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badreferencesBest ResponseYou've already chosen the best response.2
Is... that tan(x)^78? Good... Gods...
 2 years ago

badreferencesBest ResponseYou've already chosen the best response.2
I'm not doing 78 half angle identity integrals.
 2 years ago

ggrreeBest ResponseYou've already chosen the best response.0
Apparently there's a clever way to do it
 2 years ago

badreferencesBest ResponseYou've already chosen the best response.2
You should make your teacher mad and do 70 half angle identity integrals.
 2 years ago

HermeezeyBest ResponseYou've already chosen the best response.1
Holy crap! thats a lot Tangents, anways: \[\sec^4x = \sec^2x \sec^2x = (\tan^2x +1)(\sec^2x)\] now use a USub 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)
 2 years ago

HermeezeyBest ResponseYou've already chosen the best response.1
i might have made a mistake somwhere, i mean, THATS A LOT O TANGENTS haha
 2 years ago

badreferencesBest ResponseYou've already chosen the best response.2
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.
 2 years ago

HermeezeyBest ResponseYou've already chosen the best response.1
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 halfangles
 2 years ago

ggrreeBest ResponseYou've already chosen the best response.0
That's the right answer. Thanks!
 2 years ago
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