anonymous
  • anonymous
integrate6x^3(1+x^2)^(1/2) using trig substitution. I have it down to the integral of 6sec^3(theta)tan^3(theta)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
I used x=tan(theta) so dx=sec^2(theta)
anonymous
  • anonymous
and then got (1+x^2)^(1/2)=sec(theta)
anonymous
  • anonymous
plugged that all into the formula and got \[6\int\limits_{}^{}\tan ^{3}(\theta)\sec ^{3}(\theta) d(\theta)\]

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anonymous
  • anonymous
GOod.
anonymous
  • anonymous
No see a trig idendity?
anonymous
  • anonymous
How di dyou get a cubic for sec though?
anonymous
  • anonymous
\[6\int\tan^3t\sec^3t~dt=6\int\tan^2t\sec^2t~(\sec t\tan t~dt)\] \(\tan^2t=\sec^2t-1\), 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
  • anonymous
I was getting to that :3 .
anonymous
  • anonymous
But good job :) .
anonymous
  • anonymous
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
  • anonymous
For powers of sec and tan you often have to to that :) .
anonymous
  • anonymous
I'm pretty sure I can get it from here. Thank you!

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