Question

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

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

Let $I={\int }_0^{\pi /4}\frac{x.\sin x}{{\cos }^{3}x}dx$

Applying by-parts formula, we get

$={\left[\frac{x}{2{\cos }^{2}x}\right]}_0^{\pi /4}-{\int }_0^{\pi /4}\frac{1}{2{\cos }^{2}x}dx$

$=\frac{\frac{\pi }{4}}{2{\left(\frac{1}{\sqrt{2}}\right)}^2}-0-{\int }_0^{\pi /4}\frac{{\sec }^{2}x}{2}dx$

$=\frac{\pi }{4}-\frac{1}{2}{\left[\tan x\right]}_0^{\pi /4}$

$=\frac{\pi }{4}-\frac{1}{2}[\tan \frac{\pi }{4}-\tan 0]$

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

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