frac{\pi}{4} \cdot \tan \frac{\pi}{4}-0\right)-[\log |\sec t|]_0^{\frac{\pi}{4}}$
$=\frac{\pi}{4}(1)-\left[\log \left|\sec \frac{\pi}{4}\right|-\log |\sec 0|\right]$
$=\frac{p}{4}-(\log \sqrt{2}-\log 1)$
$=\frac{\pi}{4}-\left(\log 2^{\frac{1}{2}}-0\right)$
$\therefore I =\frac{\pi}{4}-\frac{1}{2} \log 2$

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