Question

The value of $I=\underset{0}{\overset{\pi /2}{\int }}\frac{{\left(sinx+cosx\right)}^2}{\sqrt{1+sin2x}}dx$ is

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (C)

$\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{{\left(sinx+cosx\right)}^2}{\sqrt{{\left(sinx+cosx\right)}^2}}dx=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(sinx+cosx\right)dx={\left|sinx+cosx\right|}_0^{\frac{\pi }{2}}=2.$