Help integrating Sin[42x]*Cos[42x]^(1/3)

[Jaskaran]

My problem is the cube root of it...so I take the entire thing to the 4th power to get

Sin^4(42x)*Cos^(4/3)(42x)

(Sin^2(42x))^2*Cos^(4/3)(42x)

(1-Cos^2(42x))^2*Cos^(4/3)(42x)

Where to from here? I thought about doing u = Cos(42x), but then du would be -(du/42) = Sin(42x) but there's no Sin(42x).

Basically, how do I figure out what power to take the entire thing to to get rid of the exponent on the Cosine?

I know what the answer is, but I need to know how they got it.

-Cos[42*x]^(4/3)/56

 

[BigGlenntheHeavy]

\(\displaystyle \int sin(42x)cos^{1/3}(42x)dx\)

\(\displaystyle Let \ u \ = \ cos(42x), \ then \ \frac{-du}{42} \ = \ sin(42x)dx\)

\(\displaystyle Can \ you \ take \ it \ from \ here?\)