int frac{sin x+cos x}{sin x-cos x} d x= | vedclass

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(D) Let $I = \int \frac{\sin x+\cos x}{\sin x-\cos x} d x$. Consider the substitution $u = \sin x - \cos x$. Then,$du = (\cos x - (-\sin x)) dx = (\co

Vedclass · Accepted Answer

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

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(D) Let $I = \int \frac{\sin x+\cos x}{\sin x-\cos x} d x$. Consider the substitution $u = \sin x - \cos x$. Then,$du = (\cos x - (-\sin x)) dx = (\cos x + \sin x) dx$. Substituting these into the integral,we get: $I = \int \frac{1}{u} du$. Integrating with respect to $u$,we have: $I = \log |u| + c$. Substituting back $u = \sin x - \cos x$,we get: $I = \log |\sin x - \cos x| + c$.

$\int \frac{\sin x+\cos x}{\sin x-\cos x} d x=$

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