Question

${\int }_0^{\infty }\frac{{\sin }^4\left({\tan }^{-1}x\right){\cos }^3\left({\tan }^{-1}x\right)}{1+x^2}$ is equal to

• A. $\frac{2}{105}$
• B. $\frac{\pi }{105}$
• C. $\frac{\pi }{35}$
• D. $\frac{2}{35}$

Answer 1

The correct option is D $\frac{2}{35}$

Consider the given integral.

$I={\int }_0^{\infty }\frac{{\sin }^4\left({\tan }^{-1}x\right){\cos }^3\left({\tan }^{-1}x\right)}{1+x^2}dx$

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

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

Therefore,

$I={\int }_0^{\pi /2}{\sin }^{4}t{\cos }^{3}tdt$

$I={\int }_0^{\pi /2}{\sin }^{4}t\cos t{\cos }^{2}tdt$

$I={\int }_0^{\pi /2}{\sin }^{4}t\cos t(1-{\sin }^{2}t)dt$

$I={\int }_0^{\pi /2}({\sin }^{4}t\cos t-{\sin }^{6}t\cos t)dt$

$I={\int }_0^{\pi /2}{\sin }^{4}t\cos tdt-{\int }_0^{\pi /2}{\sin }^{6}t\cos tdt$

Let $u=\sin t$

$du=\cos tdt$

Therefore,

$I={\int }_0^1{u}^{4}du-{\int }_0^1{u}^{6}du$

$I={\left[\frac{u^5}{5}\right]}_0^1-{\left[\frac{u^7}{7}\right]}_0^1$

$I=[\frac{1^5}{5}-0]-[\frac{1^7}{7}-0]$

$I=\frac{1}{5}-\frac{1}{7}$

$I=\frac{7-5}{35}$

$I=\frac{2}{35}$

Hence, this is the answer.