Question

Find solution in terms of indefinite integration, using substitution
${\int }_0^{1}x\cos \left({\tan }^{-1}x\right)dx$

Answer 1

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

Let,

${\tan }^{-1}x=t$

at, $x=0,t=0;x=1,t=\frac{\pi }{4}$

$dt=\frac{1}{1+x^2}dx$

$\therefore {\int }_0^{1}x\cos \left({\tan }^{-1}x\right)dx$

= ${\int }_0^{\frac{\pi }{4}}\tan t\times \cos t\times (1+{\tan }^{2}t)dt$

= ${\int }_0^{\frac{\pi }{4}}\tan t\times \cos t\times \left({\sec }^{2}t\right)dt$ ......{$\because 1+{\tan }^{2}t={\sec }^{2}t$}

= ${\int }_0^{\frac{\pi }{4}}\frac{\sin t}{{\cos }^{2}t}dt$ ......{$\because \frac{1}{\cos t}=\sec tand\tan t=\frac{\sin t}{\cos t}$}

= ${\left[\frac{{\left(\cos t\right)}^{(-2+1)}}{-2+1}\right]}_0^{\frac{\pi }{4}}$

=${\left[\frac{(\cos t{)}^{-1}}{-1}\right]}_0^{\frac{\pi }{4}}$

=$\left[\right(-\sec \frac{\pi }{4})-(-\sec 0\left)\right]$

=$-\sqrt{2}+1$