Integration Problems

**DCC** · January 25th 2009, 10:30 PM

Hi, I'm new and have a few problems that I'm beating my head over. Hopefully you all can help me understand. Thanks in advance!

Number 1:
$\int \frac {x^2} {\sqrt {9-x^2}}dx$

Here's what I did with it, and where I ended up. I'm not really sure if I've messed up or just need to keep going forward. If I am on the right track...I can't figure out where to go next.

$x=3\sin\theta$

$dx=3 \cos \theta d \theta$

$\int \frac {9\sin^2{\theta} 3 \cos{ \theta}d \theta}{ \sqrt{9(1-sin^2{ \theta})}}$

$\int \frac {9\sin^2{\theta} 3 \cos{ \theta}d \theta}{ 3\cos{\theta}}$

$\int {9\sin^2{\theta} d \theta}$

$9(\frac {1}{2} \theta - \frac {1}{4} \sin 2 \theta) +C$

Number 2:

$\int \frac {3x^2 -4x-2}{(x-1)^2(x+2)}dx$

I'm pretty sure that I've made a dumb mistake on this one, probably an obvious one. Something doesn't feel right... Used partial fractions for the first bit.

$3x^2 -4x-2=A(x+2) + B(x-1)(x+2) + C (x-1)^2$

$A=-1 \ \ \<br /> B=1 \ \ \ <br /> C=2$

$\int ( \frac {-1}{(x-1)^2} +\frac {1}{(x-1)} +\frac {2}{(x+2)})dx$

$\frac {1}{(x-1)} + \ln(x-1) + 2\ln(x+2) +C$

Number 3:

$\int \frac {x}{\sqrt {x} -1}dx$

Ok...I'm pretty sure this one starts with a substitution, but I'm totally lost with it. Help?

Again, thanks for looking over all of this, I hope it all makes sense.

**Rincewind** · January 26th 2009, 03:27 AM

Question 1

You have gone well to that point. Now what you need to do is substitute back to get the answer in terms of the variable of the question. IE $x$ .

So you have $\sin\theta = \frac{x}{3}$ and $\sin 2\theta = 2\sin\theta\cos\theta$ and $\cos\theta=\sqrt{1-\sin^2\theta}$

Follow these substitutions you should get to right answer. Note that the final answer will have an $\arcsin\frac{x}{3}$ term. There is nothing you can do about that.

Question 2

You answer looks right to me. The only thing you might want to do is collect the log terms together into

$= \frac{1}{x-1} + \ln\left[(x-1)(x+2)^2\right] +C.$

Question 3

Here you should use the substitution $u=\sqrt{x}$ This gives

$\int\frac{x}{\sqrt{x}-1}dx = 2\int\frac{u^3}{u-1}du,$

Then using partial fractions you can express the integrand as

$= 2\int\left(u^2+u+1+\frac{1}{u-1}\right)du.$

You then should be able to integrate this easily and then don't forget to substitute back $u=\sqrt{x}$ .

Hope this helps...

**mr fantastic** · January 26th 2009, 03:32 AM

Originally Posted by DCC

Hi, I'm new and have a few problems that I'm beating my head over. Hopefully you all can help me understand. Thanks in advance!

Number 1:
$\int \frac {x^2} {\sqrt {9-x^2}}dx$

Here's what I did with it, and where I ended up. I'm not really sure if I've messed up or just need to keep going forward. If I am on the right track...I can't figure out where to go next.

$x=3\sin\theta$

$dx=3 \cos \theta d \theta$

$\int \frac {9\sin^2{\theta} 3 \cos{ \theta}d \theta}{ \sqrt{9(1-sin^2{ \theta})}}$

$\int \frac {9\sin^2{\theta} 3 \cos{ \theta}d \theta}{ 3\cos{\theta}}$

$\int {9\sin^2{\theta} d \theta}$

$9(\frac {1}{2} \theta - \frac {1}{4} \sin 2 \theta) +C$
[snip]

You're nearly there. Now you have to back-substitute for x.

Note that $\sin (2 \theta) = 2 \sin \theta \cos \theta = \frac{2x}{3} \, \frac{\sqrt{9- x^2}}{3}$ .

Originally Posted by DCC

[snip]
Number 2:

$\int \frac {3x^2 -4x-2}{(x-1)^2(x+2)}dx$

I'm pretty sure that I've made a dumb mistake on this one, probably an obvious one. Something doesn't feel right... Used partial fractions for the first bit.

$3x^2 -4x-2=A(x+2) + B(x-1)(x+2) + C (x-1)^2$

$A=-1 \ \ \<br /> B=1 \ \ \ <br /> C=2$

$\int ( \frac {-1}{(x-1)^2} +\frac {1}{(x-1)} +\frac {2}{(x+2)})dx$

$\frac {1}{(x-1)} + \ln(x-1) + 2\ln(x+2) +C$
[snip]

Your answer is correct. Nicely played.

Originally Posted by DCC

[snip]
Number 3:

$\int \frac {x}{\sqrt {x} -1}dx$

Ok...I'm pretty sure this one starts with a substitution, but I'm totally lost with it. Help?

Again, thanks for looking over all of this, I hope it all makes sense.

Many approaches are possible. One way is to make the substitution $u = \sqrt{x} - 1$ . Then the integral becomes $\int \frac{2 (u + 1)^3}{u} \, du$ . Now expand the numerator and integrate term-by-term. Then back-substitute for x.

**DCC** · January 26th 2009, 08:06 AM

Thanks a bunch, to both of you. That really helped out ^_^

Thread: Integration Problems

LinkBack

Thread Tools

Display

Integration Problems

Similar Math Help Forum Discussions

Integration help please these are three small even basic integration problems

Problems with integration word problems

integration problems

Integration problems

Problems with some integration

Search Tags