I am asked to find the coefficients for $f(t)=\sin^{2}(5t)$
$$Period =\frac{\pi}{5}$$
so I wrote $$a_n\cdot\sin(\frac{n\pi{t}}{\frac{\pi}{10}})=\sin^{2}(5t)$$
$$a_n\cdot\sin(10n{t})=\sin^{2}(5t)$$
and I saw I can't go further.
and I wrote a little bit different $\sin^{2}(5t)=\frac{1}{2}-\frac{1}{2}\cos(10t)$ , I hope it's right(they have same period),so
$$b_n\cdot\cos(10nt)=\frac{1}{2}-\frac{1}{2}\cos(10t)$$
so here I say : $n=1,a_0=\frac12,b_1=-\frac12$
Did I do something wrong ?
I would gladly answer this question myself but wolfram gives : this and I don't understand it :(.
I used this as my formula, where L= Period/2 :
$$a_n=\frac{2}{\pi/5}\int_0^{\pi/5}\sin^2(5t)\cos\left(\frac{2\pi nt}{\pi/5}\right)dt=a_n=\frac{10}{\pi}\int_0^{\pi/5}\sin^2(5t)\cos\left(10 nt\right)dt=...$$
$$b_n=\frac{2}{\pi/5}\int_0^{\pi/5}\sin^2(5t)\sin\left(\frac{2\pi nt}{\pi/5}\right)dt=a_n=\frac{10}{\pi}\int_0^{\pi/5}\sin^2(5t)\sin\left(10 nt\right)dt=...$$