If x notin [2npi - frac{pi}{4}, 2npi + frac{3 | vedclass

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Question

(C) We know that $1 - \sin 2x = \cos^2 x + \sin^2 x - 2 \sin x \cos x = (\cos x - \sin x)^2$. Thus,$\int \sqrt{1 - \sin 2x} \, dx = \int |\cos x - \si

Vedclass · Accepted Answer

$-\cos x - \sin x + c$

Vedclass · Answer

(C) We know that $1 - \sin 2x = \cos^2 x + \sin^2 x - 2 \sin x \cos x = (\cos x - \sin x)^2$. Thus,$\int \sqrt{1 - \sin 2x} \, dx = \int |\cos x - \sin x| \, dx$. Given $x 
otin [2n\pi - \frac{\pi}{4}, 2n\pi + \frac{3\pi}{4}]$,the expression $\cos x - \sin x$ is negative in this interval. Therefore,$|\cos x - \sin x| = -(\cos x - \sin x) = \sin x - \cos x$. Integrating this,we get $\int (\sin x - \cos x) \, dx = -\cos x - \sin x + c$.

If $x \notin [2n\pi - \frac{\pi}{4}, 2n\pi + \frac{3\pi}{4}]$ and $n \in Z$,then $\int \sqrt{1 - \sin 2x} \, dx = $

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