Question

$\underset{0}{\overset{\pi /2}{\int }}\frac{{\sin }^{3}x}{\left({\cos }^{4}x+3{\cos }^{2}x+1\right){\tan }^{-1}(\sec x+\cos x)}dx$ is equal to

• A. $\frac{\pi }{2}-{\tan }^{-1}2$
• B. $\ln \frac{\pi }{2}-\ln \left({\tan }^{-1}2\right)$
• C. $\ln \left({\tan }^{-1}2\right)$
• D. $\ln \frac{\pi }{2}$

Answer 1

Answer: (B)

$\underset{0}{\overset{\pi /2}{\int }}\frac{{\sin }^{3}x}{\left({\cos }^{4}x+3{\cos }^{2}x+1\right){\tan }^{-1}(\sec x+\cos x)}dx$
Put $\sec x+\cos x=t\Rightarrow (\sec x\tan x-\sin x)dx=dt\Rightarrow \sin x\frac{{\sin }^{2}x}{{\cos }^{2}x}dx=dt$ $\Rightarrow {\sin }^{3}xdx={\cos }^{2}xdt$
$I=\underset{2}{\overset{\infty }{\int }}\frac{dt}{\left({\cos }^{2}x+{\sec }^{2}x+3\right){\tan }^{-1}t}=\underset{2}{\overset{\infty }{\int }}\frac{dt}{\left(t^2+1\right){\tan }^{-1}t};$ Put ${\tan }^{-1}t=z$
$I=\ln z{]}_{{\tan }^{-1}2}^{\frac{\pi }{2}}=\ln \frac{\pi }{2}-\ln \left({\tan }^{-1}2\right)$.