# Calculus AP

I'm doing trigonometric integrals
i wanted to know im doing step

∫ tan^3 (2x) sec^5(2x) dx
=∫ tan^2(2x) sec^4(2x) tan*sec(2x) dx
=∫ (sec^2(2x)-1)sec^4 tan*sec(2x) dx
let u=sec x, du= 1/2 tan*sec(2x) dx
=1/2∫ (u^2(2x)-1) u^4 du
=1/2∫ (u^8(2x)-u^4) du
=1/2 sec^9/9-sec^5/5 +c

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1. you dropped some 2's here and there, and the final integral is

1/2∫ (u^2-1) u^4 du
= 1/2 ∫ u^6 - u^4 du
= 1/2 (1/7 u^7 - 1/5 u^5)
= 1/14 sec^7(2x) - 1/10 sec^5(2x) + C

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posted by Steve

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