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Apparently there's a clever way to do it
You should make your teacher mad and do 70 half angle identity integrals.
Holy crap! thats a lot Tangents, anways:
\[\sec^4x = \sec^2x \sec^2x = (\tan^2x +1)(\sec^2x)\]
now use a U-Sub where:
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)
i might have made a mistake somwhere, i mean, THATS A LOT O TANGENTS haha
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.
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