# Integrals involving Secant & Tangent Derivation

1. May 13, 2012

### Nano-Passion

1. The problem statement, all variables and given/known data
If the power of the secand is even and positive..
$$\int sec^{2k} x tan^{n} x dx = \int (sec^2 x)^{k-1} tan ^n x sec^2 x dx$$

3. The attempt at a solution

The way I see it,

$$sec^{2k} x = sec^2 x dx * sec^k x dx$$

the next step seems to be to break down sec^k, but on closer introspection, the break down goes something like this

$$sec^k x dx = (sec^2 x dx)^{k-1} = sec^{2k-2}x dx$$

I'm having trouble intuitively accepting that something like 2k-2 = k. It doesn't seem to add up, there seems to be something a bit more complex into it that is left out of detail.

Last edited: May 13, 2012
2. May 13, 2012

### gopher_p

You may want to review your exponent rules.

3. May 13, 2012

### Nano-Passion

$$(x^n)^m = x^{nm}$$
$$sec^m x = (sec x)^m$$
$$(sec^2 x)^{k-1} = (sec x)^ {(2)(k-1)}$$

4. May 13, 2012

### gopher_p

OK. If $x^{nm}=(x^n)^m$ (which is correct), then what is $x^n\cdot x^m$?

5. May 13, 2012

### InfinityZero

This is right, note that there are two terms on the right integral.

Umm, I don't think that's right remember that $$a^{b}*a^{c}=a^{b+c}$$

I also don't know where your $$dx$$ terms are coming from there.

You seem to have multiplied your exponents correctly, but forgot that there is another $$sec^{2}x$$ term in the right integral up top.

When doing this integral

$$\int sec^{2k} x tan^{n} x dx = \int (sec^2 x)^{k-1} tan ^n x sec^2 x dx$$

From here there is a handy trig identity involving $$sec^{2}x$$ that you can use to make it into something easy to use u substitution on.

6. May 13, 2012

### Nano-Passion

$$= x^{n+m}$$

7. May 13, 2012

### Nano-Passion

I tend to do silly mistakes sometimes, just ignore them.

Oh there you go, now I get back the original expression. Thanks. ^.^

I know about u substitution, the calculation is fairly trivial to me. What I was primarily concerned with is the derivation.