Question

${\int }_0^{\pi /2}\frac{{\cos }^{2}x\sin x}{\sqrt{(1+{\cos }^{2}x)}}dx=\frac{1}{n}[\sqrt{k}-log(\sqrt{m}+1\left)\right]$.Find $k+m+n$ ?

Answer 1

Let $I={\int }_0^{\frac{\pi }{2}}\frac{{\cos }^{2}x\sin xdx}{\sqrt{(1+{\cos }^{2}x)}}$
Substitute $\cos x=\tan t\Rightarrow -\sin xdx={\sec }^{2}tdt$
$\therefore I=-{\int }_{-\frac{\pi }{4}}^0\frac{{\tan }^{2}t.{\sec }^{2}t}{\sec t}dt\Rightarrow I={\int }_0^{\frac{\pi }{4}}\tan t.\sec t.\tan tdt$
$={\int }_0^{\frac{\pi }{4}}\sqrt{{\sec }^{2}t-1}\sec t\tan tdt$
Substitute $\sec t=z$
$\therefore I={\int }_1^{\sqrt{2}}\sqrt{z^2-1}dx={[\frac{z}{2}\sqrt{z^2-1}-\frac{1}{2}\log z+\sqrt{z^2+1}]}_1^{\sqrt{2}}$
$=\frac{1}{2}\sqrt{2}-\frac{1}{2}\left[\log (\sqrt{2}+1)-0\right]=\frac{1}{2}[\sqrt{2}-\log (\sqrt{2}+1)]$