# Integration homework

I haven't done the calculation myself, but you should, according to Euler's Equation, take the real part of the sum of exponential functions.

Isn't the real part the cos2nx series?

vela
Staff Emeritus
Homework Helper
I know, you are not allowed to solve the question form me but if the student is stuck shouldn't you give some hints? :(
We've given you hints, but for some reason, you've completely ignored them.

I ignored some because Euler's formula isn't covered by the syllabus. Maybe we should use cis2x+(cis2x)^2+...+(cis2x)^n as a suitable geometric series.

I was able to do the integral without using the series but that isn't acceptable for this question.

Let $$I_n=\int_0^{\pi/2}\frac{\sin ((2n+1)x)}{\sin(x)}dx$$
we can use sum to products formula: $$\sin((2n+1)x)-\sin((2n-1)x)=2\sin(x)\cos(2nx)$$ $$\therefore I_n=\int_0^{\pi/2} \frac{\sin((2n+1)x)}{\sin(x)}dx$$
$$=\int_0^{\pi/2} \frac{\sin((2n-1)x)}{\sin(x)}dx+2\int_0^{\pi/2}\cos(2nx)dx$$
The cosine integral is zero, so we have that $$I_n=I_{n-1}$$
By induction this implies that $$I_n=I_0=\pi/2$$