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