# Integral of tan(x)^3

1. Mar 3, 2009

### tshaw101

I am having an issue with this problem.

$$\int tan(x)^3:$$

You separate a tan(x)^2, and use the identity 1+tan(x)^2 = sec(x)^2

You then end up with $$\int tan(x) sec(x)^2 dx$$ + $$\int tan(x) dx$$

$$\int tan(x) dx = ln absval(sec(x))$$

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

Here, the book says to solve the problem this way:

set u = tan(x) du = sec(x)^2dx

so $$\int u du = 1/2 u ^2 = 1/2 tan(x)^2$$

Why can't you solve this problem this way instead?

$$\int (tan(x) sec(x)) sec(x) dx$$

set u = sec(x), du = (tan(x) sec(x)) dx

so $$\int u du = 1/2 u ^2 = 1/2 sec(x)^2$$

I think I must be missing something...any help you could give me would be greatly appreciated.

2. Mar 3, 2009

### Dick

You can solve it that way. What you are missing is the +C. An indefinite integral always has an indefinite integration constant. (1/2)*tan(x)^2+C and (1/2)*sec(x)^2+D are the same thing. tan^2 and sec^2 differ by a constant. Only the C and D are different constants, right?

3. Mar 3, 2009

### praharmitra

you arent doing anything wrong. both answers are correct. they just differ by a constant. Remember

sec(x)^2 - tan(x)^2 = 1

4. Mar 3, 2009

### tshaw101

If it is a definite integral don't the constants cancel out?

5. Mar 3, 2009

### Dick

Yes, that's the point. (1/2)tan(x)^2+C and (1/2)sec(x)^2+C both differentiate to the same thing. They are both antiderivatives of tan(x)*sec(x)^2. If you try to do a definite integral, you don't care what which form you use. You get the same thing.

6. Mar 3, 2009

### tshaw101

I see what you are saying, this makes sense to me now...thanks a lot for getting back to me so quick, I really appreciate it.