# Indefinite integral problem

1. Mar 31, 2014

### Loopas

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

Evaluate the indefinite integral of x*cos(3x)^2

2. Relevant equations

Integration by parts: $\int(udv)$= uv - $\int(vdu)$

3. The attempt at a solution

Im having trouble finding the antiderivative of cos(3x)^2 (which I designated as dv when doing integration by parts). I keep getting (1/9)(sin(3x)^3) using the power and chain rules, but that's not correct.

Thanks!

2. Mar 31, 2014

### Staff: Mentor

$$\int x cos(9x^2)dx$$

If so, you might be able to do it by integration by parts, but I wouldn't do it this way. An ordinary substitution will work just fine.

3. Mar 31, 2014

### LCKurtz

Or is it$$\int x\cos^2(3x)$$in which case a double angle formula and integration by parts would work.

4. Mar 31, 2014

### Staff: Mentor

Loopas,
Obviously, what you wrote is ambiguous.

5. Mar 31, 2014

### HallsofIvy

Assuming you mean $cos^2(3x)$ use the identity $cos(2a)= cos^2(a)- sin^2(a)= cos^2(a)- (1- cos^2(a)= 2cos^2(a) -1$. So $cos^2(3x)= (cos(6a)+ 1)/2$.

Last edited by a moderator: Mar 31, 2014
6. Apr 1, 2014

### Loopas

Ok, the integral is $\int(xcos^{2}(3x))$

Im trying to use integration by parts but I can't find the antiderivative of cos$_{2}$(3x)

What exactly is the form of the double angle formula?

7. Apr 1, 2014

### Staff: Mentor

What you wrote is extremely difficult to read. I have fixed it up to make it clear.

8. Apr 1, 2014

### LCKurtz

$$\cos^2(3x)=\frac {1+\cos(6x)} 2$$

9. Apr 1, 2014

### Loopas

So would the correct antiderivative be $\frac{x}{2}$+$\frac{sin(6x)}{12}$?

10. Apr 1, 2014

### Staff: Mentor

It's easy enough to check for yourself. Just differentiate your answer and you should get the integrand.

11. Apr 1, 2014

### Loopas

Ok so I can use the double angle formula to put cos$^{2}$(3x) into a form that I can find the antiderivative more easily. But I'm confused about how to use the double angle identity to get $\frac{1+cos(6x)}{2}$

12. Apr 1, 2014

### HallsofIvy

The basic double angle formula for cosine says that $cos(2\theta)= cos^2(\theta)- sin^2(\theta)$. But $sin^2(\theta)= 1- cos^2(\theta)$ so that can be written $cos(2\theta)= cos^2(\theta)- (1- cos^2(\theta))= 2cos^2(\theta)- 1$ From that $2cos^2(\theta)= 1+ cos(2\theta)$ and so $cos^2(\theta)= \frac{1+ cos(2\theta)}{2}$.

Now replace "$\theta$" with "3x": $cos^2(3x)= \frac{1+ cos(6x)}{2}$.

13. Apr 1, 2014

### Loopas

Ahh I can see it now! Thanks guys this was all really helpful!