Question

integrate! (i think it's a trig substitute)

Answer 1

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

Answer 2

Hmm, trig sub indeed :o

Answer 3

but... i did bad on this section :(

Answer 4

If we factor out a 9, and do some fancy business,\[\Large\rm \frac{1}{3}\int\limits\limits\frac{ dx }{ \sqrt{1-\left(\frac{x}{3}\right)^{2}}}\]We get it into that form,

Should I go through that a little slower or is that step ok?

Answer 5

The `reason` I did that,
is to put in the form 1-(something)^2

Which tells us sine would be a good substitute.

Answer 6

what is it that lets us know that it is sine? ohh and im not 100 percent on your algebra you did

Answer 7

\[\Large\rm \sin^2x+\cos^2x=1 \qquad\implies\qquad 1-\sin^2x = \cos^2x\]So if we have something of the form:

1-(stuff)^2

We can replace (stuff) with sinx and the whole thing simplifies to cos^2x, right?

Answer 8

Crappp my game is starting :C
I gotta go!

I'll come back in a bit if you're still stuck >.< sozzzz

Answer 9

ok thank you!

Answer 10

also that makes sense!

Answer 11

\[\Large\rm \int\frac{1}{\sqrt{9-x^2}}~dx\]So for the algebra steps,
I started by factoring a 9 out of each term under the root,\[\Large\rm \int\frac{1}{\sqrt{9(1-\frac{1}{9}x^2)}}~dx\]Taking the root of 9 gives us:\[\Large\rm \int\frac{1}{3\sqrt{1-\frac{1}{9}x^2}}~dx\]Bring the 1/3 out front,\[\Large\rm \frac{1}{3}\int\limits\frac{1}{\sqrt{1-\frac{1}{9}x^2}}~dx\]Then rewrite 1/9 as 1/3^2 and bring it into the square with the x,\[\Large\rm \frac{1}{3}\int\frac{1}{\sqrt{1-\left(\frac{x}{3}\right)^2}}~dx\]

Answer 12

Making the substitution:\[\Large\rm \frac{x}{3}=\sin \theta\]Taking a derivative gives us:\[\Large\rm \frac{1}{3}dx=\cos \theta ~d \theta\]

Answer 13

Subbing the stuff in gives us:\[\Large\rm \int\limits\frac{1}{\sqrt{1-\sin^2\theta}}~\cos \theta ~d \theta\]Which simplifies quite a bit!\[\Large\rm \int\limits d \theta\]