Question

${\int }_0^{\pi /4}\frac{x.sin}{co{s}^{3}x}dx$ 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

The correct option is B $\frac{\pi }{4}-\frac{1}{2}$

$I={\int }_0^{\pi /4}\frac{x\sin x}{{\cos }^{3}x}dx$ $\begin{array}{c}4=x\\ 4=1\end{array}$ $\begin{array}{c}v=\frac{\sin 3}{{\cos }^{3}x}\\ \int vdx=\frac{1}{2{\cos }^{2}x}\end{array}$

$={\left[\int \frac{\sin x}{{\cos }^{3}x}dx-\int 1(\int \frac{\sin x}{{\cos }^{3}x}dx)dx\right]}^{\pi /4}$ (By parts integration)

$={[\frac{x}{2{\cos }^{2}x}-\int \frac{1}{2{\cos }^{2}x}dx]}^{\pi /4}$ $[\because \int {\sec }^{2}xdx=\tan x]$

$={[\frac{x{\sec }^{2}x}{2}-\frac{\tan x}{2}]}^{\pi /4}$

$=\frac{\pi }{4}\frac{\left(2\right)}{2}-\frac{1}{2}=\frac{\pi }{4}-\frac{1}{2}$

Option $B$ is correct