Question

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

• A. $16$
• B. $8$
• C. $4$
• D. $1$

Answer 1

The correct option is D $1$

Let

$I={\int }_{\pi /6}^{\pi /2}\left(\frac{1+\sin 2x+\cos 2x}{\sin x+\cos x}\right)dx$

$={\int }_{\pi /6}^{\pi /2}\left(\frac{1+2\sin x\cos x+2{\cos }^{2}x-1}{(\sin x+\cos x)}\right)dx$

$={\int }_{\pi /6}^{\pi /2}\frac{2\cos x(\sin x+\cos x)}{(\sin x+\cos x)}dx$

$={\int }_{\pi /6}^{\pi /2}2\cos xdx=2[\sin x{]}_{\pi /6}^{\pi /2}$

$=2(\sin \frac{\pi }{2}-\sin \frac{\pi }{6})=2(1-\frac{1}{2})=2\times \frac{1}{2}=1$