Trig. integration tan^5(x)dx

[SpicyPepper]

I'm not sure if my answer is just wrong or basically the same as the one in the back of my book.

Homework Statement

$$\int tan^5(x)dx$$

The Attempt at a Solution

My answer:

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

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

Book answer:

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

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

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

instead of

$$tan^2(x)$$

 

[Dick]

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

 

[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