$\int_0^{\pi/2} \sqrt{1 - \cos 4x}\,dx$ equals:

Question

Accepted 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 $2x \in [0,\pi]$).
$\int_0^{\pi/2} \sqrt 2 \sin 2x\,dx = \sqrt 2 [-\cos 2x/2]_0^{\pi/2} = \sqrt 2 [1/2 + 1/2] = \sqrt 2$.

Answer

$1$

Answer

$2$

Answer

$\pi$

$\int_0^{\pi/2} \sqrt{1 - \cos 4x}\,dx$ equals:

Keep practising Unit 6B: Definite Integration & Riemann Sums