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.

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

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 :)


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