_0^(/2) √(1 - 2x) dx is equal to : Option 1: 2√2 Option 2: 2(√2 + 1) Option 3: 2 Option 4: 2(√2 - 1)

Answer: 2(√2 - 1). Solution: 1- 2x=( x- x)^2, so √(1- 2x)=, x- x, . Split at /4: _0^(/4)( x- x)dx+_(/4)^(/2)( x- x)dx =(√2-1)+(√2-1)=2(√2-1).

Calculus

Integrals

Medium

core

$\int_0^{\pi/2} \sqrt{1 - \sin 2x} \, dx$ is equal to :

$2\sqrt{2}$

$2(\sqrt{2} + 1)$

$2$

$2(\sqrt{2} - 1)$

✅ Correct Option: 4

$1-\sin 2x=(\sin x-\cos x)^2$ , so $\sqrt{1-\sin 2x}=|\sin x-\cos x|$ .

Split at $\pi/4$ : $\int_0^{\pi/4}(\cos x-\sin x)dx+\int_{\pi/4}^{\pi/2}(\sin x-\cos x)dx$

$=(\sqrt{2}-1)+(\sqrt{2}-1)=2(\sqrt{2}-1)$ .

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