Question

Evaluate the integral
${\int }_0^{\pi /4}{\sec }^{3}xdx$

• A. $\frac{1}{\sqrt{2}}+\frac{1}{2}\log (\sqrt{2}+1)$
• B. $\frac{1}{\sqrt{2}}-\frac{1}{2}\log (\sqrt{2}+1)$
• C. $2\sqrt{2}\cdot \log \sqrt{2}$
• D. $\frac{1}{\sqrt{2}}log\left(\sqrt{2}\right)$

Answer 1

The correct option is A $\frac{1}{\sqrt{2}}+\frac{1}{2}\log (\sqrt{2}+1)$

Let $I=\int {\sec }^{3}xdx$

Using reduction formula $\int {\sec }^{m}xdx=\frac{\sin x.{\sec }^{m-1}x}{m-1}+\frac{m-2}{m-1}\int {\sec }^{-2+m}xdx$

We get $I=\frac{1}{2}\tan x\sec x+\frac{1}{2}\int \sec xdx$

$=\frac{1}{2}\tan x\sec x+\frac{1}{2}\int \frac{{\sec }^{2}x+\tan x\sec x}{\tan x+\sec x}dx$

Let $I_1=\frac{1}{2}\int \frac{{\sec }^{2}x+\tan x\sec x}{\tan x+\sec x}dx$

Substituting $t=\tan x+\sec x\Rightarrow dt=({\sec }^{2}x+\tan x\sec x)dx$, we get

$I_1=\frac{1}{2}\int \frac{1}{t}dt=\frac{\log t}{2}=\frac{1}{2}\log (\tan x+\sec x)$

Therefore, $I=\frac{1}{2}\tan x\sec x+\frac{1}{2}\log (\tan x+\sec x)$

Hence, ${\int }_0^{\pi /4}{\sec }^{3}xdx={[\frac{1}{2}\tan x\sec x+\frac{1}{2}\log (\tan x+\sec x)]}_0^{\pi /4}$

$=\frac{1}{\sqrt{2}}+\frac{1}{2}\log (\sqrt{2}+1)$