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

• Dec 19th 2009, 12:49 PM
james121515
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
• Dec 19th 2009, 02:03 PM
mr fantastic
Quote:

Originally Posted by james121515
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 $\displaystyle x = a \sin \theta$ then clearly
Quote:

from a constructing your triangle, you get that Cos(theta) = Sqrt(a^2 - x^2)
If you make the substitution $\displaystyle x = a \cos \theta$ then clearly
Quote:

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).
sin is usually chosen so that $\displaystyle \frac{-\pi}{2}\leq {\theta}\leq \frac{\pi}{2}$. Because then $\displaystyle cos{\theta}>0$.