# Integration using complex numbers problem.

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#1
Hi,

I am struggling with question 6 here, 6a is fine of course.

https://ibb.co/Nm0HYtV

Then in 6b I am struggling to integrate the expression using what I have created in 6a.

As you can see from my working, I have tried rewriting the trig using what I gained from 6a, but this doesn't seem to help me at all!

https://ibb.co/jGpTZrc

As always, any help is much appreciated!
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3 months ago
#2
I thought you had a representation for the integrand like
sin(x) + sin(3x) + ... + sin(7x)
? Obviously there are some coeffs to include.
https://www.wolframalpha.com/input/?...cos%5E4%28x%29
Last edited by mqb2766; 3 months ago
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#3
(Original post by mqb2766)
I thought you had a representation for the integrand like
sin(x) + sin(3x) + ... + sin(7x)
? Obviously there are some coeffs to include.
https://www.wolframalpha.com/input/?...cos%5E4%28x%29
How does that link to part 6a though?

sin^3(x)=1/4(3sin(x)-sin(3x))

cos^4(x)=1/8cos(4x)+1/2cos(2x)+3/8

So how do we get sin^3(x)cos^4(x) ?
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3 months ago
#4
(Original post by jc768)
How does that link to part 6a though?

sin^3(x)=1/4(3sin(x)-sin(3x))

cos^4(x)=1/8cos(4x)+1/2cos(2x)+3/8

So how do we get sin^3(x)cos^4(x) ?
You are supposed to rework it so you don't have any trig terms multiplied together. That is, find an expression for sin^3 x cos^4 x of the form .
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3 months ago
#5
(Original post by jc768)
How does that link to part 6a though?

sin^3(x)=1/4(3sin(x)-sin(3x))

cos^4(x)=1/8cos(4x)+1/2cos(2x)+3/8

So how do we get sin^3(x)cos^4(x) ?
Express
cos^2 = 1 - sin^2
Square both sides and multiply by sin^3.
So you get roughly
Sin^3 + sin^5 + sin^7
Then transform each to a multiple angle which should be like
sin(kx)
Where k is odd
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#6
(Original post by mqb2766)
Express
cos^2 = 1 - sin^2
Square both sides and multiply by sin^3.
So you get roughly
Sin^3 + sin^5 + sin^7
Then transform each to a multiple angle which should be like
sin(kx)
Where k is odd
Thank you
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3 months ago
#7
(Original post by jc768)
Thank you
Using part a)
You could do square and square again the cos expression (imaginary exponentials) and multiply it by the sin cubed. They're all exponential which will be easy to integrate?
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3 months ago
#8
(Original post by jc768)
Thank you
Unless I'm misreading the question (quite possible!) part (a) implies that your trig functions will be in terms of exponentials, so when you cube / take the 4th power you will end up with a set of terms that are (complex) exponentials e.g. for some integers k. You can then integrate these terms directly.
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3 months ago
#9
(Original post by davros)
Unless I'm misreading the question (quite possible!) part (a) implies that your trig functions will be in terms of exponentials, so when you cube / take the 4th power you will end up with a set of terms that are (complex) exponentials e.g. for some integers k. You can then integrate these terms directly.
Agreed.
(Original post by mqb2766)
Using part a)
You could do square and square again the cos expression (imaginary exponentials) and multiply it by the sin cubed. They're all exponential which will be easy to integrate?
As with the other very similar question posted today (that I thought was by the same poster, but actually isn't), if , then starting from (z+1/z)^4(z-1/z)^3, you can rearrange as (z+1/z)[(z+1/z)(z-1/z)]^3 = (z+1/z)(z^2-1/z^2)^3, and you can finish from here in a couple of lines.
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