InterviewSolution
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InterviewSolution
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`(sin^(-1)sqrtx-cos^(-1)sqrtx)/(sin^(-1)sqrtx+cos^(-1)sqrtx),(x in [0,1])` |
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Answer» माना `I=int(sin^(-1)sqrtx-cos^(-1)sqrtx)/(sin^(-1)sqrtx+cos^(-1)sqrtx)dx` हम जानते हैं कि `sin^(-1)sqrtx+cos^(-1)sqrtx=(pi)/(2)` `rArr cos^(-1)sqrtx=(pi)/(2)-sin^(-1)sqrtx` `therefore" "I=int(sin^(-1)sqrtx-((pi)/(2)-sin^(-1)sqrtx))/((pi)/(2))dx` `" "=int(2sin^(-1)sqrtx-(pi)/(2))/((pi)/(2))dx` `" "=(2)/(pi)int(2sin^(-1)sqrtx-(pi)/(2))dx` `" "=(4)/(pi)intsin^(-1)sqrtxdx-int1dx` `" "=(4)/(pi)intsin^(-1)sqrtx dx-x` `rArr" "I=(4)/(pi)I_(1)-x+C" ...(1)"` जहाँ `" "I_(1)=intsin^(-1)sqrtx dx` माना `sqrtx=t" "rArr" "x=t^(2)" "rArr" "dx=2tdt` `I_(1)=intsin^(-1)t 2t dt = 2 int underset("I ")(sin^(-1))underset("II")(t.t) dt ` `=2(sin^(-1)t.(t^(2))/(2)-int(1)/(sqrt(1-t^(2))).(t^(2))/(2)dt)` `=t^(2)sin^(-1)t-int(t^(2))/(sqrt(1-t^(2)))dt` `=t^(2)sin^(-1)t-int(-(1-t^(2))+1)/(sqrt(1-t^(2)))dt` `=t^(2)sin^(-1)t+intsqrt(1-t^(2))dt-int(1)/(sqrt(1-t^(2)))dt` `=t^(2)sin6(-1)t+(tsqrt(1-t^(2)))/(2)+(1)/(2)sin^(-1)t-sin^(-1)t` `=(t^(2)-(1)/(2))sin^(-1)t+(1)/(2)tsqrt(1-t^(2))` `=(1)/(2)[(2x-1)sin^(-1)sqrtx+sqrtxsqrt(1-x)]` `=(1)/(2)[(2x-1)sin^(-1)sqrtx+sqrt(x-x^(2))]` `I_(1)` का मान समीकरण (1 ) में रखने पर, `int(sin^(-1)sqrtx-cos^(-1)sqrtx)/(sin^(-1)sqrtx+cos^(-1)sqrtx)dx` `=(2)/(pi)[(2x-1)sin^(-1)sqrtx+sqrt(x-x^(2))]-x+C` |
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