integral of:
tan^78 (x) * sec^4 (x)

Question

integral of:
tan^78 (x) * sec^4 (x)

anonymous · Answer

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

anonymous · Answer

I'm not doing 78 half angle identity integrals.

ggrree · Answer

yes :P

ggrree · Answer

Apparently there's a clever way to do it

anonymous · Answer

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

anonymous · Answer

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 · Answer

i might have made a mistake somwhere, i mean, THATS A LOT O TANGENTS haha

anonymous · Answer

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 · Answer

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 · Answer

That's the right answer. Thanks!