MCQ
$\int_0^\pi \sqrt{\frac{1+\cos 2 x}{2}} d x$ is equal to
  • A
    $0$
  • 2
  • C
    1
  • D
    -1

Answer

Correct option: B.
2
(B)
$\int_0^\pi \sqrt{\frac{1+\cos 2 x}{2}} d x=\int_0^\pi|\cos x| d x$
$=\int_0^{\pi / 2} \cos x d x-\int_{\pi / 2}^\pi \cos x d x$
$\begin{array}{l}=[\sin x]_0^{\pi / 2}-[\sin x]_{\pi / 2}^\pi \\ =\left[\sin \frac{\pi}{2}-\sin 0\right]-\left[\sin \pi-\sin \frac{\pi}{2}\right]=1+1=2\end{array}$

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