+(1)/(sqrt((x-alpha)(beta-x)))dx,(beta>alpha)

Evaluate: int(1)/((x-alpha)(beta-x))dx,(beta>alpha)

int(dx)/(sqrt((x-alpha)(beta-x)))

Putting x=alphacos^(2)theta+betasin^(2)theta show that, int_(alpha)^(beta)sqrt((x-alpha)(beta-x))dx=((beta-alpha)^(2)pi)/(8)

Evaluate: int1/((x-alpha)(beta-x))dx ,(beta>alpha)

Evaluate: int1/((x-alpha)(beta-x))dx ,(beta>alpha)

The value of the integral int_alpha^beta (1)/(sqrt((x-alpha)(beta-x)))dx " for "beta gt alpha , is

The value of the inntegral int_(alpha)^(beta) (1)/(sqrt((x-alpha)(beta-x)))dx is

int (dx)/(sqrt((x-alpha)(x-beta)))

Integrate : int (dx)/(sqrt((x-alpha)(beta-x)))[ alpha lt x lt beta]

Evaluate : int(dx)/((x-alpha)(beta-x))beta gt alpha