Integration using complex numbers problem.

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/?i=Sin%5E3%28x%29cos%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) ?

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

Thank you

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. $e^{ikx}$ for some integers k. You can then integrate these terms directly.

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?