Question
Evaluate: $\int_{\frac{1}{\sqrt{2}}}^1 \frac{\left( e ^{\left.\cos ^{-1} x\right)\left(\sin ^{-1} x\right)}\right.}{\sqrt{1-x^2}} d x$
✓
Answer
$ \text { Let } I =\int_{\frac{1}{\sqrt{2}}}^1 \frac{\left( e ^{\cos ^{-1} x}\right)\left(\sin ^{-1} x\right)}{\sqrt{1-x^2}} d x$
$=\int_{\frac{1}{\sqrt{2}}}^1 \frac{\left( e ^{\left.\frac{\pi}{2}-\sin ^{-1} x\right)\left(\sin ^{-1} x\right)}\right.}{\sqrt{1-x^2}} d x \quad \ldots . .\left[\because \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\right] $
Put $\sin ^{-1} x=t$
$\therefore \frac{1}{\sqrt{1-x^2}} d x= dt$
When $x =\frac{1}{\sqrt{2}}, t =\frac{\pi}{4}$ and when $x =1, t =\frac{\pi}{2}$
$ \therefore I =\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\left( e ^{\frac{\pi}{2}- t }\right) t dt$
$=\left[ t \int e ^{\frac{\pi}{2}- t } dt \right]_{\frac{\pi}{4}}^{\frac{\pi}{2}}-\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\left[\frac{ d }{ dt }( t ) \int e ^{\frac{\pi}{2}- t } dt \right] dt$
$=\left[ t \cdot \frac{ e ^{\frac{\pi}{2}- t }}{-1}\right]_{\frac{\pi}{4}}^{\frac{\pi}{2}}-\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} 1 \cdot \frac{ e ^{\frac{\pi}{2}- t }}{-1} dt$
$=-\left(\frac{\pi}{2} e ^0-\frac{\pi}{4} e ^{\frac{\pi}{4}}\right)+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} e ^{\frac{\pi}{2}- t } dt$
$=-\left(\frac{\pi}{2}-\frac{\pi}{4} e ^{\frac{\pi}{4}}\right)+\left[\frac{ e ^{\frac{\pi}{2}- t }}{-1}\right]_{\frac{\pi}{4}}^{\frac{\pi}{2}}$
$=-\frac{\pi}{2}+\frac{\pi}{4} e ^{\frac{\pi}{4}}-\left( e ^0- e ^{\frac{\pi}{4}}\right)$
$=-\frac{\pi}{2}+\frac{\pi}{4} e ^{\frac{\pi}{4}}-1+ e ^{\frac{\pi}{4}}$
$\therefore I = e ^{\frac{\pi}{4}}\left(\frac{\pi}{4}+1\right)-\left(\frac{\pi}{2}+1\right) $
$=\int_{\frac{1}{\sqrt{2}}}^1 \frac{\left( e ^{\left.\frac{\pi}{2}-\sin ^{-1} x\right)\left(\sin ^{-1} x\right)}\right.}{\sqrt{1-x^2}} d x \quad \ldots . .\left[\because \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\right] $
Put $\sin ^{-1} x=t$
$\therefore \frac{1}{\sqrt{1-x^2}} d x= dt$
When $x =\frac{1}{\sqrt{2}}, t =\frac{\pi}{4}$ and when $x =1, t =\frac{\pi}{2}$
$ \therefore I =\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\left( e ^{\frac{\pi}{2}- t }\right) t dt$
$=\left[ t \int e ^{\frac{\pi}{2}- t } dt \right]_{\frac{\pi}{4}}^{\frac{\pi}{2}}-\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\left[\frac{ d }{ dt }( t ) \int e ^{\frac{\pi}{2}- t } dt \right] dt$
$=\left[ t \cdot \frac{ e ^{\frac{\pi}{2}- t }}{-1}\right]_{\frac{\pi}{4}}^{\frac{\pi}{2}}-\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} 1 \cdot \frac{ e ^{\frac{\pi}{2}- t }}{-1} dt$
$=-\left(\frac{\pi}{2} e ^0-\frac{\pi}{4} e ^{\frac{\pi}{4}}\right)+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} e ^{\frac{\pi}{2}- t } dt$
$=-\left(\frac{\pi}{2}-\frac{\pi}{4} e ^{\frac{\pi}{4}}\right)+\left[\frac{ e ^{\frac{\pi}{2}- t }}{-1}\right]_{\frac{\pi}{4}}^{\frac{\pi}{2}}$
$=-\frac{\pi}{2}+\frac{\pi}{4} e ^{\frac{\pi}{4}}-\left( e ^0- e ^{\frac{\pi}{4}}\right)$
$=-\frac{\pi}{2}+\frac{\pi}{4} e ^{\frac{\pi}{4}}-1+ e ^{\frac{\pi}{4}}$
$\therefore I = e ^{\frac{\pi}{4}}\left(\frac{\pi}{4}+1\right)-\left(\frac{\pi}{2}+1\right) $
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