Question

∫dx/x^2√1-x^2

Answer 1

\[\Large\bf\sf \int\limits \frac{dx}{x^2\sqrt{1-x^2}}\]Have you learned about `Trigonometric Substitutions` yet?
That would be a good approach here.\[\Large\bf\sf x=\sin \theta\]

Answer 2

Okay and then?

Answer 3

please help

Answer 4

So what would the differential dx become in terms of theta?

Answer 5

d/dx sin theta=cos theta and d/dx theta=0 right?

Answer 6

Take the derivative with respect to theta, not x.\[\Large\bf\sf x=\sin \theta, \qquad\to\qquad \frac{dx}{d \theta}=\cos \theta\]Then "multiply" dtheta to the other side,\[\Large\bf\sf dx=\cos \theta \;d \theta\]I just got a little confused by your zero there..

Answer 7

\[\Large\bf\sf x^2=\sin^2\theta\]

Answer 8

So there is the stuff we'll need to make our substitution.

Answer 9

thank you :) for the help

Answer 10

Oh you understand it from there? :o

Answer 11

yes i understood .. and i will do the other

Answer 12

oh ok cool c:

Answer 13

ok i got -cot theta +c what should i do next ?

Answer 14

@Lena772

Answer 15

@Luigi0210

Answer 16

HELP SOMEONE

Answer 17

@RANE

Answer 18

Stuck? Ya the next step is a little tricky.
To get our answer back in terms of x, we need to draw out a triangle with sides that correspond to our subsitution.

Answer 19

\[\Large\bf\sf \sin \theta=x\]
\[\Large\bf\sf \sin \theta\quad=\quad \frac{opposite}{hypotenuse}\quad=\quad \frac{x}{1}\]

Answer 20

Pythagorean Theorem to find the missing side: