Posted by **Vicky** on Saturday, September 22, 2012 at 10:54pm.

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^9/9-sec^5/5 +c

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