Question

The integral $\underset{\frac{\pi }{4}}{\overset{\frac{3\pi }{4}}{\int }}\frac{dx}{1+\cos x}$ is equal to :

• A. 2
• B. 4
• C. -1
• D. -2

Answer 1

Answer: (A)

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