Evaluating integral $\int\frac{e^{\cos x}(x\sin^3x+\cos x)}{\sin^2x}dx $

Question

$$\int\frac{e^{\cos x}(x\sin^3x+\cos x)}{\sin^2x}dx $$ The usual form $\int e^x(f(x)+f'(x))dx $ does not apply here. What substitution should I make ?

Answer 1

This is a tricky one! We spot a lot of things here that look like derivatives, so we should try to reframe our expression to reflect that, and make it useful for us. Split it up as follows:

$$[e^{\cos x}\sin x]x+e^{\cos x}[\frac{\cos x}{\sin^2 x}]$$

Let's integrate by parts here, setting $u'=e^{\cos x}\sin x, \frac{\cos x}{\sin^2 x}$ and $v=x, e^{\cos x}$ respectively. Noting that $\int u'v\ dx = uv - \int uv'\ dx$, we see:

$$\int [e^{\cos x}\sin x]xdx=(-e^{\cos x})x-\int[-e^{\cos x}]dx$$

$$\int e^{\cos x}[\frac{\cos x}{\sin^2 x}]dx=(e^{\cos x})(\frac{-1}{\sin x})-\int[-e^{\cos x}\sin x](\frac{-1}{\sin x})dx$$

We then notice that the extra integrals we pick up cancel each other out perfectly, leaving us with:

$$-(x+\frac{1}{\sin x})e^{\cos x}+C$$