What's wrong with using x = acosθ?

[flyingpig]

Homework Statement

In my Calculus class, my professor told us that to handle integrals in the forms of $$\sqrt{a^2 - x^2}$$, we must use $$x = a\sin\theta$$. But my question is why? I tried it with x = acosθ and it works, but it just gives me an "opposite" answer.

Let's say $$\int \sqrt{9 - x^2} dx$$

Using x = 3sinθ, you should get

$$\frac{1}{2}\left ( \sqrt{9-x^2} + 9\arcsin\left(\frac{x}{3} \right ) \right )$$

The Attempt at a Solution

Now, let's try using x = 3cosθ

$$x = 3\cos\theta$$
$$dx = -3\sin\theta d\theta$$

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

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

$$-9\int \sqrt{1 - \cos^2 \theta} \sin\theta d\theta$$


$$-9\int \sin^2\theta d\theta$$


$$-9\int \frac{1- \cos2\theta }{2} d\theta$$

$$-9\int \frac{1}{2}- \frac{\cos2\theta }{2} d\theta$$


$$-9\left [ \frac{\theta}{2}- \frac{\sin2\theta }{4} \right ]$$


$$-9\left [ \frac{\theta}{2}- \frac{2\sin\theta\cos\theta}{4} \right ]$$

Initially we had $$x = 3\cos\theta$$, so draw the triangle and we should get $$\sin\theta = \frac{\sqrt{3^2 - x^2}}{3}$$ and $$\theta = \arccos\left ( \frac{x}{3} \right )$$

Our final answer should be

$$-9\left [ \frac{arccos\left ( \frac{x}{3} \right )}{2}- \frac{x\sqrt{9-x^2}}{18} \right ]$$


Now clearly

$$-9\left [ \frac{arccos\left ( \frac{x}{3} \right )}{2}- \frac{x\sqrt{9-x^2}}{18} \right ] \neq \frac{1}{2}\left ( \sqrt{9-x^2} + 9\arcsin\left(\frac{x}{3} \right ) \right )$$

 

[osnarf]

I confess, I didn't read the whole thing :)

However, you very well may have a right answer. Both of them could be A right answer.

First off, your completely right that acosx should work. after all, cos^2 + sin^2 = 1,
so
1 - cos^2 = sin^2
just as
1 - sin^2 = cos^2 like your teacher told you to do. So your way definitely is valid.

Your forgetting something important. That's an indefinate integral. That means it carries with it an implicit constant of integration. IE:

F1(x) = x
F2(x) = x + 3

derivitives are:
f1(x) = 1
f2(x) = 1

so if you integrate both of these, you get the same thing. Because F1(x) =x and F2(x) = x + 3 are both primitives for the constant function f(x) = 1. So there really is a whole family of right answers to the indefinite integral of the function f(x) = 1.

Therefore, your answer could just be off by a constant. You should play around with some inverse trig identities and see if you add a constant to your answer, if it is equal to the answer you said is the "right" answer. I didn't check your work but it may be right.

 

[Mark44]

I never bothered to memorize which trig substitutions go with a² + x², a² - x², or x² + a². Instead, I draw a right triangle and label the sides appropriately.

For an integral with sqrt(a² - x²), the hypotenuse is a, and you can label either the base or altitude as x. The other side will be sqrt(a² - x²). If you label the base as x (and the acute angle as theta), then cos(theta) = x/a. From that you get all your other relationships, including dx and one that involves the square root.

If you label the altitude as x, then you have sin(theta) = x/a. It doesn't matter what you do as long as what you do is consistent with how you've defined the relationships between x and the angle.

 

[flyingpig]

I've always had the suspicion that trigonometric substitution always results in non-unique answers.

Because at the point$$-9\left [ \frac{\theta}{2}- \frac{\sin2\theta }{4} \right ]$$I could have left it as just [tex\sin2\theta[/tex] and then substitute $$\arccos\left(\frac{x}{3}\right)$$

 

[flyingpig]

I never bothered to memorize which trig substitutions go with a

Lol no, it's in my textbook under "strategies for integration", like recognize the form and choose what to substitute wisely.

