Calculating the integral $\int e^{\sin(mx+n)}\,dx$

Question

I know that: $$\int e^{mx+n}\,dx=\frac{1}{m}\cdot e^{mx+n}$$ But I was wondering if its possible to calculate $\int e^{\sin(mx+n)}\,dx$. As I understand it is not trivial (looked at the solution as WolframAlpha) but how to achieve it with Taylor series at least?

Answer 1

Do not expect a closed form.

Considering $$I=\int e^{\sin(mx+n)}\,dx$$ let $$mx+n=y \implies x=\frac{y-n}{m}\implies dx=\frac{dy}{m}$$ to make $$I=\frac 1 m \int e^{\sin(y)}\,dy=\frac 1 m \int \sum_{k=0}^\infty \frac{sin^k(y)}{k!}\,dy=\frac 1 m \sum_{k=0}^\infty \frac{1}{k!}\int sin^k(y)\,dy$$ and use the reduction formula $$\int\sin^k(y)\,dx= -\frac{\cos (y) \sin^{k-1}(y)}{k} + \frac{(k-1)}{k} \int \sin^{k-2} (y)\, dy$$

For illustration purposes, consider $m=2$, $n=3$ and integrate between $0$ and $5$ and consider the partial sums (up to $p$). You would obtain $$\left( \begin{array}{cc} p & S_p \\ 0 & 5.00000 \\ 1 & 4.05128 \\ 2 & 5.23616 \\ 3 & 5.12575 \\ 4 & 5.19944 \\ 5 & 5.19500 \\ 6 & 5.19704 \\ 7 & 5.19695 \\ 8 & 5.19698 \end{array} \right)$$