Question

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

Answer 1

\(x=\sin\theta\) I guess.

Answer 2

asin t
correct

Answer 3

x = 3sint

Answer 4

Yea. You want it so \(a^2=9\)

Answer 5

in 3 sin t in for x in the denominator?

Answer 6

Yes

Answer 7

yes. for every x in the problem

Answer 8

\[
dx=3\cos t\;dt
\]

Answer 9

so \[\frac{ \sqrt{9\cos ^{2}\theta} }{ 3\sin \theta }* 3\cos \theta d \theta \]

Answer 10

Yes

Answer 11

then \[\frac{ 9\cos ^{2}\theta }{ 3\sin \theta }d \theta \]

Answer 12

Nope. it is \[
\frac{
9|\cos\theta|\cos \theta
}{3\sin\theta}d\theta
\]

Answer 13

what are those lines there for?

Answer 14

Absolute value.

Answer 15

is it not multiplied?

Answer 16

Well if \(\cos\theta <0\) then \(|\cos\theta|\cos\theta = -\cos^2\theta\)

Answer 17

Otherwise it is \(|\cos\theta|\cos\theta = \cos^2\theta\)

Answer 18

For any definite integral, we would have to split it up for certain intervals of \(\theta\).

Answer 19

I suppose for simplicity you could just pretend it will always be positive and put a foot note.

Answer 20

so this is because it is an indefinite integral so cosine could be positive or negative?

Answer 21

Yeah. It it were a definite integral we would split it up at its roots so we know what intervals it is negative and what intervals it is positive.

Answer 22

Since it is an indefinite integral, we'd have to make certain assumptions.

Answer 23

ok that makes sense but now where do we go from here

Answer 24

Assume it is possible and then integrate.

Answer 25

positive^

Answer 26

You basically have \(\tan\theta \cos \theta\)

Answer 27

You could try parts here I guess

Answer 28

would that eventually get me to \[3\ln(\frac{ \sqrt{9-x ^{2}}-3 }{ x })+\sqrt{9-x ^{2}}+C\]