# Integration of quotient

1. Jan 11, 2012

### sharks

The problem statement, all variables and given/known data
$$\int \frac{r^3-r^5}{\sqrt{1-r^2}}\,.dr$$

The attempt at a solution

The presence of $\sqrt{1-r^2}$ suggests that i use the substitution r=sinθ

The integrand becomes: $$\frac{\sin^3\theta-\sin^5\theta}{\cos\theta}$$
$$\frac{dr}{d\theta}=\cos\theta$$
$$\int \frac{r^3-r^5}{\sqrt{1-r^2}}\,.dr=\int \sin^3\theta-\sin^5\theta\,.d\theta=\int\sin^3\theta\,.d\theta-\int\sin^5\theta\,.d\theta$$
Using the substitution u=cosθ
$$\int\sin^3\theta\,.d\theta=\frac{\cos^3\theta}{3}-\cos\theta$$
$$\int\sin^5\theta\,.d\theta=-\cos \theta+\frac{2\cos^3 \theta}{3}+\frac{\cos^5 \theta}{5}$$
$$\int\sin^3\theta\,.d\theta-\int\sin^5\theta\,.d\theta=-\frac{\cos^3 \theta}{3}-\frac{\cos^5 \theta}{5}$$

Last edited: Jan 11, 2012
2. Jan 11, 2012

### Syrus

Consider factoring the middle expression in the bottom line of your above post and then using a trigonometric identity on the resulting expression. Finally, return back from expressing your problem in terms of theta into an expression involving r. The final integral should be elementary.

3. Jan 11, 2012

### Syrus

You edited your post. My post above regards factoring the integrand sin^3 - sin^5. From there it should be easy.

4. Jan 11, 2012

### sharks

Hi Syrus

Yes, i'm still trying to work it out, step by step, slowly.

5. Jan 11, 2012

### Syrus

hey sharks. i understand. But i think you should consider my post and follow where my hints lead you. From your edited post, looks like you may be going to unneccesary lengths to solve this one!

6. Jan 11, 2012

### sharks

$$\int \sin^3\theta-\sin^5\theta\,.d\theta=\int\sin^3 \theta\cos^2 \theta\,.d\theta$$
Looks like i'll have to use substitution again to solve this.
Let u=cosθ
This gives:
$$\frac{\cos^5 \theta}{5}-\frac{\cos^3 \theta}{3}$$
Then, i have to replace r=sinθ into this integration result... It looks complicated again. I'm not sure i'm doing things right.

Last edited: Jan 11, 2012
7. Jan 11, 2012

### Syrus

My apologies sharks, when i attempted the solution I was working too quickly and overlooked the fact that cosθdθ = dr and not dr = cos^2 dθ, as I had.

8. Jan 11, 2012

### sharks

It's OK, Syrus. I'll try to figure it out.

From: $$\int\sin^3\theta\,.d\theta-\int\sin^5\theta\,.d\theta=-\frac{\cos^3 \theta}{3}-\frac{\cos^5 \theta}{5}$$
Reverting the previous substitution, r=sinθ, i get the final answer:
$$\frac{-5(1-r^2)^{\frac{3}{2}}-3(1-r^2)^{\frac{5}{2}}}{15}$$
This is the correct answer (or maybe they are equivalent, but i doubt it):
$$\frac{-1}{15}(1-r^2)^{\frac{3}{2}}(3r^2 +2)$$

Last edited: Jan 11, 2012
9. Jan 11, 2012

### Curious3141

It's easier to simplify algebraically, then integrate by parts (no trig. sub.)

Start by observing that:
$$\int \frac{r^3-r^5}{\sqrt{1-r^2}}dr = \int \frac{r^3(1-r^2)}{\sqrt{1-r^2}}dr = \int {r^3}{\sqrt{1-r^2}}dr = \int (-\frac{1}{2}r^2)(-2r){\sqrt{1-r^2}}dr$$

Now put $u = (-\frac{1}{2}r^2)$ and $dv = (-2r){\sqrt{1-r^2}}\,.dr$ into the integration by parts formula $\int udv + \int vdu = uv$ and everything should become clearer.

Last edited: Jan 11, 2012
10. Jan 11, 2012

### Curious3141

Not equivalent. Definitely a sign error and at least one coefficient error there. Please try and use the algebraic method, it's much quicker and neater, I assure you.

11. Jan 11, 2012

### sharks

Hi Curious3141

Your method isn't really orthodox but i guess it's better than nothing. So, here are my calculations:
$$\int udv = uv - \int vdu$$
$$uv=\frac{-r^2(1-r^2)^{\frac{3}{2}}}{3}$$
$$\int v.du=\int\frac{-2r}{3}(1-r^2)^{\frac{3}{2}}=\frac{2(1-r^2)^{\frac{5}{2}}}{15}$$

Last edited: Jan 11, 2012
12. Jan 11, 2012

### Curious3141

"Orthodox"?! What's that?:rofl: I would say working this out with a direct algebraic method would be considered more "orthodox" than trying complicated trig subs.

You've worked out $uv$ and $\int vdu$ correctly (except you forgot the dr element in the latter integral).

But you forgot to integrate v with respect to u to evaluate $\int vdu$!

I see you've edited your earlier post. I'm assuming it's a work in progress, so I'll let you finish. Remember to do the full algebraic simplification (gather together common factors) of the final form to make it look like the form you quoted (presumably from the online Wolfram integrator).

13. Jan 11, 2012

### sharks

$$\int udv = \frac{-r^2(1-r^2)^{\frac{3}{2}}}{3}-\frac{2(1-r^2)^{\frac{5}{2}}}{15}=(1-r^2)^{\frac{3}{2}}.-\frac{3r^2+2}{15}$$
Finally, the correct answer. I suppose that the standard recognition methods don't apply in this case, as the algebraic manipulation isn't so obvious, at least to me.

14. Jan 11, 2012

### Curious3141

[/PLAIN] [Broken]

Yes, *I* know what "orthodox" means, thank you , I'm asking what *you* mean when you say my method isn't "orthodox"?

The algebraic method is perfectly acceptable and more direct, and, ideally, you should've recognised the pattern (which is fairly obvious, to be honest). You'll probably be able to do it with more practice.

Last edited by a moderator: May 5, 2017
15. Jan 11, 2012

### sharks

In my notes, the integration examples are more directly solved, by applying the various integration formulas rather than algebraic manipulation (which seems a bit risky to me as it's like trial and error, especially in tests and exams where time is very limited). After reviewing your method, it is indeed easier. I'll need to practice more and be more confident. Thank you for your help.

16. Jan 11, 2012

### Curious3141

No worries, glad to be of help.

17. Jan 12, 2012

### tiny-tim

hi everyone!

substituting u = 1 - r2 is easier

18. Jan 12, 2012

### Curious3141

Do you mean $u^2 = 1 - r^2$? Because that one's dead easy.

Thanks for spotting it. Sharks - you should definitely use this sub. instead. Practically a four-step solution.

19. Jan 12, 2012

### tiny-tim

either will do

20. Jan 20, 2012

### sharks

Thanks for the suggestion, tiny-tim. I tested the substitution $u^2 = 1 - r^2$ and got the answer. It is indeed quicker, but since there's no set method for figuring out how to find that lucky substitution... I guess i'll stick to the safe (and more difficult) road.

Last edited: Jan 20, 2012