Back
to top
Integration using complex numbers problem.
Watch
Announcements
Page 1 of 1
Go to first unread
Skip to page:
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!
0
reply
Report
#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; 3 months ago
0
reply
(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
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) ?
0
reply
Report
#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) ?
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) ?
.
0
reply
Report
#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) ?
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
0
reply
(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
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
0
reply
Report
#7
(Original post by jc768)
Thank you
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?
0
reply
Report
#8
(Original post by jc768)
Thank you
Thank you
for some integers k. You can then integrate these terms directly.
0
reply
Report
#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.
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 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?
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?
, 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.
0
reply
X
Page 1 of 1
Go to first unread
Skip to page:
Quick Reply
Today on TSR
-
Have your say on 2021 assessmentOfqual and DfE want students' thoughts
- Who has received uni offers?
- Starting my new job - AMA!
- Win a year of Headspace
- Private candidates support thread