Indefinite integral:$\int \cos(2018x)\sin^{2016}(x)dx$

Question

Evaluate $\int \cos(2018x)\sin^{2016}(x)dx$

I could solve this using IBP, $$I=\int \cos(2017x+x)\sin^{2016}(x)dx$$ $$=\int \cos(2017x)\sin^{2016}(x) \cos(x)dx -\int \sin^{2017}(x)\sin(2017x)$$ $$=\frac{\cos(2017x)\sin^{2017}(x)}{2017} +\int \frac{2017\sin^{2017}(x)\sin(2017x)}{2017}dx - \int \sin^{2017}(x)\sin(2017x)$$ $$=\frac{\cos(2017x)\sin^{2017}(x)}{2017}+c$$

However I while trying to solve this question using complex numbers, I didn't obtain the final result. Here's what I did:

The give integral is $\int e^{2018ix} (\frac{e^{ix}-e^{-ix}}{2i})^{2016} dx$(considering real of this and in subsequent steps)

$$=\frac{1}{2^{2016}} \int e^{2ix}(e^{2ix}-1)^{2016} dx$$.

$e^{2ix}-1=t$, $e^{2ix}2idx=dt$

$$=\frac{1}{2^{2016}} \int t^{2017} dt/2i$$.

$$=\frac{t^{2017}}{2^{2017}i \cdot 2017}+c$$

So answer is $$-Im(\frac{t^{2017}}{2^{2017} \cdot 2017})$$

I'm unable to evaluate $t^{2017}=(e^{2ix}-1)^{2017}$ and get it to the form as obtained by IBP. I did the binomial expansion however, I wasn't able to get it to a nice form.

Also is there a generalisation to this problem? Can $\int \cos(mx) \sin^{n}(x) dx$ also be evaluated like this?(not by using reduction formula)

Answer 1

$$t^{2017}=(e^{2ix}-1)^{2017}=(e^{ix})^{2017}(e^{ix}-e^{-ix})^{2017}$$ So, $$t^{2017}=(e^{2017ix})(2i\sin x)^{2017}=(\cos 2017x+i\sin 2017x)(2i\sin x)^{2017}$$ If you take Im part of this expression, you get $$Im(t^{2017})=2^{2017}\cos(2017 x)(\sin x)^{2017}$$ So, the answer $$\frac{Im(t^{2017})}{2017\cdot 2^{2017}}=\frac{\cos(2017 x)(\sin x)^{2017}}{2017}$$ Coincides with your other answer.

Answer 2

Note that $$ \frac{1}{n+1}\frac{d}{dx} \sin^{n+1}(x)\cos((n+1)x) = \sin^n(x)\cos((n+1)x)\cos(x)-\sin^{n+1}(x)\sin((n+1)x) $$ $$ = \sin^n(x)\left(\cos((n+1)x)\cos(x)-\sin(x)\sin((n+1)x)\right) $$ $$ = \sin^n(x)\cos((n+2)x) $$So, the antiderivative is $\frac{1}{2017}\sin^{2017}(x)\cos(2017x)+C$. It might be possible to generalize this as well.