Home › MHT-CET › Mathematics › Definite Integration › $\int_0^{\pi/2} \sqrt{1 - \cos 4x}\,dx$ equals:
$\int_0^{\pi/2} \sqrt{1 - \cos 4x}\,dx$ equals:
A$\sqrt 2$
B$1$
C$2$
D$\pi$
Answer & Solution
Correct answer: A. $\sqrt 2$
Use $1 - \cos 4x = 2\sin^2 2x$, so $\sqrt{1-\cos 4x} = \sqrt 2|\sin 2x|$. On $[0, \pi/2]$, $\sin 2x \ge 0$ (since $2x \in [0,\pi]$). $\int_0^{\pi/2} \sqrt 2 \sin 2x\,dx = \sqrt 2 [-\cos 2x/2]_0^{\pi/2} = \sqrt 2 [1/2 + 1/2] = \sqrt 2$.
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