# Integration of Rational Functions by partial fractions

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• Feb 21st 2008, 03:02 PM
FalconPUNCH!
Integration of Rational Functions by partial fractions
I understand how to do it but I get stuck really easily. Like this for example

$\displaystyle \int_{0}^{2\sqrt{3}} \frac{x^3}{\sqrt{16-x^2}}$

Yeah it's probably an easy one but I'm not that great with limit of integration. so I get

$\displaystyle x = 4sin(o)$
$\displaystyle dx = 4cos(o)$

So here's where I am stuck. I don't know what to do now. I plug those in but I don't know what my limit of integration is supposed to be. This stuff is killing me. I'm trying really hard to understand. Please someone help me.
• Feb 21st 2008, 03:08 PM
Krizalid
Quote:

Originally Posted by FalconPUNCH!
I understand how to do it but I get stuck really easily. Like this for example

$\displaystyle \int_{0}^{2\sqrt{3}} \frac{x^3}{\sqrt{16-x^2}}$

Much better to substitute $\displaystyle u^2=16-x^2.$

Quote:

Originally Posted by FalconPUNCH!
$\displaystyle x = 4sin(o)$

For $\displaystyle x=0\implies 0=4\sin\theta.$ For $\displaystyle x=2\sqrt3\implies 2\sqrt3=4\sin\theta.$ Compute $\displaystyle \theta$ in both cases.
• Feb 21st 2008, 03:13 PM
FalconPUNCH!
Quote:

Originally Posted by Krizalid
Much better to substitute $\displaystyle u^2=16-x^2.$

For $\displaystyle x=0\implies 0=4\sin\theta.$ For $\displaystyle x=2\sqrt3\implies 2\sqrt3=4\sin\theta.$ Compute $\displaystyle \theta$ in both cases.

So in order to solve for what the new limits of integration I have to set what I'm substituting like in this example sin(o) = to the old limits of integration and solve. Thanks Krizalid.
• Feb 21st 2008, 03:17 PM
galactus
If you must use trig sub, you have the correct subs.

Did you sub them in and simplify?. Also, you have to change your limits of integration because you're now in terms of theta.

$\displaystyle x=4sin({\theta}), \;\ dx=4cos({\theta})d{\theta}$

Therefore, solve for theta using our old limits.

$\displaystyle 2\sqrt{3}=4sin({\theta}); \;\ {\theta}=\frac{\pi}{3}$

$\displaystyle \int\frac{64sin^{3}({\theta})}{\sqrt{16-(4sin({\theta}))^{2}}}\cdot{4cos({\theta})}d{\thet a}$

$\displaystyle =64\int_{0}^{\frac{\pi}{3}}sin^{3}({\theta})d{\the ta}$

Can you finish?.
• Feb 21st 2008, 03:38 PM
Soroban
Hello, FalconPUNCH!

Quote:

$\displaystyle \int_{0}^{2\sqrt{3}} \frac{x^3}{\sqrt{16-x^2}}\,dx$
You said: .$\displaystyle x \:= \:4\sin\theta\quad\Rightarrow\quad dx \:= \:4\cos\theta\,d\theta$

. . Substitute: .$\displaystyle \int\frac{64\sin^3\!\theta}{4\cos\theta}(4\cos\the ta\,d\theta) \;=\;64\!\!\int\!\sin^3\!\theta\,d\theta$

To change the limits, recall your substitution:
. . . $\displaystyle x \:=\:4\sin\theta \quad\Rightarrow\quad \sin\theta \:=\:\frac{x}{4}$

When $\displaystyle x = 0$, we have: .$\displaystyle \sin\theta \:=\:\frac{0}{4} \:=\:0\quad\Rightarrow\quad\boxed{\theta \:=\:0}$

When $\displaystyle x = 2\sqrt{3}$, we have: .$\displaystyle \sin\theta \:=\:\frac{2\sqrt{3}}{4} \:=\:\frac{\sqrt{3}}{2}\quad\Rightarrow\quad\boxed {\theta \:=\:\frac{\pi}{3}}$

. . And there are the new limits . . .

