Question

The value of the integral $\int\limits_{0}^{\pi/ 4} \frac{1+\sin^{2} }{ \cos^{3} x} dx$ is

• A. 1
• B. 2
• C. $\sqrt{2}$
• D. $2 \sqrt{2}$

Answer 1

Answer: (C)

$\sqrt{2}$

Answer 2

$\int\limits_{0}^{\frac{\pi}{4}} \frac{1+\sin^{2} x}{ \cos^{3} x} dx$
$= \int\limits_{0}^{\frac{\pi }{4}} \frac{\cos^{2} x + \sin^{2}x +\sin^{2} x}{\cos^{3} x} dx$
$= \int\limits_{0}^{\frac{\pi }{4}} \frac{\cos^{2} x +2 \sin^{2}x }{\cos ^{3} x.\cos^{3} x} dx$
$= \int\limits_{0}^{\frac{\pi }{4}} \left(\sec x +2 \tan^{2} x \sec x \right)dx$
$= \int\limits_{0}^{\frac{\pi }{4}} \sec x dx +2 \int\limits_{0}^{\frac{\pi }{4}} \tan x\left(\sec x \tan x\right)dx$
Put $\sec x =t \Rightarrow \sec x \tan x dx =dt$

If $x = 0 \Rightarrow t = 1$ and $x = \frac{\pi}{4} \Rightarrow \: t = \sqrt{2}$
$= \left[\log\left|\sec x +\tan x\right|\right]^{\pi 4}_{0} + 2\int\limits_{1}^{\sqrt{2}} \sqrt{t^{2} - 1} dt$
$= \log\left|\left(\sqrt{2} +1\right)\right|- \log\left|\left(1+0\right)\right| +2 \left[\frac{t}{2} \sqrt{t^{2} -1} - \frac{1}{2} \log\left|t+ \sqrt{t^{2} -1}\right|\right]^{\sqrt{2}}_{1}$
$= \log\left|\sqrt{2} + 1\right|+ 2\left[\frac{\sqrt{2}}{2} - \frac{1}{2} \log\left|\sqrt{2} + 1\right|\right] = \sqrt{2}$