Evaluate $$I=\int \frac{(x^2+x^2 \csc ^2 x) \:dx}{\sin ^2 x}$$
My try:
we have $$I=\int x^2 \csc ^2 x\,dx+\int x^2 \csc ^4 x\,dx=P+Q$$
where $$P=\int x^2 \csc ^2 x\,dx$$ and $$Q=\int x^2 \csc ^4 x\,dx=\int x^2 \csc^2 x \times \csc ^2 x\,dx$$
Using Parts for $P$ we get
$$P=x^2 (-\cot x)+\int 2x \cot x\,dx \tag{1}$$
Using Parts for $Q$ we get
$$Q=x^2 \csc ^2 x(-\cot x)+\int \left(-2x^2 \csc^2 x\cot x+2x\csc ^2 x\right)\cot x\,dx$$ $\implies$
$$Q=x^2 \csc ^2 x(-\cot x)+\int \left(-2x^2 \csc^2 x\cot^2 x\right)dx+\int 2x \csc^2 x\cot x\,dx$$ $\implies$
$$Q=x^2 \csc ^2 x(-\cot x)+\int \left(-2x^2 \csc^4 x\right)dx+\int 2x^2 \csc^2 x\,dx+\int 2x \csc^2 x\cot x\,dx$$ $\implies$
$$3Q=x^2 \csc ^2 x(-\cot x)+2P+\int 2x \csc^2 x\cot x\,dx \tag{2}$$ Now
$$\int 2x \csc^2 x\cot x\,dx=2x \cot x(-\cot x)+\int \left (-x \csc^2 x+\cot x \right)\cot x\,dx$$
$\implies$
$$\int 2x \csc^2 x\cot x\,dx=\frac{2x \cot x(-\cot x)+\int \left (\cot x \right)\cot x\,dx}{3}$$
$\implies$
$$\int 2x \csc^2 x\cot x\,dx=\frac{2x \cot x(-\cot x)-\cot x-x}{3}$$
$$3Q-2P=x^2 \csc ^2 x(-\cot x)+\frac{2x \cot x(-\cot x)-\cot x-x}{3}$$
But with this can we find $P+Q$?