:)Esss:)
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I don’t understand how to integrate cos(2x) I know the answer is (1/2)sin(2x)+ c

We haven’t learnt how integrate cos yet but I read the book so got a general understanding that integration of cos(x) gives sin(x).
So is it the (2x) but that’s giving rise to the 1/2?
Idk if that question made sense but yh?
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Mr M
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(Original post by :)Esss:))
I don’t understand how to integrate cos(2x) I know the answer is (1/2)sin(2x)+ c

We haven’t learnt how integrate cos yet but I read the book so got a general understanding that integration of cos(x) gives sin(x).
So is it the (2x) but that’s giving rise to the 1/2?
Idk if that question made sense but yh?
\int \cos (ax+b)  dx = \frac{1}{a} \sin (ax+b) + c

If you can't see why, try integrating by substitution with u=ax+b
Last edited by Mr M; 1 year ago
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Mona123456
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(Original post by :)Esss:))
I don’t understand how to integrate cos(2x) I know the answer is (1/2)sin(2x)+ c

We haven’t learnt how integrate cos yet but I read the book so got a general understanding that integration of cos(x) gives sin(x).
So is it the (2x) but that’s giving rise to the 1/2?
Idk if that question made sense but yh?
Okay, I’m going to try and explain this working backwards from differentiation.

So hopefully you’ve learnt that differentiating sin x = cos x.

You might also know that sin (kx) therefore is differentiated to k cos (kx).

So for example:
Differentiating sin (2x) gives 2cos(2x).

Now as differentiation and integration are sort of ‘opposites’ in a way eg if you integrate dy/dx you get back to y, then:

Integrating 2cos(2x) equals sin(2x)
So integrating cos(2x) equals 0.5sin(2x)

Hopefully this makes some sort of sense, but if not don’t worry, when you get taught it, it will make more sense!

Edit: I forgot the plus c as I’m using an example where I’m pretending they are definite integrals, but otherwise you put the plus c as normal!
Last edited by Mona123456; 1 year ago
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:)Esss:)
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(Original post by Mr M)
\int \cos (ax+b)  dx = \frac{1}{a} \sin (ax+b) + c

If you can't see why, try integrating by substitution with u=ax+b
I’m so sorry but I still don’t get it.
What do you mean by “\frac{1}{a}\sin(ax+b)” I don’t understand that notation is that meant to be a fraction?
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:)Esss:)
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(Original post by Mona123456)
Okay, I’m going to try and explain this working backwards from differentiation.

So hopefully you’ve learnt that differentiating sin x = cos x.

You might also know that sin (kx) therefore is differentiated to k cos (kx).

So for example:
Differentiating sin (2x) gives 2cos(2x).

Now as differentiation and integration are sort of ‘opposites’ in a way eg if you integrate dy/dx you get back to y, then:

Integrating 2cos(2x) equals sin(2x)
So integrating cos(2x) equals 0.5sin(2x)

Hopefully this makes some sort of sense, but if not don’t worry, when you get taught it, it will make more sense!

Edit: I forgot the plus c as I’m using an example where I’m pretending they are definite integrals, but otherwise you put the plus c as normal!
Ah ok thanks I get it
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ghostwalker
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(Original post by :)Esss:))
I’m so sorry but I still don’t get it.
What do you mean by “\frac{1}{a}\sin(ax+b)” I don’t understand that notation is that meant to be a fraction?
Some access methods for TSR don't support LaTex (the typescrpting used for mathematics.) Can't recall which one in particular. If you're accessing through a browser you should be able to see it correctly.
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Mona123456
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(Original post by :)Esss:))
Ah ok thanks I get it
You’re welcome—no problem. It’s great to see someone interested in maths who’s reading ahead!
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Physics Enemy
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(Original post by :)Esss:))
... understanding that integration of cos(x) gives sin(x). So is it the (2x) that’s giving rise to the 1/2?
Yes. Differentiation: cos(ax) -> -asin(ax) and sin(ax) -> +acos(ax). Integration: cos(ax) -> +(1/a)sin(ax) and sin(ax) -> -(1/a)cos(ax). So there's sign switch and switch in times or divide by a.

Integrating the derivative of cos(ax) i.e) -asin(ax), we get -a(-1/a)cos(ax) = cos(ax). Integration undoes the derivative, gives back original function (& vice versa). Inverse operators as expected.
Last edited by Physics Enemy; 1 year ago
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Eimmanuel
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(Original post by ghostwalker)
Some access methods for TSR don't support LaTex (the typescrpting used for mathematics.) Can't recall which one in particular. If you're accessing through a browser you should be able to see it correctly.
TSR apps on the mobile does not render the Latex. But if we access TSR forum using mobile phone web browser apps, we can see the Latex rendering properly.
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