Question

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

• A. $0$
• B. $2$
• C. $4$
• D. $-2$

Answer 1

The correct option is B

$2$

Step 1: Simplify the equation :

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

According to the trigonometric property : $1+\cos 2x=2{\cos }^{2}x$

Substitute the above value in Equation $\left(1\right)$

$$\begin{gathered} I={\int }_0^{\mathrm{\pi }}\sqrt{\frac{2{\cos }^{2}x}{2}}dx \\ I={\int }_0^{\mathrm{\pi }}\sqrt{{\cos }^{2}x}dx \\ I={\int }_0^{\mathrm{\pi }}\left|\cos x\right|dx \end{gathered}$$

In the first quadrate $\cos x$ positive from $0to\frac{\mathrm{\pi }}{2}$ and in the second quadrate $\cos x$ negative from $\frac{\mathrm{\pi }}{2}to\mathrm{\pi }$, so that we can say that

$\left|\cos x\right|=\left\{\begin{array}{ll}\cos x,& 0\le x\le \frac{\mathrm{\pi }}{2}\\ -\cos x,& \frac{\mathrm{\pi }}{2}\le x\le \mathrm{\pi }\end{array}\right.$

Step 2: Substitute the above values in $\left|\cos x\right|$

$\begin{array}{rcl}I& =& {\int }_0^{\frac{\mathrm{\pi }}{2}}\cos xd+{\int }_{\frac{\mathrm{\pi }}{2}}^{\mathrm{\pi }}-\cos xdx\\ I& =& {\left[\sin x\right]}_0^{\frac{\mathrm{\pi }}{2}}+{\left[\sin x\right]}_{\frac{\mathrm{\pi }}{2}}^{\mathrm{\pi }}\\ I& =& \left(\sin \frac{\mathrm{\pi }}{2}-\sin 0\right)+\left(\sin \mathrm{\pi }-\sin \frac{\mathrm{\pi }}{2}\right)\\ I& =& \left(1-0\right)+\left(0+1\right)\\ I& =& 1+1\\ I& =& 2\end{array}$

Hence, the correct answer is Option (B).