1. ## integral(tan^(2m)(x)*sec^(2k+1)(x),x)

Hello,

How would I find the integral of an even powered tangent function multiplied by an odd powered secant function?

integral(tan^(2m)(x)*sec^(2k+1)(x),x)

For instance, how would we find
integral(tan^2(x)*sec(x),x)?

I was informed the method would be integration by parts.

Using int(udv) = uv - int(vdu)

if i choose dv to be sec(x), v is not fun to have for the next integral.
if i choose dv to be tan(x)sec(x), i have to integrate sec^3(x).
if i choose dv to be tan(x), the next integral is not fun.

So it seems that i could solve this if would learn how to integrate sec^n(x).

Are there any better methods then what I have outlined above?

Thanks,

2. Originally Posted by n0083
For instance, how would we find
integral(tan^2(x)*sec(x),x)?
If $u=\sec(x)$ then $du=\sec(x)\tan(x)dx$ and this becomes simple.

3. i don't see how u=sec(x) helps.

we only have one secant, the du requires one too.

Originally Posted by n0083
For instance, how would we find
integral(tan^2(x)*sec(x),x)?
This is really hard to see without knowing the "trick".

$\sec(x) \times \frac{\sec(x)+\tan(x)}{\sec(x)+\tan(x)}=\frac{\sec ^2(x)+\sec(x)\tan(x)}{\sec(x)+\tan(x)}$

This is now $\frac{1}{u}du$

I think you can put it together now.

5. I'm not sure what did you choose for u and v' but if it was like this:
u = tan^2(x)
v' = sec(x) dx

You'll get a more complicated integral. Here's one way to solve $\int \tan^2{x} \sec{x} dx$:

$\int (\sec^2{x} - 1)\sec{x} dx$

$\int \sec^2{x}\sec{x}~dx - \int \sec{x}~dx$

The second integral can be solved the way Jameson told you to, so let's put all our focus to the first one, which is an obvious integration by parts.

$\int \sec^2{x} \sec{x}~dx = \int (\tan{x})'\sec{x}~dx = \sec{x}\tan{x} - \int \tan^2{x}\sec{x}~dx
$

$= \sec{x}\tan{x} - \int (\sec^3{x} - \sec{x})~dx$

Now you should notice that you can treat sec^3(x) as the unknown and then solve for it algebraically. Problem solved.

You can also use a reduction formula for sec^3(x)