∫tan^2 (θ) sec^4 (θ) dθ

∫ tan²Θ sec^4(Θ) dΘ

∫ tan²Θsec²Θsec²Θ dΘ

∫ tan²Θ(1 + tan²Θ)sec²Θ dΘ

let u = tanΘ
du = sec²Θ dΘ

∫ u²(1 + u²) du
∫ (u² + u^4) du
1/3u³ + 1/5u^5 + C

1/3tan³Θ + 1/5tan^5(Θ) + C

It's a u- substitution problem

∫tan^2 (θ) sec^4 (θ) dθ

u= tan (θ)
du= sec^2 (θ) dθ

∫ u^2 (sec^2 (θ)) du



using your identities:

sin^2 (x) + cos^2 (x) = 1

sin^2 (x) / (cos^2 (x)) + cos^2 (x) / (cos^2 (x)) = 1 / (cos^2 (x))

tan^2 (x) + 1 = sec^2 (x)



getting back to the prob:

∫ u^2 (sec^2 (θ)) du

∫ u^2 (tan^2 (θ) + 1) du

remember : u = tan (θ)

and I'll hand this one back to you

hope that helped :)

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Visit the wolframalpha page given below and select the "show steps" button.

∫tan^2 (θ) sec^4 (θ) dθ =
∫tan^2 (θ) sec^2 (θ) d(tan (θ)) =
∫tan^2 (θ) (1 + tan^2 (θ)) d(tan (θ)) =
∫tan^2 (θ) d(tan (θ)) + ∫tan^4 (θ) d(tan (θ)) =
(1/3) tan^3 (θ) + (1/5) tan^5 (θ)+ C

Or use tan(θ) = u and sec^2 (θ) d(θ) = du