Question

solve.. .. .. .. .. .. .. .. .. ..​

Answer 1

Correct Question:-

Evaluate $\displaystyle\int\dfrac{\sin^{-1}x}{(1-x^2)^{\frac{3}{2}}}\ dx.$

Solution:-

Given to evaluate,

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

Substitute,

$\longrightarrow \theta=\sin^{-1}x$

$\longrightarrow x=\sin\theta$

$\longrightarrow dx=\cos\theta\ d\theta$

Also,

$\longrightarrow\cos\theta=\sqrt{1-x^2}$

$\longrightarrow\tan\theta=\dfrac{x}{\sqrt{1-x^2}}$

Then the integral becomes,

$\displaystyle\longrightarrow I=\int\dfrac{\theta}{(1-\sin^2\theta)^{\frac{3}{2}}}\ \cos\theta\ d\theta$

$\displaystyle\longrightarrow I=\int\dfrac{\theta}{(\cos^2\theta)^{\frac{3}{2}}}\ \cos\theta\ d\theta$

$\displaystyle\longrightarrow I=\int\dfrac{\theta}{\cos^3\theta}\ \cos\theta\ d\theta$

$\displaystyle\longrightarrow I=\int\dfrac{\theta}{\cos^2\theta}\ d\theta$

$\displaystyle\longrightarrow I=\int\theta\,\sec^2\theta\ d\theta$

Performing integration by parts,

$\displaystyle\longrightarrow I=\theta\tan\theta-\int\tan\theta\ d\theta$

$\displaystyle\longrightarrow I=\theta\tan\theta+\log|\cos\theta|+C$

Undoing substitutions for $\theta,\ \tan\theta$ and $\cos\theta,$ we get,

$\displaystyle\longrightarrow\underline{\underline{I=\dfrac{x\sin^{-1}x}{\sqrt{1-x^2}}+\log\sqrt{1-x^2}+C}}$