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Math Help - Trig substituion: what is the flaw in my logic?

  1. #1
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    Trig substituion: what is the flaw in my logic?

    Hi all,

    I have a general question about trig substitution. My calculus book says if you ever have an expression dealing with Sqrt(a^2 - x^2), you let x =a Sin(theta), and from a constructing your triangle, you get that Cos(theta) = Sqrt(a^2 - x^2). Why is WRONG to switch Sine and Cosine here? For example, you could x be the adjacent side, Sqrt(a^2-x^2) be the opposite side, and a be the hyp, and you get:
    x = a Cos(theta)
    Sqrt(1-x^2) = a Sin(Theta).

    And the triangle is still correct. But I tried working a problem this way and got the wrong answer. What is the deal?

    Also, at the end of the integral, when you solve for Theta in terms of x, does it matter whether you use Arctan, Arcsin, or Arccos?

    Thanks,
    Tyler
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  2. #2
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    Quote Originally Posted by james121515 View Post
    Hi all,

    I have a general question about trig substitution. My calculus book says if you ever have an expression dealing with Sqrt(a^2 - x^2), you let x =a Sin(theta), and from a constructing your triangle, you get that Cos(theta) = Sqrt(a^2 - x^2). Why is WRONG to switch Sine and Cosine here? For example, you could x be the adjacent side, Sqrt(a^2-x^2) be the opposite side, and a be the hyp, and you get:
    x = a Cos(theta)
    Sqrt(1-x^2) = a Sin(Theta).

    And the triangle is still correct. But I tried working a problem this way and got the wrong answer. What is the deal?

    Also, at the end of the integral, when you solve for Theta in terms of x, does it matter whether you use Arctan, Arcsin, or Arccos? Mr F says: What matters is substituting the correct expression for theta based on the substitution you made.

    Thanks,
    Tyler
    If you make the substitution x = a \sin \theta then clearly
    from a constructing your triangle, you get that Cos(theta) = Sqrt(a^2 - x^2)
    If you make the substitution x = a \cos \theta then clearly
    you could x be the adjacent side, Sqrt(a^2-x^2) be the opposite side, and a be the hyp, and you get:
    x = a Cos(theta)
    Sqrt(1-x^2) = a Sin(Theta).
    Both substitutions lead to equivalent correct answers.

    If you need more help, please post a concrete question and your solution to it (note: solution = working plus answer) that illustrates your problem with the substitution method.
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  3. #3
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    sin is usually chosen so that \frac{-\pi}{2}\leq {\theta}\leq \frac{\pi}{2}. Because then cos{\theta}>0.

    It is about the restriction on theta.
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