Question

Evaluate: ${\int }_{\pi /2}^{3\pi /2}\frac{1}{1+\cos x}dx$

• A. $2$
• B. $-2$
• C. $1/2$
• D. $-1/2$

Answer 1

The correct option is B $-2$

${\int }_{\pi /2}^{3\pi /2}\frac{1}{1+\cos x}dx$

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

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

$={\int }_{\pi /2}^{3\pi /2}\frac{1-\cos x}{{\sin }^{2}x}dx$

$={\int }_{\pi /2}^{3\pi /2}(co{\sec }^{2}x-\cot x.co\sec x)dx$

$={[-\cot x+co\sec x]}_{\pi /2}^{3\pi /2}$

$=\{-\cot (3\pi /2)+co\sec (3\pi /2\left)\right\}-\{-\cot (\pi /2)+co\sec (\pi /2\left)\right\}$

$=(0-1)-(0+1)=-2$