Question

Help on Trig Subsitution! :) Thanks

Answer 1

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

Answer 2

the idea is to make the radical in the numerator disappear.
if you remember the t-identities,
\[\sin^2\theta+\cos^2\theta=1\]
this identity helps you convert a subtraction in a radical sign.
the other one
\[1+\tan^2\theta=\sec^2\theta\]
helps you convert a positive sign

Answer 3

so, you can make x=cos(t)
or x = sin(t)

both will serve the purpose. It is a convention to use the "sine" function though

Answer 4

im not realy understanding how to do the question

Answer 5

given the problem, what is the evil part that is preventing you from integrating?
ans: the radical
correct?

Answer 6

yes

Answer 7

so draw a triangle?

Answer 8

no no. no need.

Answer 9

folr any angle \(\theta\), the t-identity stands valid.. agree?

Answer 10

yes

Answer 11

does \[\sqrt{9-x^2}=3sinx\]

Answer 12

good.
now, we have to remove that radical..
but what is "under" that radical?
\(9-x^2\)
how do we eliminate this?
by remembering that
\[\sin^2\theta=1-\cos^2\theta\\\text{and}\\\cos^2\theta=1-\sin^2\theta\]

Answer 13

consider the second form
if we multiply both sides by "9", would we change anything?

Answer 14

you follow?>

Answer 15

i can't find a connection between that with the identities

Answer 16

do you understand that we gotta get rid of
\(9-x^2\)??

Answer 17

as another "square"...

Answer 18

\[\int\limits_{}^{}3sinx/x^2\]

Answer 19

yes I do

Answer 20

then do not go jumping here and there..
lets do it step by step..

Answer 21

okay sorry...

Answer 22

so, we know \[\sin^2\theta=1-\cos^2\theta\\\implies9\sin^2\theta=9-9\cos^2\theta\\\implies9\sin^2\theta=9-(3\cos\theta)^2\]
so, set \(x=3\cos\theta\).
now, since we changed "x" ot "theta", we change "dx" to "\(d\theta\)"

\[dx=3(-\sin\theta)d\theta\]

Answer 23

so I understand that I need to get rid of the radical of the integral to go furthur.

Answer 24

so, the integral now becomes,
\[
I=\int\frac{\sqrt{9\sin^2\theta}}{(3\cos\theta)^2}(3)(-\sin\theta)d\theta
\]

Answer 25

follow?

Answer 26

then what do I do next?

Answer 27

\[
I=-\int\frac{3\sin\theta}{\cancel{3}\cos^2\theta}\cancel{3}\sin\theta d\theta
\]

Answer 28

\[
I=-3\int\tan^2\theta\;d\theta
\]

Answer 29

we know \(\tan^2\theta\) is not a perfect integral but we can convert it using \(\tan^2\theta=\sec^2\theta-1\)

Answer 30

\[
I=-3\int(\sec^2\theta-1)d\theta
\]

Answer 31

now, solve this one.

Answer 32

isnt there a formula for secx^2-1?

Answer 33

note: its not "x" anymore. our variable is "theta" they are different. we have to substitute "x" at the end.

we have formula for \(\sec^2\theta\)

Answer 34

im not even sure how to start on solving the new integral you solved.

Answer 35

im really confused now... :(