1. Integration

$\displaystyle \int \sqrt{4-x^2}dx$

I tried writing this as a 1/2 power and making a u substitution of the bracket. But then what do you do with the dx? I mean you have an integration of u wrt x? I forget how to do this?

$\displaystyle \int \frac{1}{x^2-2x-15}dx$

$\displaystyle \int \frac{1}{9x^2+6x+17}dx$

These two both quadratics but I forget when to use parfrac or when to try completing the square. I can't decide how to start these? Please somebody help, thanks.

2. Originally Posted by TeaWithoutMussolini
$\displaystyle \int \sqrt{4-x^2}dx$

I tried writing this as a 1/2 power and making a u substitution of the bracket. But then what do you do with the dx? I mean you have an integration of u wrt x? I forget how to do this?
Have you tried any trig substitutions?

3. Look at this tool.
Be sure to click Show Steps

4. Originally Posted by dwsmith
Have you tried any trig substitutions?
Can't say I understand? For what and when? (I haven't, no).

Where on earth does this idea come from? This wasn't taught at any stage on the course. How would you know what to use?

5. Originally Posted by TeaWithoutMussolini
Can't say I understand? For what and when? (I haven't, no).
$\displaystyle \sin^2+\cos^2=1\Rightarrow 1-\sin^2=\cos^2$

6. See here.

7. $\displaystyle \int\sqrt{4-x^2} \dx$

Let $\displaystyle x=2\sin\theta, \ dx=2\cos\theta \ d\theta$

$\displaystyle \int\sqrt{4-(2\sin\theta)^2}(2\cos\theta \ d\theta)$

8. This is above and beyond the level expected on this course... there must be a simpler way?

How would you KNOW when to use what substitution, please?

I arrive at $\displaystyle sin(2arcsin\frac{x}{2}) + 2arcsin\frac{x}{2}.$ What are they (Wolfram) doing with the first term?

My main concern, however, is how you know to use the substitution $\displaystyle x=2sinu.$

Also, how about the other q's?

9. Originally Posted by TeaWithoutMussolini
This is above and beyond the level expected on this course... there must be a simpler way? How would you KNOW when to use what substitution, please?
That is actually a simple problem.
How can it be "beyond the level expected on this course"?
If it is, you should not have been asked to do the question.

10. Because its in the paper but it has not been taught?

11. I made the substitution for this reason (see below)

$\displaystyle \int\sqrt{4-(2\sin\theta)^2}(2\cos\theta \ d\theta)=2\int\cos\theta\sqrt{4(1-\sin^2\theta)} \ d\theta$

$\displaystyle 4\int\cos\theta\sqrt{1-\sin^2\theta} \ d\theta$

I previous mentioned $\displaystyle \sin^2+\cos^2=1$

$\displaystyle 4\int\cos\theta\sqrt{\cos^2\theta}\text{why?} \ d\theta$

$\displaystyle 4\int\cos^2\theta \ d\theta$

$\displaystyle \cos^2=\frac{1+\cos(2\theta)}{2}$

The rest should be easy enough to do.

12. Originally Posted by TeaWithoutMussolini
Because its in the paper but it has not been taught?
Then why do it?
Have you looked carefully at reply #3?

13. Plato, thanks for the link, yes, I can follow every step. The only things I don't get are:

Knowing that you have to make this substitution (I understand with hindsight, but how to know beforehand)
What they are doing with the sin(2arcsin(x/2)) to put it back in to a root?

Why do it? Because its in the paper...

14. For the second question, I got $\displaystyle \frac{1}{8}(log\frac{x-5}{x+3}).$ Stephen makes it slightly different with $\displaystyle 5-x$ instead of $\displaystyle x-5$ and I'm wondering if this matters at all? (Since technically its modulus of the log its the same thing). It also uses a completely different method (I used parfrac) but I think I have the same result?