1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ $$2^x = e^{x \log{2}}$$ $\endgroup$ – 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
0
$\begingroup$

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 $\int a^x \ \text{d}x = \frac{a^x}{\log a}$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.