Question

Integrate ${\int }_0^{\pi }\sqrt{1+\sin 2x}dx$

• A. 3
• B. 2
• C. 4
• D. 5

Answer 1

Answer: (B)

2

$I={\int }_0^{\pi }\sqrt{1+\sin 2xdx}$

we have $f\left(x\right)=\sqrt{1+\sin 2x}$

$=\sqrt{{\sin }^{2}x+{\cos }^{2}x+2\sin x\cos x}$

$=\sqrt{(\sin x+\cos x{)}^2}$

$=\sin x+\cos x$

$I={\int }_0^{\pi }(\sin x+\cos x)dx$

$=[-\cos x{]}_0^{\pi }+[\sin x{]}_0^{\pi }$

$=-[\cos \pi -\cos 0]+[\sin \pi -\sin 0]$

$=-[-1-1]+0$

$=2$