Question

The value of the integral $\underset{\pi /6}{\overset{\pi /2}{\int }}\left(\frac{1+\sin 2x+\cos 2x}{\sin x+\cos x}\right)dx$ is equal to

• A. 16
• B. 8
• C. 4
• D. 1

Answer 1

Answer: (D)

Let $I=\underset{\pi /6}{\overset{\pi /2}{\int }}\left(\frac{1+\sin 2x+\cos 2x}{\sin x+\cos x}\right)dx$
$=\underset{\pi /6}{\overset{\pi /2}{\int }}\left(\frac{1+2\sin x\cos x+2{\cos }^{2}x-1}{(\sin x+\cos x)}\right)dx$
$=\underset{\pi /6}{\overset{\pi /2}{\int }}\frac{2\cos x(\sin x+\cos x)}{(\sin x+\cos x)}dx$
$=\underset{\pi /6}{\overset{\pi /2}{\int }}2\cos xdx=2[\sin x{]}_{\pi /6}^{\pi /2}$
$=2\left(\sin \frac{\pi }{2}-\sin \frac{\pi }{6}\right)$
$=2\left(1-\frac{1}{2}\right)=2\times \frac{1}{2}=1$