## ggrree 3 years ago integral of: tan^78 (x) * sec^4 (x)

Is... that tan(x)^78? Good... Gods...

I'm not doing 78 half angle identity integrals.

3. ggrree

yes :P

4. ggrree

Apparently there's a clever way to do it

You should make your teacher mad and do 70 half angle identity integrals.

6. Hermeezey

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)

7. Hermeezey

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.

9. Hermeezey

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

10. ggrree