Mastering Integrals: Solving for cos(2x) and Other Trigonometric Functions

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Homework Help Overview

The discussion revolves around integrating trigonometric functions, specifically focusing on forms like cos(2x) and sin(2x). Participants express confusion regarding the treatment of constants in these integrals and the implications for definite integrals when applying substitution methods.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the use of substitution for integrating functions like cos(2x) and sin(2x), questioning the necessity and implications of changing boundaries in definite integrals. There is also a discussion on the relationship between differentiation and integration in the context of these functions.

Discussion Status

Some participants have provided guidance on using substitution and the treatment of boundaries in definite integrals, while others express uncertainty about the application of the substitution rule due to their current curriculum. The conversation reflects a mix of interpretations and attempts to clarify the concepts involved.

Contextual Notes

Participants note that they have not yet covered the substitution rule in their studies, which adds to their confusion regarding the integration process and the handling of constants and boundaries in definite integrals.

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


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..

[tex]\int sin( \Pi x)dx = -(1/ \Pi )cos( \Pi x)dx[/tex]

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 definite integral when using the sub rule... Would I have to do anything to the boundaries? (I've been asked to solve using FTC where [tex]\int \stackrel{b}{a} = F(b) - F(a)[/tex] ) --I found that problem in the sub rule section which is why I'm confused about the boundaries--
 
Last edited:
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.
 
Oh right.. that makes sense! Thanks for taking the time to reply :)
 
sunfleck said:
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?
I don't think there's a substitution rule, per se. Substitution is nothing more than applying the chain rule of differentiation to antidifferentiation.
sunfleck said:
Updating because I found a similar question in my text...
The text has a problem showing..

[tex]\int sin( \Pi x)dx = -(1/ \Pi )cos( \Pi x)dx[/tex]

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 definite integral when using the sub rule... Would I have to do anything to the boundaries? (I've been asked to solve using FTC where [tex]\int \stackrel{b}{a} = F(b) - F(a)[/tex] ) --I found that problem in the sub rule section which is why I'm confused about the boundaries--

To expand on what I said above, if g(x) = sin(2x), then g'(x) = 2cos(2x). Looking at things from the opposite direction,
[tex]\int 2cos(2x) dx = sin(2x) + C[/tex]
Equivalently,
[tex]\int cos(2x) dx = 1/2 sin(2x) + C_1[/tex]
where C1 = C/2.

Here's a slightly more complicated problem:
[tex]\int xcos(x^2) dx[/tex]

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
[tex]\int cos(u) du = sin(u) + C_1[/tex]
so

[tex]\int 1/2*cos(u) du = 1/2*sin(u) + C[/tex]

= 1/2*sin(x^2) + C
 

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