Question

Solve ${\int }_0^1{e}^{ln{\tan }^{-1}x}\cdot {\sin }^{-1}\left(\cos x\right)dx$

Answer 1

Let $A={\int }_0^1{e}^{ln{\tan }^{-1}x}\cdot {\sin }^{-1}\left(\cos x\right)dx.$

We know that $e^{lnx}=x$

Thus, $e^{ln\left({\tan }^{-1}x\right)}={\tan }^{-1}x$

Using this in the above integration, we get :

$A={\int }_0^1{\tan }^{-1}x\cdot {\sin }^{-1}\left(\cos x\right)dx$

We know that ${\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}$

$A={\int }_0^1{\tan }^{-1}x\cdot (\frac{\pi }{2}-{\cos }^{-1}(\cos x\left)\right)dx$

Since, we are integrating from 0 to 1, ${\cos }^{-1}\left(\cos x\right)=x$

Thus, $A={\int }_0^1{\tan }^{-1}x\cdot (\frac{\pi }{2}-x)dx$

$=\frac{\pi }{2}{\int }_0^1{\tan }^{-1}xdx-{\int }_0^{1}x{\tan }^{-1}xdx$

Integrating by parts,

$A=\frac{\pi }{2}({\left[x{\tan }^{-1}xdx\right]}_0^1-{\int }_0^1\frac{x}{1+x^2}dx)-({\left[\frac{x^2}{2}{\tan }^{-1}x\right]}_0^1-{\int }_0^1\frac{x^2}{2(1+x^2)}dx)$

$=\frac{\pi }{2}((\frac{\pi }{4}-0)-{\left[\frac{1}{2}\ln \right(1+x^2\left)\right]}_0^1)-((\frac{1}{2}\times \frac{\pi }{4}-0)-{[\frac{1}{2}-\frac{1}{2}{\tan }^{-1}x]}_0^1)$

$=\frac{{\pi }^2}{8}-\frac{\pi }{4}\ln \left(2\right)+\frac{1}{2}$