Solving Trig Integral: (sin(2x))^3(cos2x)^2dx Using Substitution

[mshiddensecret]

Homework Statement

Integrate: (sin(2x))^3(cos2x)^2dx

Homework Equations

Using substitution

Cos2x= (1-(sinx)^2)

The Attempt at a Solution

I sub u= sin2x

But then got nowhere because I had cos2x to the power of 2 and I don't know how to compensate for it with du. [/B]

 

[haruspex]

Homework Statement

Integrate: (sin(2x))^3(cos2x)^2dx

Homework Equations

Using substitution

Cos2x= (1-(sinx)^2)

The Attempt at a Solution

I sub u= sin2x

But then got nowhere because I had cos2x to the power of 2 and I don't know how to compensate for it with du. [/B]

First, an easy substitution gets rid of the 2 factors in the 2x terms.
When you have a mix of sin and cos in an integral dx, look for combining one of them with the dx, e.g. cos(x)dx = d sin(x).
In the present case, you can choose a cos or a sin. Which works better?

 

[HallsofIvy]

When you have even powers of both sine and cosine, reduce the powers using sin^2(x)= (1/2)(1- cos(2x)) and cos^2(x)= (1/2)(1+ cos(2x)).

 

[pasmith]

Homework Statement

Integrate: (sin(2x))^3(cos2x)^2dx

Homework Equations

Using substitution

Cos2x= (1-(sinx)^2)

I can't tell whether you mean \cos^2 x= 1 - \sin^2 x, which is true, or \cos 2x = 1 - \sin^2 x, which is false: \cos 2x = \cos^2 x - \sin^2 x = 1 - 2\sin^2 x.

The Attempt at a Solution

I sub u= sin2x

But then got nowhere because I had cos2x to the power of 2 and I don't know how to compensate for it with du. [/B]

You have an extra power of \sin 2x = -\frac12 \frac{d}{dx} \cos 2x. That suggests u = \cos 2x, not u = \sin 2x.

 

[haruspex]

When you have even powers of both sine and cosine, reduce the powers using sin^2(x)= (1/2)(1- cos(2x)) and cos^2(x)= (1/2)(1+ cos(2x)).

It's simpler than that. See my post #2.

 

[mshiddensecret]

I got it. You use he identity and make sin into 1-cos. and then sub u for cos 2x and work from there.

 

[HallsofIvy]

Please, please, please be more careful about what you are writing. You cannot "make sin into 1- cos"! They are not equal.