# What am I doing wrong with this simple integration?

1. Oct 2, 2013

### Turion

1. The problem statement, all variables and given/known data

$$\int { { sin }^{ 2 }x{ cos }^{ 2 }xdx }$$

2. Relevant equations

3. The attempt at a solution

$$\int { { sin }^{ 2 }x{ cos }^{ 2 }xdx } \\ =\frac { 1 }{ 4 } \int { (1-cos2x)(1+cos2x)dx } \\ =\frac { 1 }{ 4 } \int { -{ cos }^{ 2 }x+1dx } \\ =\frac { x }{ 4 } -\frac { 1 }{ 8 } \int { cos2x+1 } dx\\ =\frac { x }{ 4 } -\frac { 1 }{ 8 } [\frac { sin2x }{ 2 } +x]\\ =\frac { 2x-sin2x }{ 16 }$$

It's incredibly frustrating to do so many stupid mistakes on such simple integration. I can't believe how weak I've gotten at doing integrals after only 4 months...

Edit: never mind. I got it. I put x instead of 2x on line 3. Really dumb mistakes but I keep making them from a result of being so rusty.

2. Oct 2, 2013

### tiny-tim

Hi Turion!

Even quicker would have been sin2xcos2x = (1/4)sin22x = … ?

3. Oct 2, 2013

### jfizzix

I think the problem is in the difference of two squares
$(1-Cos(2x))(1+Cos(2x)) = 1-Cos^{2}(2x)$