Question

$\stackrel{\pi /4}{\underset{0}{\int }}\frac{x.\sin x}{{\cos }^{3}x}dx$ 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

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

$I={\int }_0^{\pi /4}\frac{xsinx}{co{s}^{3}x}dx={\int }_0^{\pi /4}xtanxse{c}^{2}xdx$

Intergating by parts,$u=x,v=tanxse{c}^{2}x$

$I=x\int \left(tanxsecx\right)secxdx-\int -\int (tanx.secx)secxdx$

$=x\frac{se{c}^{2}x}{2}-\int \frac{se{c}^{2}x}{2}dx$

$[\because \frac{d}{dx}(secx)=tanxsecx]$

$=x\frac{se{c}^{2}x}{2}-\frac{tanx}{2}+c$

Applying limits

$I={[\frac{xse{c}^{2}x}{2}-\frac{tanx}{2}]}_0^{\pi /4}=\frac{\pi }{4}\frac{\left(2\right)}{2}-\frac{1}{2}-0$

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

$\therefore$ option B is correct