Question

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

Answer 1

There are several ways to write the correct answer.

Explanation:

`intsinxcosxdx`

Solution 1
With `u = sinx` we get `sin^2x/2 +C`

Solution 2
With `u = cosx`, we get `-cos^2x/2 + c`

Solution 3
Noting that `sinxcosx = 1/2sin(2x)`, we rewrite:

`intsinxcosxdx = 1/2 int sin(2x) dx`

Now let `u = 2x` to get `1/4 cos(2x) + CC`

(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^2x/2` minus `-cos^2x/2` simplifies to:

`sin^2x/2 - (-cos^2x/2) = (sin^2x+cos^2)/2 = 1/2`