Given that $$f(t)=\begin{cases}-1 & -\frac{T}{2}\leq t \lt 0, \\ +1 & \ \ \ \ 0 \leq t \lt \frac{T}{2}\end{cases}$$
Show the steps involved such that $$\frac{2}{T}\int_{t=-\frac{T}{2}}^{\frac{T}{2}}f(t)\sin\left(\frac{2\pi kt}{T}\right)\mathrm{d}t = \frac{4}{T}\int_{t=0}^{\frac{T}{2}}\sin\left(\frac{2\pi kt}{T}\right)\mathrm{d}t$$ Where $k \in \Bbb{N}$ and $T \in \Bbb{R}$.
An attempt: $$\frac{2}{T}\int_{t=-\frac{T}{2}}^{\frac{T}{2}}f(t)\sin\left(\frac{2\pi kt}{T}\right)\mathrm{d}t = \frac{2}{T}\int_{t=-\frac{T}{2}}^{0}f(t)\sin\left(\frac{2\pi kt}{T}\right)\mathrm{d}t+\frac{2}{T}\int_{t=0}^{\frac{T}{2}}f(t)\sin\left(\frac{2\pi kt}{T}\right)\mathrm{d}t$$
If someone could please show me the steps involved to complete this proof it would be most appreciated.
Many thanks,
Blaze
The first equality comes from the fact that if $g$ is an even function, then $$ \int_{-a}^a g(t)\,dt=2\int_0^a g(t)\,dt. $$ Either draw a figure or write the integral as $\int_{-a}^0+\int_0^a$ and make the substitution $t\mapsto -t$ in the first integral to see that.
The second one is just calculating the integral using the fundamental theorem of calculus, using the facts that $$ \cos 0 = 1\quad\text{and}\quad \cos(k\pi)=(-1)^k\ \text{if}\ k\in\mathbb N. $$