Question

∫ tan^2 (x) sec^4 (x) dx

∫ [tan^2 (t) + tan^4 (t)] dt

∫ [1-tan^2 (x)] / [sec^2 (x)] dx

Trigonometric integral
Please show steps so I can understand!

Answer 1

tan^2 (sec^4) = tan^2(1+tan^2)sec^2
= (tan^2 + tan^4)(sec^2) dx
Now note that if u = tanx, you have
(u^2 + u^4) du

See what you can do with the others.

Answer 2

I got the first two, but I'm unsure with the last one.

I got the answer to be -ln |secxtanx|+C
Here's what I did:
∫ [1-tan^2 (x)] / [sec^2 (x)] dx
∫(1/secx)-[(sin^2x/cos^2x)/(1/cosx)
∫cosx-sinx(sinx/cosx)
∫cosx-∫sin^2(x)/cosx
sinx-∫(1-cos^2(x))/cosx
sinx-∫(1/cosx)-cosx
sinx-∫secx-∫cosx
sinx-sinx-∫secx
=-ln |secxtanx|+C

Can someone just verify that I did everything correctly?