InterviewSolution
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InterviewSolution
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Evaluate: `intsqrt(3-2x-2x^2) dx` |
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Answer» We have `(3-2x-2x^(2))=2*{(3)/(2)-x-x^(2)}` `=2*[(3)/(2)-(x^(2)+x+(1)/(4))+(1)/(4)]` `=2*[(7)/(4)-(x+(1)/(2))^(2)]=2*{((sqrt(7))/(2))^(2)-(x+(1)/(2))^(2)}`. `thereforesqrt(3-2x-2x^(2))=sqrt(2)*sqrt((sqrt(7)/(2))^(2)-(x+(1)/(2))^(2))` `rArrintsqrt(3-2x-2x^(2))dx=sqrt(2)*intsqrt((sqrt(7)/(2))^(2)-(x+(1)/(2))^(2))dx` `=sqrt(2)*intsqrt((sqrt(7)/(2))^(2)-t^(2))dt, "where"(x+(1)/(2))=t` and dx = dt `=sqrt(2)*{(t)/(2)*sqrt(7)/(4)-t^(2)+(7)/(8)" sin"^(-1)(t)/((sqrt(7)//2))}+C` `[becauseintsqrt(a^(2)-t^(2))dt=(t)/(2)sqrt(a^(2)-t^(2))+(a^(2))/(2)*"sin"^(-1)(t)/(a)+C]` `=sqrt(2)*{(1)/(2)(x+(1)/(2))sqrt((7)/(4)-(x+(1)/(2))^(2))+(7)/(8)"sin"^(-1)((x+(1)/(2)))/((sqrt(7)//2))}+C` `=sqrt(2){(1)/(4)(2x+1)*sqrt((3)/(2)-x^(2))+(7)/(8)sin^(-1)((2x+1)/(sqrt(7)))}+C` `=(1)/(4)(2x+1)sqrt(3-2x-2x^(2))+(7)/(4sqrt(2))sin^(-1)((2x+1)/(sqrt(7)))+C`. Integrals of the form `int(px+q)sqrt((ax^(2)+bx+c))dx` Method Let `(px+q)=A*(d)/(dx)(ax^(2)+bx+c)+B`. Find A and B . Then , we get the integrand which is easily integrable. |
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