ggrree
  • ggrree
integral of: tan^78 (x) * sec^4 (x)
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
Is... that tan(x)^78? Good... Gods...
anonymous
  • anonymous
I'm not doing 78 half angle identity integrals.
ggrree
  • ggrree
yes :P

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ggrree
  • ggrree
Apparently there's a clever way to do it
anonymous
  • anonymous
You should make your teacher mad and do 70 half angle identity integrals.
anonymous
  • anonymous
Holy crap! thats a lot Tangents, anways: \[\sec^4x = \sec^2x \sec^2x = (\tan^2x +1)(\sec^2x)\] now use a U-Sub where: \[u=tanx\] \[du=\sec^2x dx\] so now we have: u^78 (u^2 + 1) du distributing the u^78, we get: u^80 - u^78 du now just integrate and we get (1/81) (u^81) + (1/79) (u^79) now plug back in the tanx (1/81) (tan^81x) + (1/79)(tan^79x)
anonymous
  • anonymous
i might have made a mistake somwhere, i mean, THATS A LOT O TANGENTS haha
anonymous
  • anonymous
Whoa man, that's like, clever. I was thinking also of something clever, like a series representation of each new half angle integral, but your method was by far the best.
anonymous
  • anonymous
Haha I just learned how to integrate the trig stuff so its still fresh in my mind. But I think it would be pretty smart to use a Series representation for each of the half-angles
ggrree
  • ggrree
That's the right answer. Thanks!

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