Question

The value of ${\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\left(\sin \right(x)+\cos {\left(x\right))}^2}{\sqrt{1+\sin \left(2x\right)}}dx=$

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is C

$2$

Explanation for correct option

Given: ${\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\left(\sin \right(x)+\cos {\left(x\right))}^2}{\sqrt{1+\sin \left(2x\right)}}dx$

Let ${\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\left(\sin \right(x)+\cos {\left(x\right))}^2}{\sqrt{1+\sin \left(2x\right)}}dx=I$

$$\begin{gathered} I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\left(\sin \right(x)+\cos {\left(x\right))}^2}{\sqrt{1+\sin \left(2x\right)}}dx \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\left({\sin }^2\right(x)+\cos {}^2\left(x\right)+2\sin \left(x\right)·\cos \left(x\right))}{\sqrt{1+\sin \left(2x\right)}}dx\left[as{(a+b)}^2=a^2+b^2+2ab\right] \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{1+\sin \left(2x\right)}{\sqrt{1+\sin \left(2x\right)}}dx\left[as{\sin }^{2}x+{\cos }^{2}x=1and2\sin \left(x\right)·\cos \left(x\right)=\sin \left(2x\right)\right] \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\sqrt{1+\sin \left(2x\right)}dx \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\sqrt{\left({\sin }^2(x\right)+{\cos }^2\left(x\right)+2\sin \left(x\right)\cos \left(x\right)}dx \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\sqrt{{\left(\sin \left(x\right)+\cos \left(x\right)\right)}^2}dx \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\left(\sin \left(x\right)+\cos \left(x\right)\right)dx \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\sin \left(x\right)·dx+{\int }_0^{\frac{\mathrm{\pi }}{2}}\cos \left(x\right)·dx \\ ={\left|-\cos \left(x\right)\right|}_0^{\frac{\mathrm{\pi }}{2}}+{\left|\sin \left(x\right)\right|}_0^{\frac{\mathrm{\pi }}{2}} \\ =\left|\cos \left(\frac{\mathrm{\pi }}{2}\right)-\cos \left(0\right)\right|+\left|\sin \left(\frac{\mathrm{\pi }}{2}\right)-\sin \left(0\right)\right| \\ =1+1 \\ =2 \end{gathered}$$

Hence, option C is correct.