Trig Integral Question: Solving \intsin32x dx | Tips & Tricks for Calculus

[AzMaphysics]

Homework Statement

\intsin³2x dx

2. The attempt at a solution

\intsin³2x dx =
\int(1-cos²2x)sin2x dx =
(-1/2)\int(1-cos²2x)-2sin2x dx =
(-1/2)\int(1-u²)du where u= cos2x
(-1/2)(u-(u³/3)) =
(-cos 2x/2)+(cos³2x/6) + C

Book says (cos³x/3)-(cosx/2) + C
Where did I go wrong?

 

[Mark44]

It looks like you have the right idea, but you should split the integral into two separate integrals after this step:
\int (1-cos^22x)sin2x dx

 

[AzMaphysics]

Yea integrating by parts seems to be the preferred way to attack these types of problems. The chapter was focusing on u-substitution though so I tried going with that. I just can't tell if I made a mistake or if the book did (there's been a couple times in the past) or if the book's answer is just more simplified and I'm not seeing how.

 

[n!kofeyn]

Well I actually checked your work instead of assuming you made a mistake, and you made no mistakes (I got the same thing). The way to do this is definitely a u-substitution and you don't want to split it up as mentioned. Your answer is correct.

To save work, I also computed the derivative of our answer with Maple and it gave me the integrand that we started with. I did the same for the book's "answer", and it did not. So it looks like the book is incorrect on this one.