How to find the integral for $\int 2^{\sin{x}}\cos{x}\;\mathrm{d}x$? What would be the ideal approach in finding the integral for:
$$
\int 2^{\sin{x}}\cos{x}\;\mathrm{d}x
$$

Let $u=\sin x$, $du=\cos x\, dx$, and $dx=\frac{du}{\cos x}$, then
\begin{align}
\require{cancel}
\int 2^{\sin{x}}\cos{x}\;dx&=\int 2^{u}\cancel{\cos{x}}\;\frac{du}{\cancel{\cos x}}\\
&=\int 2^u\; du\\
&=\int e^{u\ln 2}\; du\qquad;\;\;a^b=e^{b\ln a}\\
&=\frac{e^{u\ln 2}}{\ln 2}+C\qquad\quad;\;\int e^{ax}\; dx=\frac{e^{ax}}{a}+C\\
&=\frac{e^{\sin x\ln 2}}{\ln 2}+C\\
&=\frac{2^{\sin{x}}}{\ln 2}+C
\end{align}