The integral becomes: . $\displaystyle 64\!\int^{\frac{\pi}{3}}_0\!\!\sin^3\!\theta\,d\th eta$

Got it?

• Feb 21st 2008, 03:39 PM
FalconPUNCH!
Quote:

Originally Posted by galactus
If you must use trig sub, you have the correct subs.

Did you sub them in and simplify?. Also, you have to change your limits of integration because you're now in terms of theta.

$\displaystyle x=4sin({\theta}), \;\ dx=4cos({\theta})d{\theta}$

Therefore, solve for theta using our old limits.

$\displaystyle 2\sqrt{3}=4sin({\theta}); \;\ {\theta}=\frac{\pi}{3}$

$\displaystyle \int\frac{64sin^{3}({\theta})}{\sqrt{16-(4sin({\theta}))^{2}}}\cdot{4cos({\theta})}d{\thet a}$

$\displaystyle =64\int_{0}^{\frac{\pi}{3}}sin^{3}({\theta})d{\the ta}$

Can you finish?.

I got up to the last part the only thing I was having trouble with was the limits of integration.
• Feb 21st 2008, 03:40 PM
FalconPUNCH!
Quote:

Originally Posted by FalconPUNCH!
I got up to the last part the only thing I was having trouble with was the limits of integration.

Quote:

Originally Posted by Soroban
Hello, FalconPUNCH!

You said: .$\displaystyle x \:= \:4\sin\theta\quad\Rightarrow\quad dx \:= \:4\cos\theta\,d\theta$

. . Substitute: .$\displaystyle \int\frac{64\sin^3\!\theta}{4\cos\theta}(4\cos\the ta\,d\theta) \;=\;64\!\!\int\!\sin^3\!\theta\,d\theta$

To change the limits, recall your substitution:
. . . $\displaystyle x \:=\:4\sin\theta \quad\Rightarrow\quad \sin\theta \:=\:\frac{x}{4}$

When $\displaystyle x = 0$, we have: .$\displaystyle \sin\theta \:=\:\frac{0}{4} \:=\:0\quad\Rightarrow\quad\boxed{\theta \:=\:0}$

When $\displaystyle x = 2\sqrt{3}$, we have: .$\displaystyle \sin\theta \:=\:\frac{2\sqrt{3}}{4} \:=\:\frac{\sqrt{3}}{2}\quad\Rightarrow\quad\boxed {\theta \:=\:\frac{\pi}{3}}$

. . And there are the new limits . . .

The integral becomes: . $\displaystyle 64\!\int^{\frac{\pi}{3}}_0\!\!\sin^3\!\theta\,d\th eta$

Got it?

yeah it's pretty clear now. Thanks guys (Yes)
• Feb 21st 2008, 03:46 PM
FalconPUNCH!
So the answer I get for this is $\displaystyle \frac{24\sqrt{3} - 128}{3}$
• Feb 21st 2008, 03:52 PM
galactus
No, I am sorry to say, that is incorrect.
• Feb 21st 2008, 04:13 PM
xifentoozlerix
You really shouldn't let the limits of integration stop you from solving the problem. If you are having trouble finding the new limits, simply forget about them until you have to evaluate your $\displaystyle u$ function. Then just "un-substitute" your original function back in place of $\displaystyle u$ and use the original limits.
• Feb 21st 2008, 04:16 PM
FalconPUNCH!
I'm really stressing out over this. I can't even solve a single integral because of this :( I'm trying to do the ones ahead but they're even more annoying. I might do what you said, xifentoozlerix but I get confused when I reach the end.

