# Trig. Integratin Suggestion

1. Jan 12, 2008

### Sparky_

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

Can I get some help with:

$$\int -\frac {sin^2(3x)}{3cos(x)} dx$$

2. Relevant equations

3. The attempt at a solution

It looks like a substitution would work but I'm striking out with:

$$u = sin(3x)$$
$$du = 3cos(3x)$$
because this is now in the denominatior

$$u = cos(3x)$$
$$du = -3sin(3x)$$

I'm left with a sin(3x) term.

I'm guessing a trig identity is now going to be involved.

I have found that

$$\int sec(x) tan(x) dx = sec(x)$$

My problem can be re-written as

$$\int - sin(3x) tan(3x) dx$$

???

Thanks for the help.

-Sparky_

Last edited: Jan 12, 2008
2. Jan 12, 2008

### mda

Just to clarify, is the argument of the cos correct?
Its just that cos(3x) will be simpler...

3. Jan 12, 2008

### Gib Z

Assuming you meant the argument of the cosine to be 3x, its easy to see that you can solve your integral if you can solve:

$$\int \frac{ \sin^2 x}{\cos x} dx$$

To do that, remember $\sin^2 x = 1- \cos^2 x$, so that the integral becomes

$$\int \left( \sec x - \cos x \right) dx$$.

Both of those are usually regarded as standard integrals, although sometimes the secant integral is not. To do that one;

$$\int \sec x dx = \int \frac{\cos x}{\cos^2 x} dx = \int \frac{\cos x}{1- \sin^2 x} dx = \int \frac{1}{(1+u)(1-u)} du$$ when u = sin x. Now do partial fractions and your home free.

4. Jan 13, 2008

### Sparky_

Yes - I meant cos(3x).

Thanks
Sparky_