# Trig functions and integration

1. Jun 24, 2008

### chaotixmonjuish

I'm having a wee bit of a problem with this

$$\int$$(tan x)^3*(sec x)^3dx

could i use a u substitution by pulling out a sec x tan x and using it as a du

2. Jun 24, 2008

### danago

How about writing it in terms of sin(x) and cos(x) by noting that tan(x)=sin(x)/cos(x) and sec(x)=1/cos(x)? And then make the sub u=sin(x)?

I think that would work.

3. Jun 24, 2008

### Defennder

That gives me an idea. Let u=tan x. Then evaluate the integral in terms of u. It appears to be a lot easier than doing the original one by parts.

4. Jun 24, 2008

### chaotixmonjuish

This may help a bit more, the original problem was:

$$\int$$x$$^{3}$$$$\sqrt{x^{2}+4}$$

5. Jun 24, 2008

### tiny-tim

Hi chaotixmonjuish!

(have a cubed: ³ and an int: ∫)

Hint: tan³x = sec²x.tanx - tanx … then it's obvious.

6. Jun 24, 2008

### chaotixmonjuish

well here is what I did

u=secx
du=tanx*secxdx

so turn tan^2 into sec^2-1

(u^2-1)*u^2du

7. Jun 24, 2008

### tiny-tim

Carry on then … ∫(u^4 -u^2)du = … ?

8. Jun 25, 2008

### Gib Z

It does help to give us the original problem, then post what you have it down to later as well. If you told us this before, we could have told you that u=x^2 +4 makes this a trivial integral =]

9. Jun 25, 2008

### chaotixmonjuish

Okay, so I figured it out

since u=sec x
and that makes tan u^2-1

(u^2-1)u^2

that became a fairly easy integral...it was just getting to that step with those clever trig functions.

10. Jun 26, 2008

### Gib Z

...Or just perhaps the obvious substitution. But you can do it the hard way if you want.

11. Jun 26, 2008

### HallsofIvy

Perhaps because I tend to be "simple", my first reaction with any trig functions is to convert to sine and cosine. Here,
$$\int tan^3(x)sec^3(x)dx= \int \frac{sin^3(x)}{cos^6(x)}dx$$
and since sine is to an odd, power, factor one out to use with the "dx"
$$= \int \frac{sin^2(x)}cos^6(x)}sin(x)dx= \int\frac{1- cos^2(x)}{cos^6(x)}sin(x)dx$$
Now, let u= cos(x) so du= -sin(x)dx and we have
$$-\int \frac{1- u^2}{u^6}du= -\int(u^{-6}- u^{-4})du$$
which is easy