Mathematics - Indefinite Integration Question with Solution | TestHub

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

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

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

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

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

$=\frac{1}{4}\ell \mathrm{n}|\sin \mathrm{x}-\cos \mathrm{x}|+\frac{1}{8}(\sin 2\mathrm{x}+\cos 2\mathrm{x})$