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

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mshiddensecret
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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]
 
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mshiddensecret said:

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?
 
mshiddensecret said:

Homework Statement



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

Homework Equations



Using substitution

Cos2x= (1-(sinx)^2)

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

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 [itex]\sin 2x = -\frac12 \frac{d}{dx} \cos 2x[/itex]. That suggests [itex]u = \cos 2x[/itex], not [itex]u = \sin 2x[/itex].
 
I got it. You use he identity and make sin into 1-cos. and then sub u for cos 2x and work from there.