What is the Indefinite Integral of x*cos(3x)^2?

[Loopas]

Homework Statement

Evaluate the indefinite integral of x*cos(3x)^2

Homework Equations

Integration by parts: \int(udv)= uv - \int(vdu)

The Attempt at a Solution

Im having trouble finding the antiderivative of cos(3x)^2 (which I designated as dv when doing integration by parts). I keep getting (1/9)(sin(3x)^3) using the power and chain rules, but that's not correct.

Thanks!

 

[Mark44]

Homework Statement

Evaluate the indefinite integral of x*cos(3x)^2

Homework Equations

Integration by parts: \int(udv)= uv - \int(vdu)

The Attempt at a Solution

Im having trouble finding the antiderivative of cos(3x)^2 (which I designated as dv when doing integration by parts). I keep getting (1/9)(sin(3x)^3) using the power and chain rules, but that's not correct.

Thanks!

Is this your integral?
$$\int x cos(9x^2)dx$$

If so, you might be able to do it by integration by parts, but I wouldn't do it this way. An ordinary substitution will work just fine.

 

[LCKurtz]

Or is it$$
\int x\cos^2(3x)$$in which case a double angle formula and integration by parts would work.

 

[Mark44]

Loopas,
Obviously, what you wrote is ambiguous.

 

[HallsofIvy]

Assuming you mean cos^2(3x) use the identity cos(2a)= cos^2(a)- sin^2(a)= cos^2(a)- (1- cos^2(a)= 2cos^2(a) -1. So cos^2(3x)= (cos(6a)+ 1)/2.

 

[Loopas]

Ok, the integral is \int(xcos^{2}(3x))

Im trying to use integration by parts but I can't find the antiderivative of cos_{2}(3x)

Assuming you mean cos2(3x) use the identity cos(2a)=cos2(a)−sin2(a)=cos2(a)−(1−cos2(a)=2cos2(a)−1. So cos2(3x)=(cos(6a)+1)/2.

What exactly is the form of the double angle formula?

 

[Mark44]

What you wrote is extremely difficult to read. I have fixed it up to make it clear.

Ok, the integral is \int(xcos^{2}(3x))dx[/color]

Im trying to use integration by parts but I can't find the antiderivative of $cos^2(3x)$

Assuming you mean cos²(3x) use the identity cos(2a)=cos²(a)−sin²(a)=cos²(a)−(1−cos²(a)=2cos²(a)−1. So cos²(3x)=(cos(6a)+1)/2.

What exactly is the form of the double angle formula?

 

[LCKurtz]

Ok, the integral is \int(xcos^{2}(3x))

Im trying to use integration by parts but I can't find the antiderivative of cos_{2}(3x)



What exactly is the form of the double angle formula?

$$\cos^2(3x)=\frac {1+\cos(6x)} 2$$

 

[Loopas]

So would the correct antiderivative be \frac{x}{2}+\frac{sin(6x)}{12}?

 

[Mark44]

So would the correct antiderivative be \frac{x}{2}+\frac{sin(6x)}{12}?

It's easy enough to check for yourself. Just differentiate your answer and you should get the integrand.

 

[Loopas]

Ok so I can use the double angle formula to put cos^{2}(3x) into a form that I can find the antiderivative more easily. But I'm confused about how to use the double angle identity to get \frac{1+cos(6x)}{2}

 

[HallsofIvy]

The basic double angle formula for cosine says that cos(2\theta)= cos^2(\theta)- sin^2(\theta). But sin^2(\theta)= 1- cos^2(\theta) so that can be written cos(2\theta)= cos^2(\theta)- (1- cos^2(\theta))= 2cos^2(\theta)- 1 From that 2cos^2(\theta)= 1+ cos(2\theta) and so cos^2(\theta)= \frac{1+ cos(2\theta)}{2}.

Now replace "\theta" with "3x": cos^2(3x)= \frac{1+ cos(6x)}{2}.

 

[Loopas]

Ahh I can see it now! Thanks guys this was all really helpful!