Trig. Integratin Suggestion

[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_

[mda]

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

[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.

[Sparky_]

Yes - I meant cos(3x).

Thanks
Sparky_