Question

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

Answer 1

`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`