int frac{sin x}{sin x - cos x} , dx = | vedclass

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(B) Let $I = \int \frac{\sin x}{\sin x - \cos x} \, dx$.Multiply and divide by $2$: $I = \frac{1}{2} \int \frac{2\sin x}{\sin x - \cos x} \, dx$.Rewri

Vedclass · Accepted Answer

$\frac{1}{2}[\log (\sin x - \cos x) + x] + c$

Vedclass · Answer

(B) Let $I = \int \frac{\sin x}{\sin x - \cos x} \, dx$.Multiply and divide by $2$: $I = \frac{1}{2} \int \frac{2\sin x}{\sin x - \cos x} \, dx$.Rewrite the numerator: $2\sin x = (\sin x - \cos x) + (\sin x + \cos x)$.So,$I = \frac{1}{2} \int \frac{(\sin x - \cos x) + (\sin x + \cos x)}{\sin x - \cos x} \, dx$.$I = \frac{1}{2} \int \left( 1 + \frac{\sin x + \cos x}{\sin x - \cos x} \right) \, dx$.$I = \frac{1}{2} \left[ \int 1 \, dx + \int \frac{\sin x + \cos x}{\sin x - \cos x} \, dx \right]$.For the second integral,let $u = \sin x - \cos x$,then $du = (\cos x + \sin x) \, dx$.$I = \frac{1}{2} [x + \log |\sin x - \cos x|] + c$.

$\int \frac{\sin x}{\sin x - \cos x} \, dx = $

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