## integration using arcsin

1. The problem statement, all variables and given/known data

3. The attempt at a solution

I'm having a tough time understanding this step. I understand that

the integral of 1/sqrt(a^2-x^2) = arcsin(u/a) + C

But sqrt(4-x^2) does not have a 1 in the numerator. It looks like they're using u substitution, if so, then why is it + 4 arcsin rather than times 4arcsin. I also don't understand where the 4 comes from.
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 Quote by robertjford80 1. The problem statement, all variables and given/known data 3. The attempt at a solution I'm having a tough time understanding this step. I understand that the integral of 1/sqrt(a^2-x^2) = arcsin(u/a) + C But sqrt(4-x^2) does not have a 1 in the numerator. It looks like they're using u substitution, if so, then why is it + 4 arcsin rather than times 4arcsin. I also don't understand where the 4 comes from.
I'm guessing it's just a typical trig sub. Set ##x = 2 \sin \theta##, use a trigonometric identity, integrate and then solve for ##\theta## in terms of x.

If you haven't seen this technique before, check out this link.
 Mentor What's the derivative of$\displaystyle \frac{3}{2}\left(x\sqrt{4-x^2\,} -4\,\sin^{-1}\left(\frac{x}{2} \right) \right)\ ?$

## integration using arcsin

 Quote by SammyS What's the derivative of$\displaystyle \frac{3}{2}\left(x\sqrt{4-x^2\,} =4\,\sin^{-1}\left(\frac{x}{2} \right) \right)\ ?$
If I knew I wouldn't have posted the thread

 Quote by spamiam I'm guessing it's just a typical trig sub. Set ##x = 2 \sin \theta##, use a trigonometric identity, integrate and then solve for ##\theta## in terms of x. If you haven't seen this technique before, check out this link.
well, if If the integrand contains x^2 − a^2, let

$$x = a \sec \theta\,$$

and use the identity

$$\sec^2 \theta-1 = \tan^2 \theta.\,$$

Then why do they use arcsin instead of secant

Mentor
 Quote by robertjford80 If I knew I wouldn't have posted the thread
How do you expect to be able to integrate, if you don't know basic differentiation ?

According to your OP, the derivative of $\displaystyle \sin^{-1}\left(\frac{x}{2} \right)$ is $\displaystyle \frac{1}{\sqrt{4-x^2}}\ .$

The rest is pretty basic.

 Quote by robertjford80 well, if If the integrand contains x^2 − a^2, let $$x = a \sec \theta\,$$ and use the identity $$\sec^2 \theta-1 = \tan^2 \theta.\,$$ Then why do they use arcsin instead of secant
There's a very important difference in whether you have x2 - a2 or a2 - x2 so that you use x = asecθ or x = asinθ, respectively.

 Quote by Bohrok There's a very important difference in whether you have x2 - a2 or a2 - x2 so that you use x = asecθ or x = asinθ, respectively.
well, is a^2-x^2 which results in asin. However, that integral only works if there is a one in the numerator and clearly there is not, so I'm still wondering where 4asin came from.

Sammy, the rest is not basic.

Mentor
 Quote by robertjford80 ... Sammy, the rest is not basic.
How is $\displaystyle \frac{d}{dx}\left(x\sqrt{4-x^2\,}\right)$ not basic ?
 It's basic alright, but it doesn't even appear on the left side of the equation. Why is the text using that derivative?
 ∫√(4 - x2) dx x = 2sinθ → x2 = 22sin2θ dx = 2cosθ dθ $$\int\sqrt{4-4\sin^2\theta}\text{ }2\cos\theta\text{ }d\theta = \int\sqrt{4(1-sin^2\theta)}\text{ }2\cos\theta\text{ }d\theta = \int\sqrt{4\cos^2\theta}\text{ }2\cos\theta\text{ }d\theta$$$$= \int 2\cos\theta\text{ }2\cos\theta\text{ }d\theta = 4\int \cos^2\theta\text{ }d\theta = 4\int\frac{1}{2}(\cos2\theta + 1)\text{ }d\theta$$The cos2θ part will turn into x√(4 - x2) and the 1 part will turn into sin-1(x/2) after substituting back for x and simplifying.

 Quote by spamiam I'm guessing it's just a typical trig sub. Set ##x = 2 \sin \theta##, use a trigonometric identity, integrate and then solve for ##\theta## in terms of x. If you haven't seen this technique before, check out this link.
 Quote by robertjford80 well, if If the integrand contains x^2 − a^2, let $$x = a \sec \theta\,$$ and use the identity $$\sec^2 \theta-1 = \tan^2 \theta.\,$$ Then why do they use arcsin instead of secant
robert, the method I outlined in my original response (which I've quoted above) is still correct. You had the right idea, but as Bohrok pointed out, the substitution should be ##x = 2 \sin \theta## (as I said originally). You can then use the identity ##1 - \sin^2 \theta = \cos^2 \theta##. What does the integral become with this substitution?

Edit: Bohrok seems to have done a lot of the work for you.

Note that he or she did leave out the factor of 3, however.

Mentor
 Quote by SammyS What's the derivative of$\displaystyle \frac{3}{2}\left(x\sqrt{4-x^2\,} -4\,\sin^{-1}\left(\frac{x}{2} \right) \right)\ ?$
The reason I asked this question is that this derivative should be equal to $\displaystyle 3\sqrt{4-x^2}\ .$

The steps involved in taking this derivative might help you understand how one finds the anti-derivative.

 Quote by Bohrok x = 2sinθ
How is this legal?

 The cos2θ part will turn into x√(4 - x2)
By what rule? I don't see how you can get cos to turn into x

 Quote by SammyS The reason I asked this question is that this derivative should be equal to $\displaystyle 3\sqrt{4-x^2}\ .$
The book says they're equal so I already know that.

 The steps involved
That's what I'm trying to figure out.

 Quote by robertjford80 How is this legal? By what rule? I don't see how you can get cos to turn into x
Legal? It's a substitution. If I said "Let ##\theta = \arcsin\left(x/2\right)##," would you still take issue with it? Certainly there is a lot of work you'd have to do to provide a rigorous proof that it works, but I'm guessing you've done substitutions before without taking a real analysis course. Take a look at the link I posted above.

The last step that Bohrok did uses a double-angle formula. See if you can work it out. Once you've integrated (in terms of ##\theta##) substitute back in for x.

 Quote by robertjford80 How is this legal?
We're driven to use x = 2sinθ since the domain of √(4 - x2) is [-2, 2], and the range of 2sinθ is [-2, 2]. Not tired just yet, so I typed up the steps (ignoring +C)

$$x = 2\sin\theta \Longrightarrow \frac{x}{2} = \sin\theta \Longrightarrow\sin^{-1}\left(\frac{x}{2}\right) = \theta$$$$\int\cos2\theta\text{ }d\theta = \frac{1}{2}\sin2\theta = \frac{1}{2}\cdot2\sin\theta\cos\theta = \sin\theta\cos\theta = \frac{x}{2}\cos\left(\sin^{-1}\frac{x}{2}\right) = \frac{x}{2}\sqrt{1 - \left(\frac{x}{2}\right)^2} = \frac{x}{2}\sqrt{1 - \frac{x^2}{4}}$$
There's a trig identity up there where cos(sin-1x) = √(1 - x2)