Question

${\int }_0^{\pi /4}\frac{x.\sin x}{{\cos }^{3}x}dx$ equals to :

• A. $\frac{\pi }{4}+\frac{1}{2}$
• B. $\frac{\pi }{4}-\frac{1}{2}$
• C. $\frac{\pi }{4}$
• D. $\frac{\pi }{4}+1$

Answer 1

Answer: (B)

$\frac{\pi }{4}-\frac{1}{2}$

${\int }_0^{\frac{\pi }{4}}\frac{x\sin x}{{\cos }^{3}x}dx$

$={\int }_0^{\frac{\pi }{4}}x\tan x{\sec }^{2}xdx$

Let $t=\tan x\Rightarrow dt={\sec }^{2}xdx$

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

When $x=0\Rightarrow t=0$

When $x=\frac{\pi }{4}\Rightarrow t=1$

$={\int }_0^1{\tan }^{-1}tdt$

Let $u={\tan }^{-1}t\Rightarrow du=\frac{1}{1+t^2}dt$

$dv=t\Rightarrow v=\frac{t^2}{2}$

$={\left[{\tan }^{-1}t\frac{t^2}{2}\right]}_0^1-{\int }_0^1\frac{t^2}{2}\times \frac{1}{1+t^2}dt$

$=\frac{1}{2}{\left[t^2{\tan }^{-1}t\right]}_0^1-\frac{1}{2}{\int }_0^1\frac{t^2}{1+t^2}dt$

$=\frac{1}{2}{\left[t^2{\tan }^{-1}t\right]}_0^1-\frac{1}{2}{\int }_0^1\frac{t^2+1-1}{1+t^2}dt$

$=\frac{1}{2}{\left[t^2{\tan }^{-1}t\right]}_0^1-\frac{1}{2}{\int }_0^1\frac{t^2+1}{1+t^2}dt+\frac{1}{2}{\int }_0^1\frac{dt}{1+t^2}$

$=\frac{1}{2}[{\tan }^{-1}1-0]-\frac{1}{2}{\left[t\right]}_0^1+\frac{1}{2}{\left[{\tan }^{-1}t\right]}_0^1$

$=\frac{1}{2}[\frac{\pi }{4}-0]-\frac{1}{2}{[1-0]}_0^1+\frac{1}{2}{[\frac{\pi }{4}-0]}_0^1$

$=\frac{\pi }{8}-\frac{1}{2}+\frac{\pi }{8}$

$=\frac{2\pi }{8}-\frac{1}{2}$

$=\frac{\pi }{4}-\frac{1}{2}$