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

In summary: You can use the identity sin^2 = 1- cos^2, so that sin^3 = sin^2 sin = (1- cos^2) sin. But that is not the same as "making sin into 1- cos"!In summary, the integral (sin(2x))^3(cos2x)^2dx can be solved using a substitution of u=cos(2x) and then using the identity sin^2(x)= (1/2)(1- cos(2x)) to simplify the expression. The resulting integral can then be solved using basic integration techniques.
  • #1
mshiddensecret
36
0

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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  • #2
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?
 
  • #3
When you have even powers of both sine and cosine, reduce the powers using [itex]sin^2(x)= (1/2)(1- cos(2x))[/itex] and [itex]cos^2(x)= (1/2)(1+ cos(2x))[/itex].
 
  • #4
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].
 
  • #5
HallsofIvy said:
When you have even powers of both sine and cosine, reduce the powers using [itex]sin^2(x)= (1/2)(1- cos(2x))[/itex] and [itex]cos^2(x)= (1/2)(1+ cos(2x))[/itex].
It's simpler than that. See my post #2.
 
  • #6
I got it. You use he identity and make sin into 1-cos. and then sub u for cos 2x and work from there.
 
  • #7
Please, please, please be more careful about what you are writing. You cannot "make sin into 1- cos"! They are not equal.
 

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

What is a multiple trigonometric integral?

A multiple trigonometric integral is an integral that involves trigonometric functions, such as sine, cosine, and tangent, raised to different powers and multiplied together.

How do I solve a multiple trigonometric integral?

To solve a multiple trigonometric integral, you can use various integration techniques such as substitution, integration by parts, or trigonometric identities.

Can a multiple trigonometric integral be solved using a calculator?

Yes, some calculators have the ability to solve simple multiple trigonometric integrals. However, for more complex integrals, it is best to use integration techniques or software.

What are some common trigonometric identities used in solving multiple trigonometric integrals?

Some common trigonometric identities used in solving multiple trigonometric integrals include the Pythagorean identities, double angle identities, and half angle identities.

Are there any special cases when solving multiple trigonometric integrals?

Yes, there are special cases when solving multiple trigonometric integrals, such as when the integral involves a trigonometric function raised to an even power, or when the integral has limits of integration that are not in terms of the variable being integrated.

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