`intsqrt(((a+x)/(a-x)))dx` का मान ज्ञात कीजिये ।

Explanation

माना `I = int sqrt(((a+x)/(a-x)))dx`
माना `x = a cos theta therefore dx = - a sin theta d theta`
`rArr dx = - a 2 sin.(theta)/(2)cos.(theta)/(2)d theta`
तथा `(a+x)/(a-x) = (a(1+cos theta))/(a(1-cos theta)) = (1+2cos^(2) theta//2-1)/(1-1+2sin^(2) theta//2)`
` = (cos^(2) theta//2)/(sin^(2) theta//2)`
`thereforesqrt((a+x)/(a-x)) = (cos theta//2)/(sintheta//2)`
`therefore I = a int (cos theta//2)/(sin theta //2) . 2sin.(theta)/(2)cos.(theta)/(2)d theta`
`= - a int 2cos^(2) theta//2 d theta`
` = -aint(1 + cos theta)d theta = - a[theta + sin theta]`
`= - a[theta + sqrt(1-cos^(2))theta]`
`because x = a cos theta therefore(x)/(a) = cos theta `तथा ` theta = cos^(-1)((x)/(a))`
`therefore " "I = - a [cos^(-1)((x)/(a)) + sqrt(1+(x^(2))/(a^(2)))]`
`= -a[cos^(-1)((x)/(a))+(1)/(a)sqrt(a^(2) -x^(2))]`
द्वितीय विधि (Second method)
`I = int sqrt((adx)/(a-x))dx = intsqrt(((a+x)(a+x))/((a-x)(a+x)))dx`
` = int(a+x)/(sqrt(a^(2)-x^(2))dx`
`= int (a dx)/(sqrt(a^(2)-x^(2)))+int(xdx)/(sqrt(a^(2) -x^(2)))`
`a sin^(-1)((x)/(a)) - sqrt((a^(2) -x^(2)))`