 

[fzero]

Let's say $$\int \sqrt{9 - x^2} dx$$

Using x = 3sinθ, you should get

$$\frac{1}{2}\left ( \sqrt{9-x^2} + 9\arcsin\left(\frac{x}{3} \right ) \right )$$

This should be

$$\frac{1}{2}\left ( x \sqrt{9-x^2} + 9\arcsin\left(\frac{x}{3} \right ) \right )$$

Our final answer should be

$$-9\left [ \frac{arccos\left ( \frac{x}{3} \right )}{2}- \frac{x\sqrt{9-x^2}}{18} \right ]$$


Now clearly

$$-9\left [ \frac{arccos\left ( \frac{x}{3} \right )}{2}- \frac{x\sqrt{9-x^2}}{18} \right ] \neq \frac{1}{2}\left ( \sqrt{9-x^2} + 9\arcsin\left(\frac{x}{3} \right ) \right )$$

You are forgetting the constant of integration. The periodicity of the trig functions means that

$$\arccos y = \frac{\pi}{2} - \arcsin y.$$

You can use this to show that the two methods give the same answer up to the integration constant.

The bottom line is that you can use any substitution that you want, as long as your work is correct. That's the whole point of substitutions.

 

[flyingpig]

$$\frac{1}{2}\left ( x \sqrt{9-x^2} + 9\arcsin\left(\frac{x}{3} \right ) \right )$$

and

$$-9\left [ \frac{arccos\left ( \frac{x}{3} \right )}{2}- \frac{x\sqrt{9-x^2}}{18} \right ]$$

Let's try to evaluate it from 0 to $$2\Pi$$

 

[osnarf]

I've always had the suspicion that trigonometric substitution always results in non-unique answers.

All indefinite integrals give you a non-unique answer. In fact, it gives you an infinite number of answers.

Definite integrals are another story, there is a unique solution. It doesn't matter what substitution you make or whatever strange method you may use, you will get the same answer. Definite integrals will give you a number, not an equation (unless there is another variable you are not integrating with respect to, but your answer will still be the same because you could have just pulled this out of the integral to begin with, leaving you with an indefinite integral of only variables that you are integrating with respect to, and hence you will get whatever you pulled out times a number.) And that number will be exactly the same every time.

 

[olivermsun]

Are you sure your sin x substitution answer is exactly correct? I think there should be no difference in the derivations except for a sign difference in the step where you apply the sin^2 -> double angle formula.

If so, then you can also use the fact that arccos x = pi/2 - arcsin x to show that the answers you get are the same (up to a constant of -9pi/4, which is perfectly okay since indefinite integrals are only specified "up to a constant").

 

[flyingpig]

All indefinite integrals give you a non-unique answer. In fact, it gives you an infinite number of answers.

I know by the constant, but this isn't the "big" problem here

Definite integrals are another story, there is a unique solution. It doesn't matter what substitution you make or whatever strange method you may use, you will get the same answer. Definite integrals will give you a number, not an equation (unless there is another variable you are not integrating with respect to, but your answer will still be the same because you could have just pulled this out of the integral to begin with, leaving you with an indefinite integral of only variables that you are integrating with respect to, and hence you will get whatever you pulled out times a number.) And that number will be exactly the same every time.

Then by why does the back page of my textbook only contains one unique form of solutions to integrals?

 

[flyingpig]

Are you sure your sin x substitution answer is exactly correct? I think there should be no difference in the derivations except for a sign difference in the step where you apply the sin^2 -> double angle formula.

I used Wolframalpha and it said they used sin²x

It could be my math that is wrong for the cosx

 

[olivermsun]

What I'm saying is that your "answer" using the sin substitution seems to be missing a factor of x next to the sqrt. Your procedure using cos x is correct.

The sin x and the cos x substitutions will lead to exactly the same steps until you get to the sin²x and cos²x, respectively, at which point the substitutions differ by a negative sign. However, the end result will have -arcsin in one case and +arccos in the other. Since arcsin x = pi/2 - arccos x, you're good up to a constant.

 

[chiro]

If you do transform one integral to another, then as long as the limits also reflect the change, the integral should still be valid even in the new form. So if you did transform sin's into cos's then as long as the limits reflect the correct transformation, the integral value should remain the same (and reflect the same measure be it length, area, volume etc)

 

[flyingpig]

Why does my book say only use x = asinθ then?

