` int _(alpha)^(beta) sqrt((x-alpha)/(beta -x))` dx is equal to

Options

A. `pi/2 (alpha-beta)`
B. `pi/2 (beta-alpha)`
C. ` pi (alpha-beta)`
D. `pi (beta - alpha)`

Answer

B

Explanation

Let `l=int_(alpha)^(beta)sqrt((x-alpha)/(beta-x))dx`
Put `x=alpha cos^(2) theta+beta sin^(2)theta`,
`therefore x-alpha=(beta-alpha)sin^(2)theta, beta-x=(beta-alpha) cos^(2) theta`
and `dx=2 (beta-alpha)sin thetacos theta d theta`
`therefore l = int_(0)^(pi//2)sqrt(((beta-alpha)sin^(2)theta)/((beta-alpha)cos^(2)theta))xx2(beta-alpha)sin theta cos theta d theta`
`=(beta-alpha) int_(0)^(pi//2)(1-cos2 theta)d theta`
`=(beta-alpha)[ theta-(sin 2 theta)/(2)]_(0)^(pi//2)=(beta-alpha)*(pi)/(2)`