Question

How do you integrate this equuation?

My question is about ithis function `int_(1/sqrt(x-x^2))` in the step where you have to integrate `int_(1/sqrt(1/4-u^2))`

Answer 1

`I=arc sin(2x-1)+c`

Explanation:

I think your question is of the type :

`I=int1/sqrt(x-x^2)dx`

To complete the square ,add and subtract `1/4`

`i.e. x-x^2=1/4+x-x^2-1/4=1/4-(x^2-x+1/4)`

`:.x-x^2=1/4-(x-1/2)^2`

So,

`I=int1/sqrt(color(red)(1/4)-(x-1/2)^2)dx`

Substitute , `color(blue)(x-1/2=u=>dx=du`

`:.I=int1/sqrt((color(red)(1/2))^2-u^2)du...toApply(3)."given below."`

`:.I=arcsin(u/(color(red)(1/2)))+c`

`:.I=arc sin(2u)+c`

Subst. back , `color(blue)(u=x-1/2` ,we get

`I=arc sin(2(x-1/2))+c`

`=>I=arc sin(2x-1)+c`

Note:
`(1)arc sinx=sin^-1x`
`(2)int1/sqrt(1-x^2)dx=arc sin(x)+c`
`(3)int1/sqrt(color(red)(a^2)-x^2)dx=arc sin(x/color(red)(a))+c`
If `a=5` then
`I=int1/sqrt(25-x^2)dx=int1/sqrt(color(red)(5^2)-x^2)dx=arc sin(x/color(red)(5))+c`