Finding the following definite and indefinite integrals

For $\int e^{\sin t}~dt$ ,

$\int e^{\sin t}~dt$

$=\int\sum\limits_{n=0}^\infty\dfrac{\sin^{2n}t}{(2n)!}dt+\int\sum\limits_{n=0}^\infty\dfrac{\sin^{2n+1}t}{(2n+1)!}dt$

$=\int\left(1+\sum\limits_{n=1}^\infty\dfrac{\sin^{2n}t}{(2n)!}\right)dt+\int\sum\limits_{n=0}^\infty\dfrac{\sin^{2n+1}t}{(2n+1)!}dt$

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.

For $n$ is any non-negative integer,

$\int\sin^{2n+1}t~dt$

$=-\int\sin^{2n}t~d(\cos t)$

$=-\int(1-\cos^2t)^n~d(\cos t)$

$=-\int\sum\limits_{k=0}^nC_k^n(-1)^k\cos^{2k}t~d(\cos t)$

$=-\sum\limits_{k=0}^n\dfrac{(-1)^kn!\cos^{2k+1}t}{k!(n-k)!(2k+1)}+C$

$\therefore\int\left(1+\sum\limits_{n=1}^\infty\dfrac{\sin^{2n}t}{(2n)!}\right)dt+\int\sum\limits_{n=0}^\infty\dfrac{\sin^{2n+1}t}{(2n+1)!}dt$

$=t+\sum\limits_{n=1}^\infty\dfrac{t}{4^n(n!)^2}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{((k-1)!)^2\sin^{2k-1}t\cos t}{4^{n-k+1}(n!)^2(2k-1)!}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!\cos^{2k+1}t}{(2n+1)!k!(n-k)!(2k+1)}+C$

$=\sum\limits_{n=0}^\infty\dfrac{t}{4^n(n!)^2}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{((k-1)!)^2\sin^{2k-1}t\cos t}{4^{n-k+1}(n!)^2(2k-1)!}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!\cos^{2k+1}t}{(2n+1)!k!(n-k)!(2k+1)}+C$

$\therefore$ For $\int_0^\frac{\pi}{2}e^{\sin t}~dt$ ,

$\int_0^\frac{\pi}{2}e^{\sin t}~dt$

$=\left[\sum\limits_{n=0}^\infty\dfrac{t}{4^n(n!)^2}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{((k-1)!)^2\sin^{2k-1}t\cos t}{4^{n-k+1}(n!)^2(2k-1)!}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!\cos^{2k+1}t}{(2n+1)!k!(n-k)!(2k+1)}\right]_0^\frac{\pi}{2}$

$=\sum\limits_{n=0}^\infty\dfrac{\pi}{2^{2n+1}(n!)^2}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!}{(2n+1)!k!(n-k)!(2k+1)}$