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`sqrt((1-sqrtx)/(1+sqrtx))` |
Answer» 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` |
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