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Trig Substitution

  • Thread starter Bashyboy
  • Start date
  • #1
1,421
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Homework Statement


[itex]\int\frac{1}{\sqrt{16-x^2}}dx[/itex]


Homework Equations


[itex]csc\theta=\frac{4}{\sqrt{16-x^2}}[/itex]

[itex]4cos\theta=x[/itex]

[itex]-4sin\theta d\theta=dx[/itex]

[itex]\theta=arccos(\frac{x}{4})[/itex]

The Attempt at a Solution


Using these facts, I concluded that the integral, after all of the substitution, was
[itex]-arccos(\frac{x}{4})[/itex] but the actual answer is the arcsin
 

Answers and Replies

  • #2
33,496
5,188

Homework Statement


[itex]\int\frac{1}{\sqrt{16-x^2}}dx[/itex]


Homework Equations


[itex]csc\theta=\frac{4}{\sqrt{16-x^2}}[/itex]

[itex]4cos\theta=x[/itex]

[itex]-4sin\theta d\theta=dx[/itex]

[itex]\theta=arccos(\frac{x}{4})[/itex]

The Attempt at a Solution


Using these facts, I concluded that the integral, after all of the substitution, was
[itex]-arccos(\frac{x}{4})[/itex] but the actual answer is the arcsin
You don't show your complete answer, but what I think you have is probably correct. There is an identity that involves inverse trig functions:
arcsin(x) + arccos(x) = ##\pi/2##

So arcsin(x) = -arccos(x) + ##\pi/2##

Your answer (-arccos(x/4)) and the book's answer (arcsin(x/4)) differ by a constant.

Also, you can always check your answer to an integration problem by differentiating it. If the result is your original integrand, then your answer is correct.
 
  • #3
1,421
5
Oh, okay. It certainly makes perfect sense that there would be a inverse trig identity. Thank you.
 

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