How would one integrate $e^{\sin x}\csc x$? I was trying to develop a random integral transform, but things happened, and I'm kinda confused as to what I can do.
Anyway, here's an example of my in-development transform:
$$\int e^{\sin t}\csc(t)dt$$
I just need to place the transform bounds, but here's the main idea as to what I wish to accomplish:
$$\int^b_a f(t)e^{\sin t}dt$$
where $f(t)$ can be any function, and I need to add the boundaries, and maybe I could change it up later. Of course, it would evaluate to $e^{\sin t}$ if $f(t)=\cos(t)$.
Anyway, I would really like if someone could help out with the development of this transform. I just need help with stuff like this.
 A: $\int e^{\sin t}\csc t~dt$
$=\int\csc t~dt+\int\sum\limits_{n=1}^\infty\dfrac{\sin^{2n-1}t}{(2n)!}~dt+\int\sum\limits_{n=0}^\infty\dfrac{\sin^{2n}t}{(2n+1)!}~dt$
$=\int\csc t~dt+\int\sum\limits_{n=0}^\infty\dfrac{\sin^{2n+1}t}{(2n+2)!}~dt+\int\left(1+\sum\limits_{n=1}^\infty\dfrac{\sin^{2n}t}{(2n+1)!}\right)~dt$
For $n$ is any natural number,
$\int\sin^{2n}t~dt=\dfrac{(2n)!t}{4^n(n!)^2}-\sum\limits_{k=1}^n\dfrac{(2n)!((k-1)!)^2\sin^{2k-1}t\cos t}{4^{n-k+1}(n!)^2(2k-1)!}+C$
This result can be done by successive integration by parts.
$\int\sin^{2n+1}t~dt$
$=-\int\sin^{2n}t~d(\cos t)$
$=-\int(1-\cos^2t)^n~d(\cos t)$
$=-\int\sum\limits_{k=0}^nC_k^n(-1)^k\cos^{2k}t~d(\cos t)$
$=-\sum\limits_{k=0}^n\dfrac{(-1)^kn!\cos^{2k+1}t}{k!(n-k)!(2k+1)}+C$
$\therefore\int\csc t~dt+\int\sum\limits_{n=0}^\infty\dfrac{\sin^{2n+1}t}{(2n+2)!}dt+\int\left(1+\sum\limits_{n=1}^\infty\dfrac{\sin^{2n}t}{(2n+1)!}\right)~dt$
$=-\ln(\csc t+\cot t)-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!\cos^{2k+1}t}{(2n+2)!k!(n-k)!(2k+1)}+\sum\limits_{n=0}^\infty\dfrac{(2n)!t}{4^n(2n+1)!(n!)^2}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(2n)!((k-1)!)^2\sin^{2k-1}t\cos t}{4^{n-k+1}(2n+1)!(n!)^2(2k-1)!}+C$
$=-\ln(\csc t+\cot t)+\sum\limits_{n=0}^\infty\dfrac{t}{4^n(n!)^2(2n+1)}-\sum\limits_{k=1}^\infty\sum\limits_{n=k}^\infty\dfrac{((k-1)!)^2\sin^{2k-1}t\cos t}{4^{n-k+1}(n!)^2(2n+1)(2k-1)!}-\sum\limits_{k=0}^\infty\sum\limits_{n=k}^\infty\dfrac{(-1)^kn!\cos^{2k+1}t}{(2n+2)!k!(n-k)!(2k+1)}+C$
$=-\ln(\csc t+\cot t)+\sum\limits_{n=0}^\infty\dfrac{t}{4^n(n!)^2(2n+1)}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{(k!)^2\sin^{2k+1}t\cos t}{4^{n+1}((n+k+1)!)^2(2n+2k+3)(2k+1)!}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{(-1)^k(n+k)!\cos^{2k+1}t}{(2n+2k+2)!n!k!(2k+1)}+C$
