Solving ∫cos⁴ 6x sin³ 6x dx with trig substitution

[duki]

Homework Statement

$$\int{cos^4 6x sin^3 6x dx}$$

Homework Equations

The Attempt at a Solution

I've gotten this far but now I'm stuck:

$$\int{cos^4 6x sin 6x sin^2 6x dx} = \int{cos^4 6x}*(\frac{1-cos 12x}{2})sin 6x dx$$

 

[JG89]

From you having the integrand as this: cos^4 6x sin^3 6x dx

Use a substitution u = cos^4 6x

 

[duki]

I don't understand exactly. Are you saying use substitution after using the power-reducing formula? Or do I not use power-reducing at all here? I briefly saw him explaining it to one person, and he mentioned something about reducing it from sin^3 to sin^2... I could have misunderstood though.

 

[JG89]

Sorry, that substitution was wrong. Tomorrow when I'm more awake (it's 3:40 AM here) I'll have another look at it.

 

[danago]

I don't understand exactly. Are you saying use substitution after using the power-reducing formula? Or do I not use power-reducing at all here? I briefly saw him explaining it to one person, and he mentioned something about reducing it from sin^3 to sin^2... I could have misunderstood though.

Use the fact that $$sin^3 (6x) =sin^2 (6x) . sin(6x)$$. Then perhaps make use of the pythagorean identity, and you will be 1 simple substitution away from a solution