Hello, I just have two problems. I have used integration by parts, but I get stuck. The work you show will be greatly appreciated.
$\displaystyle \int sin \sqrt x~dx$
$\displaystyle \int e^{6x} sin(e^{2x})~dx$
Thanks!
Do you have to use parts?. I think a sub is easier.
$\displaystyle \int e^{6x}sin(e^{2x})dx$
Now, let $\displaystyle u=e^{2x}, \;\ \frac{1}{2}du=e^{2x}dx$
Make the subs and we get:
$\displaystyle \frac{1}{2}\int u^{2}sin(u)du$
Now, let's use tabular integration:
This works well when we have a something such as e or sin that repeats as we take derivatives. It's based off parts, though. So, I reckon we are using parts.
$\displaystyle \text{\underline{sign}}, \;\ \;\ \text{\underline{u and its derivatives}} \;\ \text{\underline{v' and its antiderivatives}}$
$\displaystyle + \;\ \;\ \;\ \;\ \rightarrow \;\ \;\ \;\ \;\ u^{2} \;\ \;\ \;\ \;\ \searrow \;\ \;\ \;\ \;\ \;\ \;\ \;\ v'=sin(u)$
$\displaystyle - \;\ \;\ \;\ \;\ \;\ \rightarrow \;\ \;\ \;\ \;\ 2u, \;\ \;\ \;\ \;\ \searrow \;\ \;\ \;\ \;\ \;\ -cos(u)$
$\displaystyle + \;\ \;\ \;\ \;\ \;\ \rightarrow \;\ \;\ \;\ \;\ 2 \;\ \;\ \;\ \;\ \searrow \;\ \;\ \;\ \;\ \;\ \;\ -sin(u)$
$\displaystyle - \;\ \;\ \;\ \;\ \;\ \;\ \rightarrow \;\ \;\ \;\ 0, \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ cos(u)$
Add up the signed products diagonally and alternate signs.
$\displaystyle -u^{2}cos(u)+2usin(u)+2cos(u)+C$
Resub:
$\displaystyle \boxed{\frac{1}{2}\left[-(e^{2x})^{2}cos(e^{2x})+2e^{2x}sin(e^{2x})+2cos(e^ {2x})\right]}$
Someone will come by saying, "this way is easier". Oh well. I wanted to show you tabular integration in the event you have not seen it. The more one knows the better.
incidentally, a substitution works nicely with the first one also
$\displaystyle \int \sin \sqrt{x}~dx$
Let $\displaystyle u = \sqrt{x}$, that is, $\displaystyle u^2 = x$
$\displaystyle \Rightarrow 2u~du = dx$
So our integral becomes
$\displaystyle 2 \int u \sin u~du$
which is easy to do by parts. you can also use the tabular method as galactus showed you, but it would be overkill here, i think
$\displaystyle \int {\sin \sqrt x dx = \int {\sqrt x \frac{{\sin \sqrt x }}{{\sqrt x }}} } dx = - 2\int {\sqrt x d\left( {\cos \sqrt x } \right)} =$
$\displaystyle = - 2\sqrt x \cos \sqrt x + 2\int {\cos \sqrt x d\left( {\sqrt x } \right)} = - 2\sqrt x \cos \sqrt x + 2\sin \sqrt x + C.$