Here's what I'm doing:

$\displaystyle 64\int_{0}^{\frac{pi}{3}} sin^{2}0 sin(0) d0$

$\displaystyle u = cos(o), du = -sin(o)do$

$\displaystyle 64\int_{0}^{\frac{pi}{3}} (1-cos^{2}x) sin(0) d0$

$\displaystyle 64\int_{0}^{\frac{pi}{3}} (1 - u^2) -du$

$\displaystyle -64( u - \frac{1}{3}u^{3})$

$\displaystyle -64( cos(o) - \frac{1}{3}cos(o)^{3})$

I then plugged in pi/3 and 0 and the answer I posted before is what I got. I'm not sure where I messed up.
• Feb 21st 2008, 04:30 PM
galactus
Perhaps try the advice that was given about waiting 'til the end.

You can always use the reduction formula for sine. Save you all the grunt work.

$\displaystyle \int{sin^{n}(u)}du=\frac{-1}{n}sin^{n-1}(u)cos(u)+\frac{n-1}{n}\int{sin^{n-2}}du$
• Feb 21st 2008, 04:42 PM
FalconPUNCH!
Okay I haven't used to reduction formula a lot so I'm not sure if I did it correctly

$\displaystyle 64( \frac{-1}{3}sin^{2}(u)cos(u) + \frac{2}{3} \int sin du)$

$\displaystyle 64( \frac{-1}{3}sin^{2}(\frac{pi}{3})cos(\frac{pi}{3}) - sin^{2}(0)cos(0) + \frac{2}{3} ( \frac{1}{2} - \frac{pi}{2} )$

$\displaystyle 64(( (\frac{-1}{3}) (\frac{3}{4})(\frac{1}{2}) - \frac{pi}{2}) + \frac{2}{3} (\frac{1}{2} - \frac{pi}{2})$

I then simplify it to

$\displaystyle 64( \frac{-8 - 23pi}{24})$

not sure if it's right...(Worried)
• Feb 21st 2008, 04:50 PM
mr fantastic
Quote:

Originally Posted by FalconPUNCH!
Okay I haven't used to reduction formula a lot so I'm not sure if I did it correctly

$\displaystyle 64( \frac{-1}{3}sin^{2}(u)cos(u) + \frac{2}{3} \int sin du)$

$\displaystyle 64( \frac{-1}{3}sin^{2}(\frac{pi}{3})cos(\frac{pi}{3}) - sin^{2}(0)cos(0) + \frac{2}{3} ( \frac{1}{2} - \frac{pi}{2} )$

$\displaystyle 64(( (\frac{-1}{3}) (\frac{3}{4})(\frac{1}{2}) - \frac{pi}{2}) + \frac{2}{3} (\frac{1}{2} - \frac{pi}{2})$

I then simplify it to

$\displaystyle 64( \frac{-8 - 23pi}{24})$

not sure if it's right...(Worried)

$\displaystyle I = 64 \int_{0}^{\pi/3} \sin^3 \theta \, d\theta = 64 \int_{0}^{\pi/3} \sin^2 \theta \, \sin \theta \, d\theta = 64 \int_{0}^{\pi/3} (1 - \cos^2 \theta) \sin \theta \, d\theta$.

Substitute $\displaystyle u = \cos \theta$. Then:

$\displaystyle I = -64 \int_{1}^{1/2} 1 - u^2 \, du = ......$

I get $\displaystyle I = \frac{40}{3}$.
• Feb 21st 2008, 04:56 PM
FalconPUNCH!
Quote:

Originally Posted by mr fantastic
$\displaystyle I = 64 \int_{0}^{\pi/3} \sin^3 \theta \, d\theta = 64 \int_{0}^{\pi/3} \sin^2 \theta \, \sin \theta \, d\theta = 64 \int_{0}^{\pi/3} (1 - \cos^2 \theta) \sin \theta \, d\theta$.

Substitute $\displaystyle u = \cos \theta$. Then:

$\displaystyle I = -64 \int_{1}^{1/2} 1 - u^2 \, du = ......$

I get $\displaystyle \frac{40}{3}$

hmmm...seem to be getting the limits of integration wrong which is happening in my homework a lot.

for this one $\displaystyle \int_{1}^{2} \frac{\sqrt{x^{2} - 1}}{x}dx$ is it best to use u substitution from the start?
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