$$\int \frac{x^2}{(x\cos x -\sin x)(x\sin x +\cos x)}\ dx$$
My approach :
Dividing the denominator by $\cos^2x$ we get $\dfrac{x^2\sec^2x }{(x -\tan x)(x\tan x +1)}$ then
$$\int \frac{x^2\sec^2x}{x^2\tan x -x\tan^2x+x-\tan x}\ dx=\int \frac{x^2(1+\tan^2x)}{x^2\tan x -x\tan^2x+x-\tan x}dx$$
But I am not getting any relation between numerator and denominator so that I will take any substitution and solve further please suggest whether it is correct and how to proceed in this. Thanks.
Best Answer
HINT :
Rewrite the integrand $$ \frac{x^2}{(x\cos x -\sin x)(x\sin x +\cos x)} $$ as $$ \frac{x\cos x}{x\sin x +\cos x}+\frac{x\sin x}{x\cos x -\sin x} $$ then $$ \frac{\color{red}{\sin x}+x\cos x-\color{red}{\sin x}}{x\sin x +\cos x}+\frac{\color{blue}{\cos x}+x\sin x-\color{blue}{\cos x}}{x\cos x -\sin x}. $$ Now let $u=x\sin x +\cos x$ and $v=x\cos x -\sin x$.