Question about how to solve this problem

[ksdhart]

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.

 

[Deleted member 4993]

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: \(\displaystyle \int \:\frac{1}{4}\left(1-\left(1+cos\left(12x\right)\right)\right)dx = \int \:\frac{1}{4}\left(-cos\left(12x\right)\right)dx\) ...... right here

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.

cos²(Θ) = [1+ cos(2Θ)]/2 → 1 - cos²(6x) = 1/2- cos(12x)/2 → 1/4[1 - cos²(6x)] = 1/8 - cos(12x)/8 .... leading to the answer in the book

 

[ksdhart]

Oh dear. I did miss the 1/2 in the identity. That makes everything work out. Thanks.

 

[stapel]

Oh dear. I did miss the 1/2 in the identity. That makes everything work out. Thanks.

By the way, of all the identities you had to memorize back in trig (or pre-calc), these "1/2" identities are the ones you'll be needing all the time in calculus.

 

[pka]

I'm having some difficulties solving a problem from my Calculus II course. The problem is: \(\displaystyle \int \:sin^2\left(3x\right)cos^2\left(3x\right)dx\).

Here is a less complicated way.
\(\displaystyle \sin^2\left(3x\right)\cos^2\left(3x\right)=\left[\frac{1}{2}\sin(6x)\right]^2\).

Use this.