Question

$\int\limits_0^{\pi / 2} \frac{\sin ^3 x}{\left(\cos ^4 x+3 \cos ^2 x+1\right) \tan ^{-1}(\sec x+\cos x)} d x$ 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)

$\int\limits_0^{\pi / 2} \frac{\sin ^3 x}{\left(\cos ^4 x+3 \cos ^2 x+1\right) \tan ^{-1}(\sec x+\cos x)} d x$
Put $\sec x+\cos x=t \Rightarrow(\sec x \tan x-\sin x) d x=d t \Rightarrow \sin x \frac{\sin ^2 x}{\cos ^2 x} d x=d t$ $\Rightarrow \sin ^3 xdx =\cos ^2 xdt$
$I=\int\limits_2^{\infty} \frac{d t}{\left(\cos ^2 x+\sec ^2 x+3\right) \tan ^{-1} t}=\int\limits_2^{\infty} \frac{d t}{\left(t^2+1\right) \tan ^{-1} t} ;$ Put $\tan ^{-1} t=z$
$I =\ln z ]_{\tan ^{-1} 2}^{\frac{\pi}{2}}=\ln \frac{\pi}{2}-\ln \left(\tan ^{-1} 2\right)$.