Question

Evaluate: ${\int }_{\frac{\pi }{4}}^{\frac{3\pi }{4}}\frac{dx}{1+\cos x}$

• A. $-2$
• B. $2$
• C. $4$
• D. $-1$

Answer 1

Answer: (B)

$2$

Let $I={\int }_{\frac{\pi }{4}}^{\frac{3\pi }{4}}\frac{dx}{1+\cos x}$

$I={\int }_{\frac{\pi }{4}}^{\frac{3\pi }{4}}\frac{dx}{1+2{\cos }^2\frac{x}{2}-1}$

$I={\int }_{\frac{\pi }{4}}^{\frac{3\pi }{4}}\frac{dx}{2{\cos }^2\frac{x}{2}}$

$I=\frac{1}{2}{\int }_{\frac{\pi }{4}}^{\frac{3\pi }{4}}{\sec }^2\left(\frac{x}{2}\right)dx$

$I=\frac{1}{2}\times \frac{1}{\frac{1}{2}}{\left[{\tan }^2\left(\frac{x}{2}\right)\right]}_{\frac{\pi }{4}}^{\frac{3\pi }{4}}$

$I=[{\tan }^2\left(\frac{3\pi }{8}\right)-{\tan }^2\left(\frac{\pi }{8}\right)]$

$I=\sqrt{2}+1-(\sqrt{2}-1)I=2$

Hence, ${\int }_{\frac{\pi }{4}}^{\frac{3\pi }{4}}\frac{dx}{1+\cos x}=2$