# $\int \frac{2^{\sin \left(\sqrt{x}\right)} \cos \left(\sqrt{x}\right)}{\sqrt{x}} \, dx$

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.

• $$2^x = e^{x \log{2}}$$ – Ron Gordon Nov 6 '14 at 22:25
• I see why we divide by Log, but I do not see why u+1 is coming from. – ALEXANDER Nov 6 '14 at 22:28
• $2\cdot 2^u=2^{u+1}$. – vadim123 Nov 6 '14 at 22:28

## 2 Answers

You're almost there actually. $$2 \int 2^u \ \text{d}u = \int 2^{u+1} \ \text{d}u = \frac{2^{u+1}}{\log 2}$$ where we use the fact that $$\int a^x \ \text{d}x = \frac{a^x}{\log a}$$

$$I=\int\frac{2^{\sin\sqrt{x}}\cos\sqrt{x}}{\sqrt{x}}dx$$ $$u=\sin\sqrt{x}\Rightarrow \frac{du}{dx}=\frac1{2\sqrt{x}}\cos\sqrt{x}$$ 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 :)