Trig functions and integration

[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

 

[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.

 

[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.

 

[chaotixmonjuish]

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

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

 

[tiny-tim]

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

Hi chaotixmonjuish!

(have a cubed: ³ and an int: ∫)

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

 

[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

 

[tiny-tim]

well here is what I did

u=secx
du=tanx*secxdx

so turn tan^2 into sec^2-1


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

Carry on then … ∫(u^4 -u^2)du = … ?

 

[Gib Z]

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

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

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 =]

 

[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.

 

[Gib Z]

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

 

[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