# Trigo Integral

1. Aug 22, 2009

### Mentallic

1. The problem statement, all variables and given/known data
Find $$\int{tan^2xsec^2xdx}$$

2. Relevant equations
$$tan^2x=sec^2x-1$$ (1)

3. The attempt at a solution
Using (1): $$\int{(sec^4x-sec^2x)dx}$$

Now, $$\int{sec^4xdx}-\int{sec^2xdx}=\int{sec^4xdx}-tanx+c$$

I can't figure out how to solve $$\int{sec^4xdx}$$ though.

2. Aug 22, 2009

### Dick

The derivative of tan(x) is sec^2(x). Mentallic, this is a simple substitution. u=tan(x).

Last edited: Aug 22, 2009
3. Aug 22, 2009

### Mentallic

Oh yeah... ugh I feel like such an idiot.

$$u=tanx$$

$$\int{sec^4xdx}=\int{(1+u^2)du}$$

Thanks Dick.

4. Aug 22, 2009

### Dick

Sure, but why don't you use that substitution in the original integral?

5. Aug 22, 2009

### Mentallic

Oh, you mean like $$\int{u^2du}$$ ? Yeah because I love to make things harder for myself :yuck:

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