how to evaluate $\int_{0}^{+\infty} e^{-at}(sin(t))^{n} dt$ 
How to evaluate the following definite integral : 
  $$\int_{0}^{+\infty} e^{-at}\left(\sin(t)\right)^{n}\,dt$$

 A: Use De Moivre's formula to compute the Fourier sine/cosine series of $\sin(t)^n$, then exploit:
$$ \int_{0}^{+\infty}e^{-at}\sin(bt)\,dt = \frac{b}{a^2+b^2},\qquad \int_{0}^{+\infty}e^{-at}\cos(bt)\,dt = \frac{a}{a^2+b^2}.$$
A: I give you some hints:
Let
$$
I_n=\int_0^{+\infty}e^{-at}\sin^n t\,dt=\mathcal L(\sin^n t)(a),
$$
i.e. $I_n$ is the Laplace transform of $\sin^n(t)$. Using the fact that
$$
D^2(\sin^nt)=n(n-1)\sin^{n-2}t-n^2\sin^n t.
$$
we find that, for $n>1$, (using the rule for differentiation and Laplace transforms)
$$
a^2I_n=n(n-1)I_{n-2}-n^2I_n,
$$
or
$$
I_n=\frac{n(n-1)}{n^2+a^2}I_{n-2}.
$$
Here, we have also used the fact that $\sin^nt$ is zero for $t=0$ and that its derivative is zero for $t=0$.
Now, as is easily calculated
$$
I_0=\frac{1}{a},\quad\text{and}\quad I_1=\frac{1}{1+a^2}.
$$
From this it follows that
$$
I_n=\frac{n!}{a(2^2+a^2)(4^2+a^2)\cdots(n^2+a^2)},\quad \text{if $n$ is even}
$$
and
$$
I_n=\frac{n!}{(1^2+a^2)(3^2+a^2)\cdots(n^2+a^2)},\quad\text{if $n$ is odd.}
$$
