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The integral of dx/x^2 sqrt(x^2 + 9) from sqrt(3) to cube root of 3Using trigonometric...

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beniceandrea | Student, Undergraduate | (Level 1) eNoter

Posted July 7, 2012 at 3:10 AM via web

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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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embizze | High School Teacher | (Level 1) Educator Emeritus

Posted July 7, 2012 at 5:13 AM (Answer #1)

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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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