# Homework Help: Antiderivative of product of trig functions

1. Dec 11, 2008

### snipez90

1. The problem statement, all variables and given/known data
Find the antiderivative of $$sin^4(x)tan^2(x)$$

2. Relevant equations
Trig identities I may have overlooked.

3. The attempt at a solution
I tried writing the integrand in terms of sin and cos but that didn't seem to lead anywhere. I tried integration by parts since the antiderivative of $$tan^2(x)$$ is $$tanx - x + C$$, but that approach lead to a more difficult integral that might have been solved using a messier integration by parts.

This isn't really homework since I'm in a theoretical calc course and we begin the theory of integration next quarter (in fact, we're on break). Anyways I used to be pretty good with elementary terms integration but I'm stuck. Could someone suggest a good first step? Thanks.

2. Dec 11, 2008

### Staff: Mentor

I think this will go with integration by parts.
Try u = tan^2(x)sin^3(x) and dv = sin(x)dx.

This gave me an unwieldy-looking integral, but after simplifying to sines and cosines, I was left with $$\int \frac{sin x}{cos^2 x} dx$$ and $$\int sin^4 x dx$$. There were constant multipliers that I have omitted.

The first integral is an easy one and the second isn't too hard.
Hope that helps.

3. Dec 11, 2008

### Tedjn

Another potential way to do this integral is by converting everything to cosines because there is a cosine downstairs:

$$\int \sin^4x\,\tan^2x\,dx = \int\frac{(1-\cos^2 x)^3}{\cos^2 x}\,dx = \int \frac{1 - 3\cos^2 x + 3\cos^4 x - \cos^6 x}{\cos^2 x}\,dx$$​

Each of the resulting integrals should be fairly simple to solve.

4. Dec 11, 2008

### snipez90

Thank you both for the suggestions.

Mark44, did you arrive at those two integrals after applying integration by parts once? I got sin sin^4(x) term but the the one before that (I think it was sin^4(x)/cos^2(x) or something). I'll try it again though.

Tedjn, I thought of the very first equality you posed but did not think to expand it. That is quite a neat trick, i.e., decrease the powers by expansion. A few expansions and some computation and it worked out well. Thanks.

5. Dec 12, 2008

Yes.