How to find the integral for $\int 2^{\sin{x}}\cos{x}\;\mathrm{d}x$?

Question

What would be the ideal approach in finding the integral for:

$$ \int 2^{\sin{x}}\cos{x}\;\mathrm{d}x $$

Answer 1

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}

Answer 2

Hint: let $u = sinx$ , then transfer it to integral with respect to $u$.

Answer 3

Put $\cos{x}$ under derivative:

$$\int 2^{\sin{x}}\cos{x}\;\mathrm{d}x = \int 2^{\sin{x}}\;\mathrm{d}{\sin{x}} = \int 2^u\;\mathrm{d}{u} = \dfrac{2^u}{\ln{2}} = \dfrac{2^{\sin{x}}}{\ln{2}}$$

Answer 4

Hint

Rewrite $2^{\sin(x)}$ as $e^{\sin(x) \log(2)}$ and recognize that $\cos(x)$ is almost the derivative of the exponent.