Question

${\int }_0^{\pi }\sqrt{\frac{1+Cos2x}{2}}dx=$

• A. $-2$
• B. $2$
• C. $0$
• D. $-3$

Answer 1

Answer: (B)

$2$

${\int }_0^{\pi }\sqrt{\frac{1+\cos 2x}{2}}dx={\int }_0^{\pi }\sqrt{\frac{1+2{\cos }^{2}x-1}{2}}dx$
$={\int }_0^{\pi }\sqrt{{\cos }^{2}x}dx={\int }_0^{\frac{\pi }{2}}\cos xdx-{\int }_{\frac{\pi }{2}}^{\pi }\cos xdx$
$={\left[\sin x\right]}_0^{\frac{\pi }{2}}-{\left[\sin x\right]}_{\frac{\pi }{2}}^{\pi }=1-0-0+1=2$