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))

1 Answer
Aug 2, 2018

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