Integral with Trigonometric Substitution
I'm having some difficulties solving a problem from my Calculus II course. The problem is: \(\displaystyle \int \:sin^2\left(3x\right)cos^2\left(3x\right)dx\)
I began by rewriting the problem as the equivalent integral: \(\displaystyle \int \frac{1}{2}\left(1-cos\left(6x\right)\right)\cdot \frac{1}{2}\left(1+cos\left(6x\right)\right)dx\)
That simplified down to: \(\displaystyle \int \:\frac{1}{4}\left(1-cos^2\left(6x\right)\right)dx\)
I applied another identity to get to: \(\displaystyle \int \:\frac{1}{4}\left(1-\left(1+cos\left(12x\right)\right)\right)dx = \int \:\frac{1}{4}\left(-cos\left(12x\right)\right)dx\)
I pulled out the constants and evaluated the integral: \(\displaystyle -\frac{1}{4}\int \:cos\left(12x\right)dx = -\frac{1}{48}sin\left(12x\right)+C\)
And that doesn't match the answer in the back of the book. They have: \(\displaystyle -\frac{1}{8}x-\frac{1}{96}sin\left(12x\right)+C\)
As is often the case, I know I messed up somewhere along the line, but can't for the life of me figure out where. Every step I did seems to check out to me... so if anyone could help me figure it out, I'd be very grateful.
I'm having some difficulties solving a problem from my Calculus II course. The problem is: \(\displaystyle \int \:sin^2\left(3x\right)cos^2\left(3x\right)dx\)
I began by rewriting the problem as the equivalent integral: \(\displaystyle \int \frac{1}{2}\left(1-cos\left(6x\right)\right)\cdot \frac{1}{2}\left(1+cos\left(6x\right)\right)dx\)
That simplified down to: \(\displaystyle \int \:\frac{1}{4}\left(1-cos^2\left(6x\right)\right)dx\)
I applied another identity to get to: \(\displaystyle \int \:\frac{1}{4}\left(1-\left(1+cos\left(12x\right)\right)\right)dx = \int \:\frac{1}{4}\left(-cos\left(12x\right)\right)dx\)
I pulled out the constants and evaluated the integral: \(\displaystyle -\frac{1}{4}\int \:cos\left(12x\right)dx = -\frac{1}{48}sin\left(12x\right)+C\)
And that doesn't match the answer in the back of the book. They have: \(\displaystyle -\frac{1}{8}x-\frac{1}{96}sin\left(12x\right)+C\)
As is often the case, I know I messed up somewhere along the line, but can't for the life of me figure out where. Every step I did seems to check out to me... so if anyone could help me figure it out, I'd be very grateful.
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