Question

How do you find the indefinite integral of `int sin2xcos2xdx`?

Answer 1

Here are three solutions.

Explanation:

Substitution 1

`int sin2xcos2xdx`

Let `u = sin2x`, so that `du = 2cos2x dx`and the integral becomes

`1/2intudu = 1/4u^2 +C`

`int sin2xcos2xdx = 1/4sin^2 2x +C`

Substitution 2

`int sin2xcos2xdx`

Let `u = cos2x`, so that `du = -2sin2x dx`and the integral becomes

`-1/2intudu = -1/4u^2 +C`

`int sin2xcos2xdx = -1/4cos^2 2x +C`

Solution 3

`int sin2xcos2xdx = int 1/2sin4x dx`

Let `u = 4x`, so `du = 4dx` and the integral becomes

`1/8 int sinu du = -1/8 cos u +C`

`int sin2xcos2xdx = -1/8cos4x +C`

The challenge is to see why these answers are the same.

Hint find the difference between the answers.
Hint 2 "difference" means subtract.