# Homework Help: Trig. integration tan^5(x)dx

1. Oct 3, 2009

### SpicyPepper

I'm not sure if my answer is just wrong or basically the same as the one in the back of my book.
1. The problem statement, all variables and given/known data
$$\int tan^5(x)dx$$

3. The attempt at a solution

$$\int tan^5(x)dx = \frac{tan^4(x)}{4} - \int(sec^2(x)tan(x) - tan(x)) dx$$

$$\int tan^5(x)dx = \frac{tan^4(x)}{4} - \frac{tan^2(x)}{2} + ln|sec(x)| + C$$

$$\int tan^5(x)dx = \frac{sec^4(x)}{4} - tan^2(x) + ln|sec(x)| + C$$

If I understand correctly,

$$\frac{tan^4(x)}{4} + C = \frac{sec^4(x) - 1}{4} + C = \frac{sec^4(x)}{4} + C$$

If that's wrong let me know please.

However, I don't get why I keep getting

$$\frac{tan^2(x)}{2}$$

$$tan^2(x)$$

2. Oct 3, 2009

### Dick

Ooops. tan(x)^4=(sec(x)^2-1)^2. Multiply the square out.

3. Oct 3, 2009

### SpicyPepper

Ah! So that means my initial answer was right (assuming I don't keep screwing up my trig), since:

= tan^4(x)/4 - tan^2(x)/2 + ... + C

= [sec^4(x) - 2sec^2(x) + 1]/4 - tan^2(x)/2 + ... + C

= sec^4(x)/4 - sec^2(x)/2 - tan^2(x)/2 + ... + C

= sec^4(x)/4 - [tan^2(x) + 1]/2 - tan^2(x)/2 + ... + C

= sec^4(x)/4 - tan^2(x) + ... + C

Thanks