Question

Evaluate the integral.

Answer 1

\[\int\limits \frac{ 1 }{ x^2\sqrt{1-x^2} }dx\]

Answer 2

I know I'm supposed to use trig sub, Do I use a triangle and pythagorean theorem ?

Answer 3

not sure what my first step should be

Answer 4

Let's recall our Pythagorean Identities real quick:\[\large\rm 1-\sin^2\theta=\cos^2\theta\]\[\large\rm 1+\tan^2\theta=\sec^2\theta\]\[\large\rm 1+\cot^2\theta=\csc^2\theta\]
If you look at the left sides of these equations...
Which one looks like it might have the form we need?

Answer 5

Notice our square root has a 1-x^2,
which Pythagorean Identity looks similar?

Answer 6

i believe its sin theta ? 1-sin^2theta = cos^2 theta

Answer 7

I'm using that as a reference :)

Answer 8

Bahh reference, just use brain :)
lol

1-x^2 is "similar" to 1-sin^2(theta) ya?
They have the same form.

We would like to get rid of the `subtraction` from under the root,
that's why making use of a trig substitution.
Trig identity will turn 2 terms into 1 term.

Answer 9

\[\large\rm x=\sin \theta\]Ok sine sounds good.

Answer 10

What do you get for your dx?
Remember how to differentiate sine?

Answer 11

so I would replace the bottom part with x^2 (sin theta) ?

Answer 12

and then i would have \[1/x^{2}(\sin \theta) ...

Answer 13

sorry it didn't come out right

Answer 14

Hoooold your giraffes there a sec buddy! :O
What do you get for your du?
Let's find that piece, then we'll plug in all the nonsense.

Answer 15

\[1/x^2\sin \theta \]

Answer 16

dx I mean*

Answer 17

ok lol i mean.. do i differentiate that equation i have up there?

Answer 18

by doing u sub?

Answer 19

This isn't a u-sub,
it's a special type of substitution.
it'll feel a little weird compared to u-sub.
:)

Answer 20

ohh ok... so the integral of sin.. is -cos

Answer 21

Lemme just list the pieces we'll need so it's all in one place.

This is the substitution we're making:\[\large\rm x=\sin \theta\]Squaring each side gives us:\[\large\rm \color{orangered}{x^2=\sin^2\theta}\]Going back to the substitution we made, if we differentiate with respect to theta,\[\large\rm \frac{dx}{d \theta}=\cos \theta\qquad\to\qquad \color{royalblue}{dx=\cos \theta~d \theta}\]

Answer 22

And we're going to plug the colored pieces into our integral to make the change,\[\large\rm \int\limits\frac{1}{\color{orangered}{x^2}\sqrt{1-\color{orangered}{x^2}}}\color{royalblue}{dx}\]

Answer 23

ohhh i see! so for every x^2 we plug in sin theta. which makes it sin^2 theta!

Answer 24

for every x^2 we plug in sin^2theta*

Answer 25

are you squaring beforehand to remove the square root?

Answer 26

And keep in mind, you're "replacing" these x's.
So your new integral shouldn't have ANY x business in it.

Answer 27

Hmm I'm not sure what you're asking :[

Answer 28

is our goal currently to get rid of the square root on the bottom of the equation? sorry lol

Answer 29

i see! i will try it on paper right now

Answer 30

We're using the colored pieces I described earlier to make our substitution like this:
\[\large\rm \int\limits\limits\frac{1}{\color{orangered}{x^2}\sqrt{1-\color{orangered}{x^2}}}\color{royalblue}{dx}=\int\limits\limits\frac{1}{\color{orangered}{\sin^2\theta}\sqrt{1-\color{orangered}{\sin^2\theta}}}\color{royalblue}{(\cos \theta~d \theta)}\]Try to follow the colors! :)

Answer 31

ok i see, now for the dx. i know its not a u sub but is it correct to say that we take the trig identity which in our case is sin theta, and take the derivative and thats all we do to find dx? so in this case it becomes cos theta which you just plug into dx

Answer 32

yes, similar process to u-sub, taking the derivative of your "u"

Answer 33

ok:)

Answer 34

now should i get rid of the square root on the bottom of the fraction?

Answer 35

\[\large\rm \int\limits\limits\limits\frac{1}{\sin^2\theta\sqrt{1-\sin^2\theta}}(\cos \theta~d \theta)\]Apply your Pythagorean Identity under the root.
Keep in mind that: \(\large\rm \sqrt{1^2-\sin^2\theta}\ne 1-\sin\theta\)
This is a common mistake that students make, try to avoid that.

Answer 36

\[\large\rm \int\limits\frac{1}{\sin^2\theta\sqrt{\color{green}{1-\sin^2\theta}}}(\cos \theta~d \theta)=\int\limits\frac{1}{\sin^2\theta\sqrt{\color{green}{\cos^2\theta}}}(\cos \theta~d \theta)\]Understand that step?

Answer 37

yes, you are using the identity cos^theta = 1-sin^2 theta?

Answer 38

cos^2 theta i mean

Answer 39

Now we can cancel the square root!

Answer 40

so we get

Answer 41

\[\frac{ 1 }{ \sin^2 \theta \cos \theta }\]

Answer 42

the integral of that

Answer 43

Woops, don't forget about the cosine with your differential

Answer 44

\[\large\rm \int\limits \frac{ 1 }{ \sin^2 \theta \cos \theta }\cos \theta~d \theta\]

Answer 45

cool, now cancel some stuff out :)

Answer 46

yes of course! thank you

Answer 47

can we cancel the cos thetas out?

Answer 48

so we would get 1/sin^2 theta

Answer 49

and this would turn into ln(sin^2theta) ?

Answer 50

Applying another trig identity,
\[\large\rm =\int\limits\frac{1}{\sin^2\theta}d \theta\quad=\quad \int\limits \csc^2\theta~d \theta\]natural log? Hmm that doesn't look quite right.

Answer 51

oh i forgot about trig identities lol. so we would have to integrate csc^2 theta instead

Answer 52

so we get -cot(theta)

Answer 53

Ok great! :)
And now we need this back in terms of x.
So it's triangle time.