`sqrt((1-sqrtx)/(1+sqrtx))`

Explanation

माना `I=intsqrt((1-sqrtx)/(1+sqrtx))`
माना `sqrtx=cost`
`rArr" "x=cos^(2)t rArr dx=2cos t (-sin t)dt`
`therefore" "I=intsqrt((1-cost)/(1+cost))(-2sint cost)dt`
`I=-2intsqrt((2sin^(2).(t)/(2))/(2cos^(2).(t)/(2))).2sin.(t)/(2)cos.(t)/(2).cost dt`
`=-4intsin^(2).(t)/(2).cost dt`
`=-4int(1-cost)/(2).costdt`
`=-2int(cost-cos^(2)t)dt`
`=-2int(cost-(1+cos2t)/(2))dt`
`=-2sint+t+(1)/(2)sin2t+C`
`=-2sqrt(1-cos^(2)t)+t+(1)/(2)(2sqrt(1-cos^(2)t).cost)+C`
`=-2sqrt(1-x)+cos^(-1)sqrtx+sqrtxsqrt(1-x)+C`
`=-2sqrt(1-x)+cos^(-1)sqrtx+sqrt(x-x^(2))+C`