Question

$\int \frac{x^2}{{\left(x\sin x+\cos x\right)}^2}\text{dx}$

Answer 1

$I=\int \frac{x^2}{(x\sin x+\cos x{)}^2}dx$

$=\int \frac{x^2\cos x}{\cos x(x\sin x+\cos x{)}^2}$

$=\int \frac{x}{\cos x}\cdot \frac{x\cos x}{(x\sin x+\cos x{)}^2}dx$

Let $\frac{x}{\cos x}=I$ function $\frac{x\cos x}{(x\sin x+\cos x{)}^2}=$II function

Now,

$\int \frac{x\cos xdx}{(x\sin x+\cos x{)}^2}$

Let $x\sin x+\cos x=t$

$\Rightarrow (x\cos x+\sin x-\sin x)=dt/dx$

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

Now,

$I=\frac{x}{\cos x}\int \frac{x\cos x}{(x\sin x+\cos x{)}^2}+\int \frac{(\cos x1+x\sin x)}{(\cos x{)}^2}\times \frac{1}{(x\sin x+\cos x)}dx$

$=\frac{x}{\cos x}\int \frac{x\cos x}{(x\sin x+\cos x{)}^2}dx+\int {\sec }^{2}xdx$

$=\frac{x}{\cos x}\times (-\frac{1}{x\sin x+\cos x})+\tan x+c$

$=-\frac{x}{\cos x(x\sin x+\cos x)}+\tan x+c$.