Important integrals for fourier series

Question

To understand fourier series I have to understand a few integrals. However the problem is that the book does not explain how to solve the integrals, there are excercises which request to solve those integrals however the solutions are not provided in that chapter.

I want to show that:

$\int_0^{2\pi}\cos(kx)\sin(lx)dx=0$ for all natural numbers $n,l$

$\int_0^{2\pi}\cos(kx)\cos(lx)dx=\int_0^{2\pi}\sin(kx)\sin(lx)dx=0$ for $k\neq l$

$\int_0^{2\pi}\cos^2(kx)dx=\int_0^{2\pi}\sin^2(kx)dx=\pi$ for all $k\geq 1$

Answer 1

We have \begin{align} \int_0^{2\pi}\cos kx\sin\ell x\ \mathsf dx &= \frac1{4i}\int_0^{2\pi} \left(e^{ikx}+e^{-ikx} \right)\left(e^{i\ell x}- e^{-i\ell x} \right)\ \mathsf dx\\ &= \frac1{4i} \int_0^{2\pi} \left(e^{i(k+\ell)x}- e^{-i(k+\ell)x} + e^{-i(k-\ell)x}- e^{i(k-\ell)x} \right)\ \mathsf dx\\ &=\frac1{4i} \left[\frac 1{i(k+\ell)} (e^{i(k+\ell)x} +e^{-i(k+\ell)x}) -\frac1{i(k-\ell)}(e^{-i(k-\ell)x} + e^{i(k-\ell)x}) \right]_0^{2\pi}\\ &= \frac1{4i}\left(\frac 1{i(k+\ell)}(1 + 1 - (1+1)) -\frac1{i(k-\ell)}(1+1-(1+1)) \right)\\ &= 0. \end{align} The other relations may be computed in a similar manner.

Answer 2

$$\int_0^{2\pi}\cos^2(kx)dx=\int_0^{2\pi}\sin^2(kx)dx=\pi$$ For this one you can use: $$\cos ^2 (x)=\frac 1 2 (\cos(2x)+1)$$ $$I=\int_0^{2\pi}\cos^2(kx)dx=\int_0^{2\pi}\frac 1 2 (\cos(2kx)+1)dx$$ $$I=\frac 1 2 \left |\dfrac {\sin(2kx)}{2k}+x\right |_0^{2\pi}=\pi$$ You can then deduce : $$I=\int_0^{2\pi}\sin^2(kx)dx$$ $$I=\int_0^{2\pi}(1-\cos^2(kx))dx$$ $$I=\int_0^{2\pi}dx-\pi=2\pi-\pi=\pi$$