Question

∫dx/x^2√1-x^2
let,x=sinz
=>dx/dz=d/dz sinz
=>dx/dz=cosz
=>dx=cosz dz
Now,∫dx/x^2√1-x^2=∫cosz dz/sin^2z√1-sin^2z
=∫cosz dz/sin^2z√cos^2z=∫cosz dz/sin^2z.cosz=∫dz/sin^2z=∫dz/1/cosec^2z=∫dz cosec^2z=-cotz+c then what? @ganeshie8

Answer 1

\(\large \mathbb {\int \frac{dx}{x^2\sqrt{1-x^2}} }\)

Answer 2

its like that ?

Answer 3

yes ! :)

Answer 4

?

Answer 5

please help..i really need help :(

Answer 6

so far looks good,
u need to given answer in x's..
not z..

Answer 7

yeah its like -cotz=-cosz/sinz=-cosz/x

Answer 8

yes, wat u gona do wid numerator cosz ?

Answer 9

there cosz=dz/dx

Answer 10

if you see up

Answer 11

sorry dx/dz=cosz

Answer 12

dx/dz/x=dx/dz.1/x

Answer 13

-cotz+c

your answer is right so far.
u just need to get rid of z's ok

Answer 14

let me show u quick how to get rid of z :)

Answer 15

:) okay

Answer 16

I find this interesting

Answer 17

u have x = sin z

draw the right triangle representing this..