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#