int frac{cos 2 x}{(sin x+cos x)^{2}} d x is | vedclass

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(B) Let $I = \int \frac{\cos 2x}{(\sin x + \cos x)^2} dx$. Using the identity $\cos 2x = \cos^2 x - \sin^2 x$,we have: $I = \int \frac{\cos^2 x - \sin

Vedclass · Accepted Answer

$\log |\sin x+\cos x|+C$

Vedclass · Answer

(B) Let $I = \int \frac{\cos 2x}{(\sin x + \cos x)^2} dx$. Using the identity $\cos 2x = \cos^2 x - \sin^2 x$,we have: $I = \int \frac{\cos^2 x - \sin^2 x}{(\sin x + \cos x)^2} dx$. Factorizing the numerator as a difference of squares: $I = \int \frac{(\cos x - \sin x)(\cos x + \sin x)}{(\sin x + \cos x)^2} dx$. Simplifying the expression: $I = \int \frac{\cos x - \sin x}{\sin x + \cos x} dx$. Let $t = \sin x + \cos x$. Then $dt = (\cos x - \sin x) dx$. Substituting these into the integral: $I = \int \frac{1}{t} dt = \log |t| + C$. Substituting back $t = \sin x + \cos x$: $I = \log |\sin x + \cos x| + C$. Thus,the correct option is $B$.

$\int \frac{\cos 2 x}{(\sin x+\cos x)^{2}} d x$ is

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