Evaluate the following integral:(i)∫dx/√x-x² from 1/4 to 1/2?

1 Answer
Feb 20, 2018

pi/12

Explanation:

int_(1/4)^(1/2) dx/sqrt(x-x^2)

Let x= sin^2 u. This gives dx= 2sinu cos u du
When x= 1/4, sin u= 1/2 ->u=sin^(-1) (1/2) = pi/6

and when x= 1/2, sin u=1/sqrt2 -> u= pi/4

int_(1/4)^(1/2) dx/sqrt(x-x^2)=int_(1/4)^(1/2) dx/(sqrtx sqrt (1-x)

= int_(u=pi/6)^ (u= pi/4) (2 sin u cos u) (du)/((sin u ) (cos u))

=int_(pi/6)^(pi/4) du = [u]_(pi/6) ^(pi/4)

(pi/4- pi/6) =pi/12