Question

If $\int \frac{{\cos }^{2}x\sin x}{\sin x-\cos x}dx=\frac{A}{16}\log |\sin x-\cos x|+\frac{1}{8}(\sin 2x+\cos 2x)+C$, then $A$ is equal to.

Answer 1

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

$=\frac{1}{2}\int \frac{(1-{(\sin x-\cos x)}^2)}{\sin x-\cos x}\cos xdx$

$=\frac{1}{2}\int \frac{\cos x}{\sin x-\cos x}dx-\frac{1}{2}\int (\sin x-\cos x)\cos xdx$

$=\frac{1}{4}\int \frac{-(\sin x-\cos x)+(\cos x+\sin x)}{\sin x-\cos x}dx-\frac{1}{4}\int (\sin 2x-1-\cos 2x)dx$

$=\frac{1}{4}(-x+\log |\sin x-\cos x|)+\frac{1}{8}\cos 2x+\frac{1}{4}x+\frac{1}{8}\sin 2x+c$

$=\frac{1}{4}\log |\sin x-\cos x|+\frac{1}{8}(\sin 2x+\cos 2x)+c$

Therefore $A=4$