# Integration using complex numbers problem.

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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!

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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#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

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; 1 month ago

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(Original post by

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

**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

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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#4

(Original post by

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) ?

**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) ?

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#5

**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) ?

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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(Original post by

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

**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

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#7

(Original post by

Thank you

**jc768**)Thank you

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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#8

(Original post by

Thank you

**jc768**)Thank you

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#9

(Original post by

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.

**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.

(Original post by

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?

**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?

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