Integration by Parts

**Mr. Engineer** · January 23rd 2009, 03:37 PM

Hello, I just have two problems. I have used integration by parts, but I get stuck. The work you show will be greatly appreciated.

$\int sin \sqrt x~dx$

$\int e^{6x} sin(e^{2x})~dx$

Thanks!

**Mush** · January 23rd 2009, 03:48 PM

Originally Posted by Mr. Engineer

Hello, I just have two problems. I have used integration by parts, but I get stuck. The work you show will be greatly appreciated.

$\int sin \sqrt x~dx$

$\int e^{6x} sin(e^{2x})~dx$

Thanks!

For the first one, do you mean $\sqrt{x} \sin(x)$ ??

**Mr. Engineer** · January 23rd 2009, 03:50 PM

no, its the way it is, i trued just using u-substitution, but the answer was not what I got, we were supposed to use integration by parts to solve that one

**TKHunny** · January 23rd 2009, 03:57 PM

Originally Posted by Mr. Engineer

Hello, I just have two problems. I have used integration by parts, but I get stuck. The work you show will be greatly appreciated.

$\int sin \sqrt x~dx$

$\int e^{6x} sin(e^{2x})~dx$

Thanks!

Well, what are you waiting for? Do it!

$<br /> \int{sin(\sqrt{x})}\;dx\;=\;x \cdot sin(\sqrt{x})\;-\;\int x \;d\left(sin (\sqrt{x}) \right)\;=\;x \cdot sin(\sqrt{x})\;-\;\int{x \cdot \frac{cos(\sqrt{x})}{2 \cdot \sqrt{x}}}\;dx<br />$

Now what?

**galactus** · January 23rd 2009, 04:17 PM

Do you have to use parts?. I think a sub is easier.

$\int e^{6x}sin(e^{2x})dx$

Now, let $u=e^{2x}, \;\ \frac{1}{2}du=e^{2x}dx$

Make the subs and we get:

$\frac{1}{2}\int u^{2}sin(u)du$

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.

$\text{\underline{sign}}, \;\ \;\ \text{\underline{u and its derivatives}} \;\ \text{\underline{v' and its antiderivatives}}$

$+ \;\ \;\ \;\ \;\ \rightarrow \;\ \;\ \;\ \;\ u^{2} \;\ \;\ \;\ \;\ \searrow \;\ \;\ \;\ \;\ \;\ \;\ \;\ v'=sin(u)$

$- \;\ \;\ \;\ \;\ \;\ \rightarrow \;\ \;\ \;\ \;\ 2u, \;\ \;\ \;\ \;\ \searrow \;\ \;\ \;\ \;\ \;\ -cos(u)$

$+ \;\ \;\ \;\ \;\ \;\ \rightarrow \;\ \;\ \;\ \;\ 2 \;\ \;\ \;\ \;\ \searrow \;\ \;\ \;\ \;\ \;\ \;\ -sin(u)$

$- \;\ \;\ \;\ \;\ \;\ \;\ \rightarrow \;\ \;\ \;\ 0, \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ \;\ cos(u)$

Add up the signed products diagonally and alternate signs.

$-u^{2}cos(u)+2usin(u)+2cos(u)+C$

Resub:

$\boxed{\frac{1}{2}\left[-(e^{2x})^{2}cos(e^{2x})+2e^{2x}sin(e^{2x})+2cos(e^ {2x})\right]}$

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.

**Jhevon** · January 23rd 2009, 04:34 PM

incidentally, a substitution works nicely with the first one also

$\int \sin \sqrt{x}~dx$

Let $u = \sqrt{x}$ , that is, $u^2 = x$

$\Rightarrow 2u~du = dx$

So our integral becomes

$2 \int u \sin u~du$

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

**DeMath** · January 23rd 2009, 04:34 PM

$\int {\sin \sqrt x dx = \int {\sqrt x \frac{{\sin \sqrt x }}{{\sqrt x }}} } dx = - 2\int {\sqrt x d\left( {\cos \sqrt x } \right)} =$

$= - 2\sqrt x \cos \sqrt x + 2\int {\cos \sqrt x d\left( {\sqrt x } \right)} = - 2\sqrt x \cos \sqrt x + 2\sin \sqrt x + C.$

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