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Reply 20
there are just some identities for cos2xcos^{2}x and sin2xsin^{2}x that i've got here on a formula sheet and i've substituted them in, that's all. but i'm sure how to 'recognise' or what to do next, really.
Reply 21
cdeu12
there are just some identities for cos2xcos^{2}x and sin2xsin^{2}x that i've got here on a formula sheet and i've substituted them in, that's all. but i'm sure how to 'recognise' or what to do next, really.


Hehe and my advice is wrong as usual, ignore me. God I'm rusty.
Reply 22
notnek
If you have something of the form:

f(x)(f(x))n\displaystyle f'(x)(f(x))^{n}

Then it's integral dx is (f(x))n+1n+1+c\displaystyle \frac{(f(x))^{n+1}}{n+1} + c


thanks. what if it's to integrate something like (e.g.) sin4xcos34xsin4x cos^{3}4x?
Reply 23
I should've added a 'k':

kf(x)(f(x))n=k(f(x))n+1n+1\displaystyle \int kf'(x)(f(x))^n = k\frac{(f(x))^{n+1}}{n+1}

IN this case, f(x)=cos(4x) and kf'(x) = -1/4 sinx
Reply 24
notnek
I should've added a 'k':

kf(x)(f(x))n=k(f(x))n+1n+1\displaystyle \int kf'(x)(f(x))^n = k\frac{(f(x))^{n+1}}{n+1}

IN this case, f(x)=cos(4x) and kf'(x) = -1/4 sinx


Sorry I'm in and out and not really paying attention to be frank, but trying to get this stuff back after a long sojourn? Would parts work? Or was I barking up entirely the wrong tree there? I mean on the original equation he gave?
Reply 25
Sidhe
Sorry I'm in and out and not really paying attention to be frank, but trying to get this stuff back after a long sojourn? Would parts work? Or was I barking up entirely the wrong tree there? I mean on the original equation he gave?

I think so but it requires a lot more work. It is the wrong method to use in this case.
Reply 26
notnek
I think so but it requires a lot more work. It is the wrong method to use in this case.


Ah thanks notnek. Just trying to get my eye back in.
Reply 27
help please!

doing a question now - there's a diagram showing part of a curve y=4sinxcos3x y = 4sinxcos^{3}x.

I need to find the area enclosed by the curve and the x-axis between x=0 and x=pi. I seem to have lost the plot a bit (excuse me cos I'm ill as well), but I'm confused about what to do!! thanks. :smile:
Reply 28
anyone at all please?
Reply 29
cdeu12
help please!

doing a question now - there's a diagram showing part of a curve y=4sinxcos3x y = 4sinxcos^{3}x.

I need to find the area enclosed by the curve and the x-axis between x=0 and x=pi. I seem to have lost the plot a bit (excuse me cos I'm ill as well), but I'm confused about what to do!! thanks. :smile:

Can you integrate 4sinxcos3x4sinxcos^{3}x dx ?
Reply 30
would it be 4/4cos4x+C-4/4cos^{4}x + C? so then cos4x+C- cos^{4}x + C... or have I done that wrong?

I'm also unsure about putting the limits in, would I have to convert it to the actual value and then put them in the calculator in radians? :s-smilie:
Reply 31
bump
Reply 32
cdeu12
would it be 4/4cos4x+C-4/4cos^{4}x + C? so then cos4x+C- cos^{4}x + C... or have I done that wrong?

I'm also unsure about putting the limits in, would I have to convert it to the actual value and then put them in the calculator in radians? :s-smilie:

Yes it is cos4(x)-cos^4(x). I don't know what you mean about the limits. Can you explain further?
Reply 33
notnek
Yes it is cos4(x)-cos^4(x). I don't know what you mean about the limits. Can you explain further?


I need to find the area enclosed by the curve and the x-axis between x=0 and x=pi.
Reply 34
cdeu12
I need to find the area enclosed by the curve and the x-axis between x=0 and x=pi.

I know.

So the limits are 0 and pi ...?
Reply 35
Actually, wait... that's not right.

By the sketch, you should be able to see that you need to integrate with limits 0-pi/2 and pi/2-pi and then add them together.
Reply 36
ohhh i didn't notice that. i assumed the limits were 0 and pi! hopefully if i try that it should work then.

and one LAST question now! :p: - i've got to integrate cosx(1+sin3x)cos x (1 + sin^{3}x), limits of pi/2 and 0.

how would i go about starting to integrate the above... it seems a bit more complex. even when multiplying it out i end up with cosx+cosxsin3xcos x + cos xsin^{3}x and that resembles something i've not gone over yet. help appreciated, thank you!
Reply 37
cdeu12
ohhh i didn't notice that. i assumed the limits were 0 and pi! hopefully if i try that it should work then.

and one LAST question now! :p: - i've got to integrate cosx(1+sin3x)cos x (1 + sin^{3}x), limits of pi/2 and 0.

how would i go about starting to integrate the above... it seems a bit more complex. even when multiplying it out i end up with cosx+cosxsin3xcos x + cos xsin^{3}x and that resembles something i've not gone over yet. help appreciated, thank you!

You should know the integral of cos(x) and you've already integrated cos(x)sin^3(x)...
Reply 38
Is there a rule for integrating something like 1/2cos2x -1/2 cos2x?

I'm not too good with latex... the -1/2 means -0.5. and then followed by cos2x.
Reply 39
cdeu12
Is there a rule for integrating something like 1/2cos2x -1/2 cos2x?

I'm not too good with latex... the -1/2 means -0.5. and then followed by cos2x.

kcos(ax) dx=kasin(ax)\displaystyle \int kcos(ax) \ dx = \frac{k}{a}sin(ax)

There's a similar result with sin(2x). You could use a substitution of u=2x to see where this comes from.

It's best to use \frac{-1}{2} for fractions in LaTeX.

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