Question

Solve: ${\int }_0^{\frac{\pi }{4}}x.{\sec }^{2}xdx$

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

Answer 1

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

Consider, $I={\int }_0^{\frac{\pi }{4}}x{sec}^{2}xdx$,

Solving the indefinite integral,

$I_1=\int x{sec}^{2}xdx$,

Applying integral by parts,

$\Rightarrow I_1=x\tan x-\int \tan xdx$

$\Rightarrow I_1=x\tan x+\ln cosx+C$

The value of the indefinite integral,

$\Rightarrow I={[x\tan x+\ln cosx]}_0^{\frac{\pi }{4}}$

$\Rightarrow I=[\frac{\pi }{4}\tan \frac{\pi }{4}+\ln \cos \frac{\pi }{4}]-[0-\ln (\cos 0\left)\right]$

$\Rightarrow I=\frac{\pi }{4}+\ln \frac{1}{\sqrt{2}}$