- #1

tony873004

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\int {\frac{{\sqrt {9 - x^2 } }}{{x^2 }}dx}

[/tex]

I understand all the steps that get me up to [tex] = - \cos \theta \, - \theta \, + C[/tex]

Then the book goes on to explain:

"Since this is an indefinate integral, we must return to the original variable x. This can be done either by using trig identities to express cot theta in terms of sin theta=x/3 or by drawing a diagram, as in Fig. 1, where theta is intrepreted as an angle of a right triangle. Since sin theta = x/3....."

Where does this x/3 come from? The steps that follow make sense, but how did we know to just pull an x/3 out of the air and start using it?

Applying this to a homework question [tex]\int_{}^{} {x^3 \sqrt {9 - x^2 } } dx[/tex] I come up with [tex]243\left( {\frac{{\cos ^5 \theta }}{5} - \frac{{\cos ^3 \theta }}{3}} \right) + C[/tex]

Now, switching it back to x, I'm going to guess that [tex]{\sqrt {9 - x^2 } }[/tex] becomes the adjacent side of my triangle, but for no good reason other than the example of the book pulled the sqrt() out of the original equation and did the same.

I just don't get this triangle stuff. The lecture didn't do it for me, and the book's explanation doesn't do it for me either >:-(