Question

What is the integral of sin(8x)cos(8x)^(1/3)dx?

Answer 1

It looks like this in equation form:
\[\int\limits_{}^{} \sin(8x)\sqrt[3]{\cos(8x)}dx\]

Answer 2

My answer was \[-6\cos ^{\frac{ 4 }{ 3 }}(8x)+c\] and i tried it twice but couldn't find my mistake...

Answer 3

integral by u-sub
let u = cos(8x)
du = -8sin(8x) dx
-1/8 du = sin(8x) dx

see, the integral becomes
int (-1/8 u^(1/3)) du
= - 1/8 int u^1/3 du
= .... (this should be easy for you now)

Answer 4

omg that was such a small mistake, i had -1/8 as my coefficient on my du instead of -8 -.- thanks for pointing that out

Answer 5

you're welcome :D