$\int \frac{2^{\sin \left(\sqrt{x}\right)} \cos \left(\sqrt{x}\right)}{\sqrt{x}} \, dx$ I have been asked to integrate:
$$\int \frac{2^{\sin \left(\sqrt{x}\right)} \cos \left(\sqrt{x}\right)}{\sqrt{x}} \, dx$$
In such a small integration you dont have to write it down but to see where I am struggling I have provided a step by step approach:
$$u=\sin \left(\sqrt{x}\right)$$
$$2 \text{du}=\frac{\cos \left(\sqrt{x}\right)}{\sqrt{x}}dx$$
$$2 \int 2^u \, du$$
Know this is where I get stuck cause I do not see that the answer from this should be:
$$\frac{2^{u+1}}{\log (2)}$$ Is there systematic approach to solving this and if not how do you reason? 
Please notice it s not the substitution I am struggling with. 

$$I=\int\frac{2^{\sin\sqrt{x}}\cos\sqrt{x}}{\sqrt{x}}dx$$
$u=\sin\sqrt{x}\Rightarrow \frac{du}{dx}=\frac1{2\sqrt{x}}\cos\sqrt{x}$ and so: $dx=\frac{2\sqrt{x}}{\cos\sqrt{x}}du$ which converts our integral into:
$$I=2\int2^{u}du$$
now to integrate this notice that:
$$2^{u}=e^{\ln(2^{u})}=e^{u\ln(2)}$$
now if we make the substitution: $v=u\ln(2)\Rightarrow du=\frac{dv}{\ln(2)}$ and so:
$$I=\frac{2}{\ln 2}\int e^vdv$$
now this should be easy, just remember the constant of integration and then back-substitute :)
