integrals by substitution

• Nov 11th 2008, 08:05 PM
vinson24
integrals by substitution
$\displaystyle \int (1/ \sqrt{x})\ sin\sqrt{x}\ dx\ u=\sqrt{x}\$

the answer is $\displaystyle -2cos(\sqrt{x})+C$ but cant seem to get there
• Nov 11th 2008, 08:18 PM
Chris L T521
Quote:

Originally Posted by vinson24
$\displaystyle \int (1/ \sqrt{x})\ sin\sqrt{x}\ dx\ u=\sqrt{x}\$

the answer is $\displaystyle -2cos(\sqrt{x})+C$ but cant seem to get there

Since they let $\displaystyle u=\sqrt{x}$, this implies that $\displaystyle \,du=\frac{\,dx}{2\sqrt{x}}$

Now, we see that $\displaystyle \int\frac{1}{\sqrt{x}}\sin\sqrt{x}\,dx\implies 2\int\frac{1}{2\sqrt{x}}\sin\sqrt{x}\,dx\implies 2\int\sin u\,du$

Can you take it from here?

--Chris
• Nov 11th 2008, 08:22 PM
tester85
let u = sqrt x
du = dx / 2sqrt x

substitute to the equation you will get, the following

2 integrate sin u du

I think you should be able to take it from here.
• Nov 11th 2008, 08:22 PM
vinson24
thanks see it now
• Nov 11th 2008, 08:25 PM
tester85
You need to put 2 outside of the integration to take care of the 1/2 that is inside the equation. If you do not put the 2 outside. The original equation would have been changed. The 2 is to prevent that from happening Hope it helps.