# Integral involving sec and tan

1. Feb 24, 2004

### tandoorichicken

How do I perform this integral?
$$\int \sec^2{3x} \tan^5{3x} \,dx$$

2. Feb 24, 2004

### NateTG

Re: Transcendentals

Try $$u=\tan{3x}$$

3. Feb 24, 2004

### ShawnD

$$\frac{tan^6(3x)}{18}$$

Now lets see if we can get that.
First of all, the derivative of tan is sec^2. So that leads to me belive the process would be the substitution method.

I'll assign "U" as tan(3x)

$$\int sec^2(3x)tan^5(3x) dx$$

$$\int sec^2(3x)U^5 dx$$

Now take the derivative of U with respect to x

$$U = tan(3x)$$

$$\frac{dU}{dx} = 3sec^2(3x)$$

$$dx = \frac{dU}{3sec^2(3x)}$$

Now fill that back into what we had above

$$\int sec^2(3x)U^5 \frac{dU}{3sec^2(3x)}$$

$$\frac{1}{3} \frac{U^6}{6}$$

$$\frac{U^6}{18}$$

$$\frac{tan^6(3x)}{18}$$

Right on

4. Feb 25, 2004

### tandoorichicken

Awesome thats what I got. Thanks guys.