InterviewSolution
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InterviewSolution
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Evaluate : (i) `intsqrt(9-x^(2))dx` (ii) `intsqrt(1-4x^(2))dx` (iii) `intsqrt(16-9x^(2))dx` |
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Answer» We know that `intsqrt(a^(2)-x^(2))dx=(x)/(2)sqrt(a^(2)-x^(2))+(a^(2))/(2)"sin"^(-1)(x)/(a)+C`. `therefore` (i) `intsqrt(9-x^(2))dx=intsqrt(3^(2)-x^(2))dx` `=(x)/(2)sqrt(9-x^(2))+(9)/(2)"sin"^(-1)(x)/(3)+C`. (ii) `sqrt(1-4x^(2))dx=2intsqrt(((1)/(4)-x^(2)))dx=2int{sqrt(((1)/(2))^(2)-x^(2))}dx` `=2[(x)/(2)sqrt((1)/(4)-x^(2))+(1)/(8)sin^(-1)((x)/((1//2)))]+C` `=(x)/(2)sqrt(1-4x^(2))+(1)/(4)sin^(-1)(2x)+C`. (iii) `intsqrt(16-9x^(2))dx=3int{sqrt(((16)/(9)-x^(2)))}dx=3int{sqrt(((4)/(3))^(2)-x^(2))}dx` `=3[(x)/(2)sqrt((16)/(9)-x^(2))+(8)/(9)"sin"^(-1)(x)/((4//3))]+C` `(x)/(2)sqrt(16-9x^(2))+(8)/(3)sin^(-1)((3x)/(4))+C`. |
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