Finding indefinite integral $\int \frac{x^2}{(x \sin x + \cos x)^2} dx $

Question

I need hint in finding the integral of $$\int \frac{x^2}{(x \sin x + \cos x)^2} dx $$ I tried dividing the term by $x^2\cos^2x$ and then substituting $\tan x$.

Answer 1

with the hint from above we get $$\frac{\sin (x)-x \cos (x)}{x \sin (x)+\cos (x)}+C$$

Answer 2

From the formula $$ D\frac fg=\frac{f'g-fg'}{g^2}, $$ we might guess that we have a quotient $f/g$ with $g=x\sin x+\cos x$, but what is then $f$?

First, we note that $$ g'=x\cos x. $$

Well, with the trig-one, we can write $$ \begin{aligned} \frac{x^2}{(x\sin x+\cos x)^2} &=\frac{x^2\sin^2x+x^2\cos^2x}{(x\sin x+\cos x)^2} =\frac{x\sin x(x\sin x+\cos x)-x\cos x(\sin x-x\cos x)}{(x\sin x+\cos x)^2}\\ &=\frac{x\sin x\cdot g-g'\cdot (\sin x-x\cos x)}{(x\sin x+\cos x)^2}. \end{aligned} $$ By pattern matching, $f=\sin x-x\cos x$ (note that, then $f'=x\sin x$).

We conclude that our guess was lucky, and that $$ \int \frac{x^2}{(x\sin x+\cos x)^2}\,dx = \frac{\sin x-x\cos x}{x\sin x+\cos x}+C. $$

Answer 3

Consider the following $$f(x)=\frac{\sin x-x\cos x}{x\sin x+\cos x}$$

$$\implies f'(x)=\frac{d}{dx}\left(\frac{\sin x-x\cos x}{x\sin x+\cos x}\right)=\frac{(x\sin x+\cos x)\frac{d}{dx}(\sin x-x\cos x)-(\sin x-x\cos x)\frac{d}{dx}(x\sin x+\cos x)}{(x\sin x+\cos x)^2}$$ $$ =\frac{(x\sin x+\cos x)(x\sin x)-(\sin x-x\cos x)(x\cos x)}{(x\sin x+\cos x)^2}$$ $$ =\frac{x^2\sin^2 x+x\sin x\cos x-(x\sin x\cos x-x^2\cos^2 x)}{(x\sin x+\cos x)^2}$$ $$ =\frac{x^2\sin^2 x+x\sin x\cos x-(x\sin x\cos x-x^2\cos^2 x)}{(x\sin x+\cos x)^2}$$ $$ =\frac{x^2\sin^2 x+x\sin x\cos x-x\sin x\cos x+x^2\cos^2 x)}{(x\sin x+\cos x)^2}$$ $$ =\frac{x^2(\sin^2 x+\cos^2 x)}{(x\sin x+\cos x)^2}$$ $$f'(x) =\frac{x^2}{(x\sin x+\cos x)^2}$$ Now, integrating both the sides we get $$\int(f'(x))dx=\int \frac{x^2}{(x\sin x+\cos x)^2}$$ $$\implies \int \frac{x^2}{(x\sin x+\cos x)^2}=f(x)+C$$ $$ \int \frac{x^2}{(x\sin x+\cos x)^2}=\frac{\sin x-x\cos x}{x\sin x+\cos x}+C$$