 

[Char. Limit]

I believe it's mainly simply because people are more used to the sine function than the cosine function, for whatever reason. Realistically, as long as you restrict the domain on whichever you're using (-pi/2 to pi/2 for sine, and 0 to pi for cosine), either one is perfectly valid.

 

[olivermsun]

No idea. Maybe they want the answers to be derived the same way so they're easier to grade?

 

[flyingpig]

Evaluate it from 0 to 1

$$-9\left [ \frac{arccos\left ( \frac{x}{3} \right )}{2}- \frac{x\sqrt{9-x^2}}{18} \right ]_{0}^{1} \approx 2.94658$$$$\frac{1}{2}\left [ \sqrt{9-x^2} + 9\arcsin\left(\frac{x}{3} \right ) \right ]_{0}^{1} \approx 1.433$$

They give different answers.

 

[Mark44]

Which suggests that there is an error somewhere. The trig substitution to use is connected with a right triangle, with a hypotenuse (in this case) of 3. If you label the base as x, and the altitude as sqrt(9 - x^2), then your substitution would be cosθ = x/3, or 3cosθ = x, from which you get dx = -3sinθ dθ.

If instead, you label the altitude as x and the base as sqrt(9 - x^2), then your substitution would be sin θ = x/3, or 3 sin θ = x, from which you get dx = 3 cos θ dθ.

It doesn't matter which you choose. If you do it both ways and get different answers, you made a mistake somewhere.

 

[fzero]

I pointed out your error a week ago in post #6.

 

[flyingpig]

Thanks, it has been resolved now. But interestingly Wolframalpha gives an "exact" solution for one substitution but not the other.

http://www.wolframalpha.com/input/?i=[+%281%2F2%29%281*sqrt%289-1^2%29+%2B+9arcsin%281%2F3%29%29+]+-+[+%281%2F2%29%280*sqrt%289-0^2%29+%2B+9arcsin%280%2F3%29%29]

http://www.wolframalpha.com/input/?i=[%28-9%2F2%29%28arccos%281%2F3%29+-+%281*sqrt%289-1^2%29%29%2F9%29]+-+[%28-9%2F2%29%28arccos%280%2F3%29+-+%280*sqrt%289-0^2%29%29%2F9%29]

Interesting.

 

[HallsofIvy]

Why does my book say only use x = asinθ then?

Please check your book again. Does it actually say "only use x= asinθ" or is that just the only example the give?

 

[flyingpig]

Please check your book again. Does it actually say "only use x= asinθ" or is that just the only example the give?

No, according to Mr. James Stewart, that's all we need to know. He didn't really make a comment on cosine

I got a pdf file, but I don' t know how to upload it so I will just give you the pages in image link

[PLAIN]http://img402.imageshack.us/img402/2497/unlediee.jpg

[PLAIN]http://img27.imageshack.us/img27/8296/unledxg.jpg

 

[SammyS]

Evaluate it from 0 to 1

$$-9\left [ \frac{\arccos\left ( \frac{x}{3} \right )}{2}- \frac{x\sqrt{9-x^2}}{18} \right ]_{0}^{1} \approx 2.94658$$


$$\frac{1}{2}\left [ \sqrt{9-x^2} + 9\arcsin\left(\frac{x}{3} \right ) \right ]_{0}^{1} \approx 1.433$$

They give different answers.

These look as though they should give the same answer. The first one is:
$$-9\left [ \frac{\arccos\left ( \frac{x}{3} \right )}{2}- \frac{x\sqrt{9-x^2}}{18} \right ]_{0}^{1}=\frac{1}{2}\left [-9 \arccos\left ( \frac{x}{3} \right )+ x\sqrt{9-x^2} \right ]_{0}^{1}$$

Then as pointed out previously: You are missing an x in the second one. It should be:

$$\frac{1}{2}\left [ x\sqrt{9-x^2} + 9\arcsin\left(\frac{x}{3} \right ) \right ]_{0}^{1}$$

Since arcsin(u) + asccos(u) = π/2, the two expressions should give the same result.

So, in answer to the question in the title of this thread, There's nothing wrong with using x = a cos(θ)

 

[flyingpig]

But there is something wrong with Mr.Stewart. Thank you everyone.

 

[olivermsun]

Well I'm not sure Mr. Stewart is exactly wrong as he is simply giving one method that works. If you noticed that there's basically no difference between sin and cos for all intents and purposes, then hats off to you! ;)