You have only 3 minutes for a better solution to this integral

[Lorena_Santoro]

 

[skeeter]

$\cos^3(2x) = [1-\sin^2(2x)]\cos(2x)$

use the substitution $u = \sin(2x) \implies du = 2\cos(2x) \, dx$

$\displaystyle \dfrac{1}{2} \int_0^1 1-u^2 \, du$

you can finish up from here

 

[Lorena_Santoro]

$\cos^3(2x) = [1-\sin^2(2x)]\cos(2x)$

use the substitution $u = \sin(2x) \implies du = 2\cos(2x) \, dx$

$\displaystyle \dfrac{1}{2} \int_0^1 1-u^2 \, du$

you can finish up from here

Thank you!