1. Assume $f(t)$ has a fundamental period $2 \pi$ and defined as \[ f(t) \cong a_{0}+\sum_{n=1}^{\infty}\left(a_{n} \cos n t+b_{n} \sin n t\right) \] Solve $\int_{-\pi}^{\pi} f(t) \sin k t d t$ to find a Fourier coefficient $b_{n}$.

Step 1 ：Assume $f(t)$ has a fundamental period $2 \pi$ and defined as $f(t) \cong a_{0}+\sum_{n=1}^{\infty}\left(a_{n} \cos n t+b_{n} \sin n t\right)$

Step 2 ：Solve $\int_{-\pi}^{\pi} f(t) \sin k t d t$ to find a Fourier coefficient $b_{n}$

Step 3 ：The integral $\int_{-\pi}^{\pi} f(t) \sin k t d t$ is used to find the Fourier coefficient $b_{n}$. The Fourier series of a function $f(t)$ with period $2\pi$ is given by the equation in the question. The coefficient $b_{n}$ is given by the formula $b_{n} = \frac{1}{\pi} \int_{-\pi}^{\pi} f(t) \sin n t dt$. So, to find $b_{n}$, we need to compute this integral.

Step 4 ：The integral is not simplified because it contains an infinite sum. However, we can simplify it by using the orthogonality properties of sine and cosine functions. The integral of the product of two sine or cosine functions over a period is zero unless the functions have the same frequency. Therefore, the only term in the sum that will contribute to the integral is the one where $n=k$. In this case, the integral of $\sin^2(n t)$ over a period is $\pi$. Therefore, the coefficient $b_n$ is just the coefficient of $\sin(n t)$ in the function $f(t)$ when $n=k$.

Step 5 ：The integral is simplified to $b_k = \pi b_k$ when $k \neq 0$ and $b_k = 0$ when $k = 0$. This means that the coefficient $b_k$ is zero when $k = 0$ and is equal to 1 when $k \neq 0$. Therefore, the Fourier coefficient $b_n$ is 1 when $n=k$ and $n \neq 0$, and is zero when $n=0$.

Step 6 ：Final Answer: The Fourier coefficient $b_n$ is given by \[\boxed{b_{n} = \begin{cases} 1 & \text{if } n=k \text{ and } n \neq 0 \\ 0 & \text{if } n=0 \end{cases}}\]