General Question About Trig Substitutions (integration)

[Saladsamurai]

So I am trying to get a section ahead in my calsulus text and I am at Trig substitutions.

It says, "To start we will be concerned with integrals that contain expressions of the form $$\sqrt {a^2-x^2}$$ where a is positive and real...etc"

The idea is to eliminate the radical. For the above example they start by saying "we can make the substitution $$x=a\sin\theta$$ " ...and then they give

absolutely no justification for using $$a\sin\theta$$. To me that is like saying "well instead of building that house out of wood, let's use cheese instead."

Now this is what I have reasoned out. Would someone please let me know if I am on the right track:

Since it is the sqrt function, then the term x^2 must be less than or equal to a^2 in order to have a real solution. Since a is a positive real number, than the product a*sin(theta) must equal x for some angle theta.

Thanks,
Casey

 

[Dick]

If |x|<=|a| then x can be written as a*sin(theta) for some angle theta. I don't see what is bugging you.

 

[Saladsamurai]

If |x|<=|a| then x can be written as a*sin(theta) for some angle theta. I don't see what is bugging you.

Well, for starters, ^^^that's exactly what I wrote, isn't it so that is what's bugging me right now.

Casey

p.s. I don't like texts who just say "this is the way it is and that's that."

 

[Dick]

If you want to know why, then sin(theta) assumes all values between -1 and 1. So if |x|<=|a|, you can find a value of theta. It's just a change of variables.

 

[Saladsamurai]

So I had it when I said:

Since it is the sqrt function, then the term x^2 must be less than or equal to a^2 in order to have a real solution. Since a is a positive real number, than the product a*sin(theta) must equal x for some angle theta.

Thanks,
Casey

I assume this same concept will apply to the other trig subs as well.

Thanks,
Casey

 

[Dick]

Yes, I guess I wasn't quite sure what the question was.

 

[Saladsamurai]

Yes, I guess I wasn't quite sure what the question was.

There isn't one.

Casey