# How do you find the integral of (sinx)(cosx)dx?

Jul 5, 2015

There are several ways to write the correct answer.

#### Explanation:

$\int \sin x \cos x \mathrm{dx}$

Solution 1
With $u = \sin x$ we get ${\sin}^{2} \frac{x}{2} + C$

Solution 2
With $u = \cos x$, we get $- {\cos}^{2} \frac{x}{2} + c$

Solution 3
Noting that $\sin x \cos x = \frac{1}{2} \sin \left(2 x\right)$, we rewrite:

$\int \sin x \cos x \mathrm{dx} = \frac{1}{2} \int \sin \left(2 x\right) \mathrm{dx}$

Now let $u = 2 x$ to get $\frac{1}{4} \cos \left(2 x\right) + \mathbb{C}$

(I love this problem and use it every time I teach Calulus I.)

Note 1: It is not an accident that i used different notations for the constants in each solution.

Note 2:
The difference (literally -- that is, the subtraction) between the apparently different solutions are constants.
For example: ${\sin}^{2} \frac{x}{2}$ minus $- {\cos}^{2} \frac{x}{2}$ simplifies to:

${\sin}^{2} \frac{x}{2} - \left(- {\cos}^{2} \frac{x}{2}\right) = \frac{{\sin}^{2} x + {\cos}^{2}}{2} = \frac{1}{2}$