I don't understand why my book is claiming the following for any $n$ or $w_0$ this is always the case over one period. I think it depends on the $w_o$ really. I have proof too, but I just want another person's opinion that my thought process is correct. So my book claims the following:
$$\int^T_0 \sin(nw_0t)\,dt = 0 $$
$$\int^T_0 \cos(nw_0t)\,dt = 0 $$
I have done a couple of fourier series examples and exercise in my book and it seems that they don't always hold (usually they do, maybe once or twice it doesn't depending on the $w_o$).
Okay, so here is the example where that claim doesn't hold. It's letter e) and the problem asks us to find the trigonometric fourier series coefficients of that waveform.
Okay so, I'm calculating my $b_n$ and I get the following integral I have to solve based on the graph e and problem e:
So I get for my fundamental period $T_o = 2T$ and I get my fundamental angular frequency to be $w_o = 2\pi f_o = 2\pi \frac{1}{2T} = \frac{\pi}{T}$
$$ b_n = \frac{2}{T_o} \int^{T_o}_0 x_5(t)\sin(nw_0t)\,dt = \frac{1}{T} \int^{T}_{0} (1+ -\frac{1}{2}\sin(\frac{\pi t}{T}))\sin(nw_ot)\,dt$$
We break this up into two integrals, so we just solve these two integrals to find our coefficients for $b_n$:
$$b_n = \frac{1}{T} \int^{T}_0 \sin(nw_0t)\,dt + -\frac{1}{2T} \int^{T}_0 \sin(\frac{\pi}{T}t)\sin(n\frac{\pi}{T}t)\,dt $$
Okay let's mark the first integral as $b_{n1}$ and the other integral $b_{n2}$ and we solve seperately:
Let's calculate $b_{n1}$:
$$ b_{n1} = \frac{1}{T} \int^{T}_0 \sin(nw_0t)\,dt = \frac{1}{T} \left[\left.-\frac{1}{nw_0}\cos(nw_ot)\right]\right|_0^T = \frac{1}{T} (-\frac{1}{nw_0})\left[\left.\cos(nw_ot)\right]\right|_0^T $$
$$ b_{n1} = -\frac{1}{n\pi}(cos(n\pi)-1)$$
Now to calculate $b_{n2}$ which was pretty straight forward after a product to sum trig identity and is easily seen to be zero.
$$ b_{n2} = -\frac{1}{4\pi} \int^{T}_0 \cos(\frac{\pi(1-n)}{T}t) - \cos(\frac{\pi(1+n)}{T}t) = 0 $$
So my main issue is with $b_{n1}$. So is my math correct or is the book correct.
A change of variables may help. For example, $$ \int_{0}^{T}\sin(n\omega t)\,dt = \frac{1}{n\omega}\int_{0}^{T}\sin(n\omega t)d(n\omega t)=\frac{1}{n\omega}\int_{0}^{n\omega T}\sin(u)\,du. $$ Therefore, if $n\omega T$ is a multiple of $2\pi$, you'll get $0$ for the integral of $\sin(n\omega t)$ and of $\cos(n\omega t)$. Otherwise, one of the integrals (sin or cos) may not be $0$.
So you are looking at $x5(t)$ with a period of $T_{0}=2T$. That means computing $$ \begin{align} a_{n}& =\frac{1}{T}\int_{0}^{2T}x5(t)\cos(2\pi n t/2T)\,dt \\ & =\frac{1}{T}\int_{0}^{T}\{1-0.5\sin(\pi t/T)\}\cos(\pi n t/T)\,dt \\ b_{n}& =\frac{1}{T}\int_{0}^{2T}x5(t)\sin(2\pi n t/2T)\,dt \\ & =\frac{1}{T}\int_{0}^{T}\{1-0.5\sin(\pi t/T)\}\sin(\pi n t/T)\,dt. \end{align} $$ I like being explicit about the constants instead of trying to unravel them later. The expression of interest to you is $$ \begin{align} b_{n1} & = \frac{1}{T}\int_{0}^{T}\sin(\pi nt/T)\,dt \\ & = \left.-\frac{1}{n\pi}\cos(n\pi t/T)\right|_{0}^{T} \\ & = -\frac{1}{n\pi}(\cos(n\pi)-1)=\frac{1}{n\pi}(1-(-1)^{n}). \end{align} $$ It looks like you and I agree on $b_{n1}$. For the other part, you have to reduce $$ \begin{align} \sin(\pi t/T)\sin(\pi n t/T) &=\frac{1}{2}\left[\cos(\pi t/T-\pi nt/T)-\cos(\pi t/T+\pi nt/T)\right] \\ & = \frac{1}{2}\left[\cos((n-1)\pi t/T)-\cos((n+1)\pi t/T)\right]. \end{align} $$ Your $b_{n2}$ has the front constant out front. There's a $0.5$ from the original problem, another $1/2$ from the trig identity, and a division by $T$ from the Fourier normalization. So, $$ b_{n2}=\frac{1}{4T}\int_{0}^{T}\left\{\cos((n-1)\pi t/T)-\cos((n+1)\pi t/T)\right\}\,dt $$ You have to be careful for the case where $n=1$ because $\cos((n-1)\pi t/T)$ is really the constant function $1$ in that case.