int frac{cos 2x}{(cos x + sin x)^2} , dx = | vedclass

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Question

(C) We know that $\cos 2x = \cos^2 x - \sin^2 x = (\cos x - \sin x)(\cos x + \sin x)$.Substituting this into the integral:$\int \frac{(\cos x - \sin x

Vedclass · Accepted Answer

$\log (\cos x + \sin x) + c$

Vedclass · Answer

(C) We know that $\cos 2x = \cos^2 x - \sin^2 x = (\cos x - \sin x)(\cos x + \sin x)$.Substituting this into the integral:$\int \frac{(\cos x - \sin x)(\cos x + \sin x)}{(\cos x + \sin x)^2} \, dx = \int \frac{\cos x - \sin x}{\cos x + \sin x} \, dx$.Let $t = \cos x + \sin x$. Then $dt = (-\sin x + \cos x) \, dx = (\cos x - \sin x) \, dx$.The integral becomes $\int \frac{1}{t} \, dt = \log |t| + c$.Substituting back $t = \cos x + \sin x$,we get $\log |\cos x + \sin x| + c$.

$\int \frac{\cos 2x}{(\cos x + \sin x)^2} \, dx = $

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