Question

The integral $\int\limits^{\frac{3\, \pi}{4}}_{\frac{\pi}{4}} \frac{dx}{ 1 + \cos \, x}$ is equal to :

• A. 2
• B. 4
• C. -1
• D. -2

Answer 1

Answer: (A)

$\int\limits^{\frac{3\, \pi}{4}}_{\frac{\pi}{4}} \frac{dx}{2 \cos^{2} \frac{x}{2}}dx = \frac{1}{2}\int\limits^{\frac{3\, \pi}{4}}_{\frac{\pi}{4}}\sec^{2} \frac{x}{2}dx$
$= \frac{1}{2}\left[\frac{\tan \frac{x}{2}}{\frac{1}{2}}\right]^{{\frac{3\pi}{4}}}_{{\frac{\pi}{4}}}$
$= \tan \frac{3\pi}{8}-\tan \frac{\pi}{8}$
$\left[\tan \frac{\pi}{8} = \sqrt{\frac{1-\cos \frac{\pi}{4}}{1+\cos \frac{\pi}{4}}} = \sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}} = \frac{\sqrt{2}-1}{1} \tan \frac{3\pi}{8} = \sqrt{\frac{1-\cos \frac{3\pi}{4}}{1+\cos \frac{3\pi}{4}}}= \sqrt{\frac{\sqrt{2}+1}{\sqrt{2}-1}} = \sqrt{2} + 1\right]$
$= \left(\sqrt{2}+1\right)-\left(\sqrt{2}-1\right)$
$= 2$