ज्ञात करें `int (dx)/sqrt((x-a)(b-x))`

Explanation

`x=a cos^2 theta+ b sin^(2) theta`, रखे, तो `dx = ( b-a) 2 sin theta cos theta d theta`
तथा `x-a=a cos^2 theta+b sin^(2)theta-a = b sin^(2) theta-a sin^2 theta =(b-a)sin^(2) theta`
`b-x=b-a cos^(2) theta -b sin^(2) theta = (b-a) cos^2 theta`
अब, `I = int(2(b-a)sin theta cos theta)/((b-a)sin thetacos theta)d theta =2 int d theta = 2 theta +c`
`=cos^( -1)((2x-a-b)/(a-b ))+c" "{{:(because,x=(a(1+cos2 theta))/2+(b(1-cos 2 theta))/2),(therefore,2x -a-b=(a-b)cos 2 theta):}`