Calculus AP
posted by Vicky .
I'm doing trigonometric integrals
i wanted to know im doing step
is my answer right?
∫ 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^9/9sec^5/5 +c

Calculus AP 
Steve
you dropped some 2's here and there, and the final integral is
1/2∫ (u^21) 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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