Question

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

Answer 1

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

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

$\frac{{\sin }^{2}x}{{\sin }^2-{\cos }^{2}x}dx+\frac{1}{2}\int \frac{2\sin x\cos x}{{\sin }^{2}x-{\cos }^{2}x}dx$

${\cos }^{2}x-si{n}^{2}x=\cos 2x$ and $2\sin x\cos x=\sin \left(2x\right)$

$-\int \frac{1-\cos 2x}{2\cos 2x}dx-\int \frac{\sin 2x}{2\cos 2x}dx$ $[\because \int \tan xdx=ln|\sec x|+c]$

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$-\int \frac{\sec 2x}{2}dx+\int \frac{1}{2}dx-\int \frac{\tan 2x}{2}dx[\because \int \sec xdx=ln|\sec x+\tan x|+c]$

$-\frac{1}{2}\times \frac{1}{2}ln|\sec 2x+\tan 2x|+\frac{x}{2}-\frac{1}{2}\times \frac{1}{2}ln|\sec 2x|+C$

$-\frac{ln|\sec 2x+\tan 2\times 1}{4}+\frac{x}{2}-\frac{ln|\sec 2x|}{4}+C$