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)

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

GOod.

Answer 5

No see a trig idendity?

Answer 6

How di dyou get a cubic for sec though?

Answer 7

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

Answer 8

I was getting to that :3 .

Answer 9

But good job :) .

Answer 10

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

Answer 11

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

Answer 12

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