Thursday

September 29, 2016
Posted by **Vicky** on Saturday, September 22, 2012 at 10:54pm.

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

- Calculus AP -
**Steve**, Sunday, September 23, 2012 at 7:33amyou 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