Question

How do you find the integral of `sin(x)cos(x)`?

Answer 1

`int sin(x)cos(x) dx = 1/2 sin^2(x) + C`

Explanation:

Notice that using the power and chain rules we find:

`d/(dx) sin^2(x) = 2 sin(x) cos(x)`

So:

`int sin(x)cos(x) dx = 1/2 sin^2(x) + C`

Answer 2

This is one of my favorite integrals for introductory courses. Here are three solutions.

Explanation:

Solution 1

`I = int sinxcosx dx`

Substitute `u=sinx`, so that `du = cosx dx`.

`I = int u du = 1/2u^2 +C`.

Undo the substitution to get

`I = 1/2 sin^2x +C`

Solution 2

`I = int sinxcosx dx`

Substitute `u=cosx`, so that `du = -sinxx dx`.

`I = int -u du = -1/2u^2 +C`.

Undo the substitution to get

`I = -1/2 cos^2x +C`

Solution 3

`I = int sinxcosx dx`.

Observe that `sin2x = 2sinxcosx`, so `sinxcosx = 1/2sin(2x)`.

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

Substitute `u = 2x`, so that `du = 2 dx` and

`I = 1/4 int sinu du = 1/4 cosu +C`.

Undo the substitution to get

`I = 1/4 cos(2x) +C`

All three answers are correct.

Can you explain?

Hint: What are the differences between the answers? (In the mathematical sense of "difference".)