Sunday

October 26, 2014

October 26, 2014

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

**Answer this Question**

**Related Questions**

calculus (check my work please) - Not sure if it is right, I have check with the...

Integration - Intergrate ¡ì sec^3(x) dx could anybody please check this answer. ...

calculus II - ∫ tan^2 x sec^3 x dx If the power of the secant n is odd, ...

calculus - find dy/dx y=ln (secx + tanx) Let u= secx + tan x dy/dx= 1/u * du/dx ...

Calculus 12th grade (double check my work please) - 2- given the curve is ...

Calculus - Question - Am I allowed to do this? for the integral of ∫ sec^4...

calculus - So I am suppose to evaulate this problem y=tan^4(2x) and I am ...

Calculus - could anybody please explain how sec x tan x - ¡ì sec x tan^2(x) dx...

calculus - Use integration by parts to evaluate the integral of x*sec^2(3x). My ...

Calculus PLEASE check my work , - 1.) which of the following represents dy/dx ...