InterviewSolution
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InterviewSolution
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फलन `intsqrt((3-2x-2x^(2)))dx` का x के सापेक्ष समाकलन कीजिए । |
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Answer» माना `" "I=intsqrt((3-2x-2x^(2)))dx` `=intsqrt({3-2(x^(2)+x+(1)/(4)-(1)/(4))})dx" "` (पूर्ण वर्ग बनाने पर ) `=intsqrt(3-2(x+(1)/(2)))dx=intsqrt((7)/(2)-2(x+(1)/(2))^(2))dx` `=sqrt2 int sqrt((sqrt7)/(2)^(2)-(x+(1)/(2))^(2))dx` माना `(sqrt7)/(2)=a` तथा `x+(1)/(2)= t therefore dx=dt` `I=sqrt2 int sqrt((a^(2)-t^(2)))dt` `=sqrt2[(t)/(2)sqrt((a^(2)-t^(2)))+(a^(2))/(2)sin^(-1)((t)/(a))]` `=sqrt2[((x+(1)/(2))/(2))sqrt((7)/(4)-(x+(1)/(2))^(2))+(7)/(8)sin^(_1)((x+(1)/(2))/((sqrt7)/(2)))]` `=(2x+1)/(2sqrt2)sqrt(((7)/(4)-x^(2)-x-(1)/(4)))+(7)/(4sqrt2)sin^(-1)((2x+1)/(sqrt7))` `=(2x+1)/(2sqrt2)[(3)/(2)-x^(2)-x]^((1)/(2))+(7)/(4sqrt2)sin^(-1)((2x+1)/(sqrt7))` `=(2x+1)/(4)[3-2x^(2)-2x]^((1)/(2))+(7)/(4sqrt2)sin^(-1)((2x+1)/(sqrt7))` |
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