# Math Help - Integration Problems

1. ## Integration Problems

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.

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

$A=-1 \ \ \
B=1 \ \ \
C=2$

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

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

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.

2. 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...

3. 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.

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

$A=-1 \ \ \
B=1 \ \ \
C=2$

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

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

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.

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