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.
Best Answer
This integral tolarates almost any substitution. In general, let $\sin x = f(u)$, where the substitution function $f(u)$ is of suitable range, but otherwise arbitrary. Then, $\cos x \> dx =f’(u)\>du$ and $$\int \sin x \cos x \>dx =\int f(u)f’(u)du =\frac12f^2(u)=\frac12\sin^2 x+C\tag1 $$
Thus, whatever form of $f(u)$ to be used, regardless of its complexity, invariably leads to the result $\frac12\sin^2x+C$, as shown in (1). As an example, the substitution $u=\sec x$ listed in OP corresponds to $$f(u)=\sin(\sec^{-1} u)=\frac{\sqrt{u^2-1}}u,\>\>\> f’(u)= \frac1{u^2\sqrt{u^2-1}}$$ Hence, there are countless number of ways to integrate $\int \sin x \cos x \>dx$, all because of unlimited choices of $f(u)$.