Question

The value of $I={\int }_0^{\pi /2}\frac{(sinx+cosx{)}^2}{\sqrt{1+sin2x}}dx$ is

• A. 3
• B. 1
• C. 2
• D. 5

Answer 1

The correct option is C 2

Consider the problem

$\begin{array}{c}{\int }_0^{\frac{\pi }{2}}\frac{{\left(\sin x+\cos x\right)}^2}{\sqrt{1+\sin 2x}}dx\\ ={\int }_0^{\frac{\pi }{2}}\frac{\left({\sin }^{2}x+{\cos }^{2}x+2\sin x\cos x\right)}{\sqrt{1+\sin 2x}}dx\\ ={\int }_0^{\frac{\pi }{2}}\frac{1+\sin 2x}{\sqrt{1+\sin 2x}}dx\left(\begin{array}{c}\because {\sin }^{2}x+{\cos }^{2}x=1\\ \&\sin 2x=2\sin x\cos x\end{array}\right)\\ ={\int }_0^{\frac{\pi }{2}}\sqrt{1+\sin 2x}dx\\ ={\int }_0^{\frac{\pi }{2}}\sqrt{{\sin }^{2}x+{\cos }^{2}x+2\sin x\cos x}dx\\ ={\int }_0^{\frac{\pi }{2}}\sqrt{{\left(\sin x+\cos x\right)}^2}dx\\ ={\int }_0^{\frac{\pi }{2}}\left(\sin x+\cos x\right)dx\\ ={{\left(-\cos x+\sin x\right)}_0}^{\frac{\pi }{2}}\\ =\left(0+1\right)-\left(-1+0\right)\\ =1+1\\ =2\end{array}$

Hence, Option $C$ is the correct answer.