# Integrals: cos(2x)

1. Jan 14, 2009

### sunfleck

Hello :)

I'm not sure how to deal with an integral of the form cos(2x), sin(2x), tan(3x), etc. Its the constant that is throwing me off - I know what the antideriv of cosx, sinx or tanx is. My text doesn't seem to explain what to do with the constant and I can't seem to find the info using a direct internet search so I was wondering if someone would please tell me what to do with the constant?

Help would be greatly appreciated

2. Jan 14, 2009

### Dick

Substitute u=2x, du=2*dx. Do the u integral and substitute back.

3. Jan 14, 2009

### sunfleck

Thanks, but we actually haven't covered the sub rule yet so I'm wondering if maybe there is a different way to get the answer?

Updating because I found a similar question in my text...
The text has a problem showing..

$$\int sin( \Pi x)dx = -(1/ \Pi )cos( \Pi x)dx$$

So I take it if f(x) = cos(2x)dx then F(x) = (1/2)sin(2x)... but the chpt explaining the sub rule says something about changing the boundaries of a definate integral when using the sub rule.... Would I have to do anything to the boundaries? (I've been asked to solve using FTC where $$\int \stackrel{b}{a} = F(b) - F(a)$$ ) --I found that problem in the sub rule section which is why I'm confused about the boundaries--

Last edited: Jan 14, 2009
4. Jan 14, 2009

### Dick

Ok. The integral of sin(2x) is -(1/2)*cos(2x). You can check that just be differentiating -(1/2)*cos(2x) and seeing that you get sin(2x). The antiderivative -(1/2)*cos(2x) is expressed in terms of the original variable x. So the limits are the same as the original limits. You don't have to change them.

5. Jan 15, 2009

### sunfleck

Oh right.. that makes sense! Thanks for taking the time to reply :)

6. Jan 15, 2009

### Staff: Mentor

I don't think there's a substitution rule, per se. Substitution is nothing more than applying the chain rule of differentiation to antidifferentiation.
To expand on what I said above, if g(x) = sin(2x), then g'(x) = 2cos(2x). Looking at things from the opposite direction,
$$\int 2cos(2x) dx = sin(2x) + C$$
Equivalently,
$$\int cos(2x) dx = 1/2 sin(2x) + C_1$$
where C1 = C/2.

Here's a slightly more complicated problem:
$$\int xcos(x^2) dx$$

For this problem, I can see that part of the integrand is cos(x^2), so I think of this as being cos(u(x)), where u(x) = x^2. I see also that du/dx = 2x, so it's helpful that the other factor in the integrand is x.

So x*cos(x^2)*dx = 1/2 * du/dx * cos(u) * dx = 1/2 * cos(u)*du

I know that
$$\int cos(u) du = sin(u) + C_1$$
so

$$\int 1/2*cos(u) du = 1/2*sin(u) + C$$

= 1/2*sin(x^2) + C