Math Is Fun Forum

Cheesyc

Member

Registered: 2006-12-02

Posts: 1

Trig Functions Subtitutions in ∫√ (a^2-x^2) , x = asinθ

I have recently just learned about substitutions with trig functions for the function in the title. I'm not sure as to how to get the function back to a state of X instead of a cosθ  or sin θ . If anyone knows anything about that, Please help me understand it. Thanks

Actual problem in my book.

∫√ (2^2-x^2)

x = asinθ
dx = acosθDθ

= ∫2cosθDθ√ (4-4sin^2θ)dθ
= ∫2cosθDθ√ (4[1-sin^2θ])dθ
= ∫2cosθDθ√ (4cos^2θ)dθ
= ∫2cosθDθ2cos θdθ
= ∫4cos^2θDθ

fgarb

Member

Registered: 2006-03-03

Posts: 89

Re: Trig Functions Subtitutions in ∫√ (a^2-x^2) , x = asinθ

Well, you already showed that you went from x to theta using x=2sin(theta), so you can get theta in terms of x by simply inverting this if you like: theta = sin^-1[(1/2)x]. It's probably a bad idea to do this before integration, however, because you have an easy function to integrate here and the substitution would make it quite tricky. If you're doing an indefinite integral, you can just integrate this and then sub in for x, remembering to put in the unknown constant. If it's definite, you just need to remember to use the theta limits while you have the expression in terms of theta, and the x limits while you have the expression in terms of x. Does that make sense?

Also, you should be careful about your d's. I see that through most of your work you have both D(theta) and d(theta). First off, in introductory calculus, everyone thinks in terms of lower case d's, so you probably should too to avoid confusion. Second, at this level you should always have the same number of d's in your equation as integrals. Each integral gets rid of one d, so if you have two d's and one integral, you will be left with a d in the final state. Then, since d(theta) is infinitely small, your answer is zero! (actually, it would represent a sort of "density" in theta as I understand it). It might be a good idea to go back and try to figure out how you got the second d and where your mistake is. Again, ask if you're confused and I'm sure someone will explain.