Integrate the function (indefinite)

Question

See the thing is that even putting $\sin^2x + \cos^2x$ in the numerator and simplifying it, I am unable to proceed further after 2 simplification: $(x\sin x/(x\sin x+ \cos x)^2) + ((x^2\cos^2x - x\sin x\cos x/(x\sin x+\cos x)^2)$ $$ \int\frac{(x^2)}{(x\sin x + \cos x)^2}$$ I think ILATE would be use but don't know at what point. Can anyone provide a better view. Many thanks.

Answer 1

The fact that $(x\sin x+\cos x)'=x\cos x$ strongly suggests to multiply the numerator and denominator of the integrand in $\cos x$ as follows $$ \int\frac{x^2}{(x\sin x + \cos x)^2}dx { = \int\frac{x^2\cos x}{(x\sin x + \cos x)^2\cos x}dx. } $$ Integration by parts with $u=-\frac{x}{\cos x}$ and $v=\frac{1}{x\sin x + \cos x}$ as $\int udv=uv-\int vdu$ gives $$ \int\frac{x^2\cos x}{(x\sin x + \cos x)^2\cos x}dx {= -\frac{x}{\cos x}\frac{1}{x\sin x+\cos x}+\int\left[\frac{1}{\cos x}+\frac{x\sin x}{\cos^2 x}\right]\frac{1}{x\sin x+\cos x}dx \\=-\frac{x}{\cos x}\frac{1}{x\sin x+\cos x}+\int\frac{1}{\cos^2 x}dx \\=-\frac{x}{\cos x}\frac{1}{x\sin x+\cos x}+\tan x+C. } $$ Therefore,

$$\int\frac{x^2}{(x\sin x + \cos x)^2}dx=-\frac{x}{\cos x}\frac{1}{x\sin x+\cos x}+\tan x+C.$$