Question

Find the integral of sqrt{1-x^2}/x^2 dx ?

Answer 1

\[\int\limits_{}^{} \sqrt{1-x^2}/x^2 dx\]

Answer 2

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

Answer 3

If so you can do a u sub \[u= \sqrt{1-x^2}\] and it seems fair to use by parts :)

Answer 4

Okay I'll try that...

Answer 5

Why would we not use trig substitution here?

Answer 6

You can try, I just thought this may be easier.

Answer 7

Okay, can we try both ways? I want to get as much practice as possible :)

Answer 8

Sure, try both and see if they work, good way to check your answer s:)

Answer 9

\[\sqrt{a^2-x^2} \implies x = a \sin(t)\]

Answer 10

Did you try it out?

Answer 11

Well I'm stuck on finding the integral of 1/sqrt{1+x^2}

Answer 12

OH wait, I got it it's sin^(-1) (x)

Answer 13

\[\checkmark\]

Answer 14

Wait so is it since 1/sinx is equal to sin^(-1) x

Answer 15

No, inverse does not = 1/x

Answer 16

\[f^{-1}(x) \neq \frac{ 1 }{ f(x) }\] haha I just showed this to the question below

Answer 17

Okay, then I'm confused on how to get that...

Answer 18

how to receive the integral of 1/sqrt{1-x^2}

Answer 19

You have to use by parts

Answer 20

\[u = \sqrt{1-x^2} ~~~du = -\frac{ x }{ \sqrt{1-x^2} }dx~~~,v = -\frac{ 1 }{ x } ~~~dv = \frac{ 1 }{ x^2 }dx\]

Answer 21

\[\int\limits u dv = uv - \int\limits v du\]

Answer 22

\[ \implies \sqrt{1-x^2}(-\frac{ 1 }{ x }) - \int\limits \left(- \frac{ x }{ \sqrt{1-x^2} } \right) (-\frac{ 1 }{ x })dx\]

Answer 23

\[ \implies -\frac{ \sqrt{1-x^2} }{ x } - \int\limits \frac{ 1 }{ \sqrt{1-x^2} }dx \implies -\frac{ \sqrt{1-x^2} }{ x }-\sin^{-1}(x)+C\]

Answer 24

Does that make sense, it may look confusing, just let me know if you don't understand something, try to redo it on paper as well.

Answer 25

How did you find out that the integral of 1/sqrt{1-x^2} is the negative inverse of sin?

Answer 26

What is the derivative of arcsin?