integrate6x^3(1+x^2)^(1/2) using trig substitution. I have it down to the integral of 6sec^3(theta)tan^3(theta)

Question

integrate6x^3(1+x^2)^(1/2) using trig substitution.  I have it down to the integral of 6sec^3(theta)tan^3(theta)

anonymous · Answer

I used x=tan(theta) so dx=sec^2(theta)

anonymous · Answer

and then got (1+x^2)^(1/2)=sec(theta)

anonymous · Answer

plugged that all into the formula and got  $$6\int\limits_{}^{}\tan ^{3}(\theta)\sec ^{3}(\theta) d(\theta)$$

anonymous · Answer

GOod.

anonymous · Answer

No see a trig idendity?

anonymous · Answer

How di dyou get a cubic for sec though?

anonymous · Answer

$$6\int\tan^3t\sec^3t~dt=6\int\tan^2t\sec^2t~(\sec t\tan t~dt)$$
$\tan^2t=\sec^2t-1$, so you have
$$6\int\left(\sec^2t-1\right)\sec^2t~(\sec t\tan t~dt)$$
Substitute $v=\sec t$, so $dv=\sec t\tan t~dt$:
$$6\int\left(v^2-1\right)v~dv$$

anonymous · Answer

I was getting to that :3 .

anonymous · Answer

But good job :) .

anonymous · Answer

ah, I see what I was doing.  I was leaving tan^3 and not pulling one off so I messed up the u sub

anonymous · Answer

For powers of sec and tan you often have to to that :) .

anonymous · Answer

I'm pretty sure I can get it from here.  Thank you!