The integral of dx/x^2 sqrt(x^2 + 9) from sqrt(3) to cube root of 3 Using trigonometric substitution
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Eric Bizzell
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Evaluate `int_(sqrt(3))^(root(3)(3))(dx)/(x^2sqrt(x^2+9))` :
Let `x=3tanu` . Then `dx=3sec^2udu` .
Now `sqrt(x^2+9)=sqrt(9tan^2u+9)=sqrt(9(tan^2u+1))=3secu` ; also `u=tan^(-1)(x/3)` .
So ignoring the limits of integration for the moment we have:
`int (dx)/(x^2sqrt(x^2+9))=int(3sec^2udu)/(9tan^2u(3secu))`
`=int(secudu)/(9tan^2u)`
`=int1/9cotucscudu`
This can be evaluated as:
`=-1/9cscu`
Substituting for u and evaluating at the limits of integration:
`=(-sqrt(x^2+9))/(9x)|_sqrt(3)^(root(3)(3))` ** `-csc(tan^(-1)(x/3))=sqrt(x^2+9)/x` **
`=(-sqrt(3^(2/3)+9))/(9root(3)(3))-(-sqrt(3+9))/(9sqrt(3))`
`~~-.0342`
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