# Thread: I dont know how he got this...

1. ## I dont know how he got this...

1. $\int(sin^2(x))^2 cos^2(x) dx$
2. $\int(\frac{1-cos(2x)}{2})^2(\frac{1+cos(2x)}{2}dx$
3. $\frac{1}{8}\int1-cos(2x)-cos^2(2x)+cos^3(2x)dx$
4. $\frac{1}{8}\int1dx-\frac{1}{8}\int cos(2x)dx-\frac{1}{8}\int cos^2(2x)dx+\frac{1}{8}\int cos^3(2x)dx$
5. $\frac{1}{16}x-\frac{1}{64}sin(4x)-\frac{1}{48}sin^3(2x)+C$

What I dont understand is how he got from 2 to 3 and from 4 to 5. For three, I had this:
$\frac{1}{8}\int-c^3-2c^2+c+1dx$ Where c is a place holder for cos 2x.

2. ## Re: I dont know how he got this...

Check your calculation from step 2 to step 3, since by my reckoning

$(1-\cos(2x))^2(1+\cos(2x))=1-\cos(2x)-\cos^2(2x)+\cos^3(2x)$

For the final step Wikipedia is certainly your friend:

List of trigonometric identities - Wikipedia, the free encyclopedia

You can use the half angle, double angle and triple angle identities to make the last two integrations in step 4 manageable.

3. ## Re: I dont know how he got this...

Wow, I wrote down the problem wrong, I had sine and cosine switched. I copied the above straight from the book, so it's correct. I wrote it wrong in my note book. I swear, I need to postpone calculus until I get down basic reading and writing...