Various ways to calculate $\int \sin(x) \cos(x) \, \mathrm{d}x$ Consider the integral
$$\mathcal{I} := \int \sin(x) \cos(x) \, \mathrm{d} x$$
$
\newcommand{\II}{\mathcal{I}}
\newcommand{\d}{\mathrm{d}}
$
This is one of my favorite basic integrals to think about as an instructor, because on the face of it, there are a lot of different ways to solve it, many are accessible to Calculus I students, and can give some insights into the nature of integration and to some trigonometry identities. For instance:


*

*Substitution with $u = \sin(x)$ gives $$\II =  \frac{\sin^2(x)}{2} + C$$


*

*Substitution with $u = \cos(x)$ gives $$ \II = - \frac{\cos^2(x)}{2} + C$$


*

*(Noted in comments by Koro) Make the substitution
$$
u = \sec(x) \implies \d u = \sec(x) \tan(x) \, \d x= \frac{\sin(x)}{\cos^2(x)} \, \d x$$
Then
$$\begin{align*}
\II &= \int \sin(x) \cos(x) \frac{\cos^2(x)}{\sin(x)} \, \d u\\
&= \int u^{-3} \, \d u\\
&= - \frac{1}{2\sec^2(x)} + C\\
&= - \frac{\cos^2(x)}{2}  + C
\end{align*}$$
A similarly motivated substitution:
$$
u = \csc(x) \implies \d u = -\cot(x) \csc(x) \, \d x = - \frac{\cos(x)}{\sin^2(x)} \, \d x
$$
yields
$$\begin{align*}
\II &= -\int \sin(x) \cos(x) \frac{\sin^2(x)}{\cos(x)} \, \d u\\
&= -\int u^{-3} \, \d u\\
&=  \frac{1}{2\csc^2(x)} + C\\
&= \frac{\sin^2(x)}{2}  + C
\end{align*}$$


*

*Using $\sin(2x) = 2 \sin(x) \cos(x)$ readily leads to $$\II = -\frac{\cos(2x)}{4} + C$$


*

*Integration by parts differentiating $\sin(x)$ yields $$\II = \sin^2(x) - \II$$ which will yield a previous solution on solving for $\II$.



*

*Integration by parts differentiating $\cos(x)$ yields $$ \II = -\cos^2(x)- \II$$ which, similarly, yields a previous solution once we solve for $\II$.



*

*Using the Weierstrass substitution $t = \tan(x/2)$ gives
$$
\II = \int \frac{2t}{1+t^2} \frac{1-t^2}{1+t^2} \frac{2}{1+t^2} \, \mathrm{d}t = \frac{2t^2}{(1+t^2)^2} + C=\frac{2 \tan^2(x/2)}{(1 + \tan^2(x/2))^2} + C
$$


*

*We can use the complex sine and cosine:
$$
\sin(x) = \frac{e^{ix} - e^{-ix}}{2i} \qquad
\cos(x) = \frac{e^{ix} + e^{-ix}}{2}
$$
Then
$$
\II = \int \frac{e^{2ix} - e^{-2ix}}{4i} \, \mathrm{d} x = - \frac 1 8 \left( e^{2ix} + e^{-2ix} \right) + C = -\frac 1 4 \cos(2x) + C
$$

(Warning for Novices: The $C$ constant in each expression may not be the same as in others. These answers are equivalent ones for the integral, but the solutions without the $+C$ terms are not equal expressions. These hint at certain trigonometry identities, which is why I find them interesting.)

This has given us a set of solutions to $\II$ via a few basic methods, and a few less-basic but accessible methods.
My question is, what other solutions can you come up with for $\II$?
I'm particularly interested in answers which:

*

*are obviously not functionally equivalent to those already given

*give answers other than those already expressed (perhaps a hint at other identities or concepts of note?)

*use methods that you don't see in a typical calculus class, or methods that are rarely used

*use unusual but slick and effective techniques

or any combination thereof! I have no real motivation for this but my own curiosity, but I'm curious to see what you guys can think of.

Let's make a series out of integration by parts. Let
$$I = \int \frac{1}{2}\sin 2x\:dx = \frac{1}{2}x\sin2x-\int x\cos 2x\:dx$$
Then keep going
$$I = \frac{1}{2}x\sin 2x - \frac{1}{2}x^2\cos 2x - \int x^2\sin 2x \;dx$$
$$= \frac{1}{2}x\sin 2x - \frac{1}{2}x^2\cos 2x - \frac{1}{3}x^3\sin 2x + \int\frac{2}{3}x^3\cos 2x\:dx$$
$$= \frac{1}{2}x\sin 2x - \frac{1}{2}x^2\cos 2x - \frac{1}{3}x^3\sin 2x + \frac{1}{6}x^4\cos2x + \int \frac{1}{3}x^4\sin 2x\:dx$$
or in other words the sum of the following two series
$$I = \sin 2x\left(\frac{1}{2}x-\frac{1}{3}x^3+\frac{1}{15}x^5-\cdots\right)-\cos 2x\left(\frac{1}{2}x^2-\frac{1}{6}x^4+\frac{1}{45}x^6-\cdots\right)$$
