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Math Help - trigonometry substitution

  1. #1
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    trigonometry substitution

    \int \frac {146}{x^2 \sqrt{36x^2 - 121}} dx

    I draw the 146 outside the integral and set the substitution as;

    6x = 11 sec t
    x = 11/6 sec t
    dx = 11/6 sec t tan t

    I have a feeling this is not the right approach because the result is an overly complicated integral once you apply the substitution . . .

    146 \int \frac {11 sec t tan t}{6x^2 \sqrt{36((11/6)sect)^2}}

    Is this correct?
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    Quote Originally Posted by Archduke01 View Post
    \int \frac {146}{x^2 \sqrt{36x^2 - 121}} dx

    I draw the 146 outside the integral and set the substitution as;

    6x = 11 sec t
    x = 11/6 sec t
    dx = 11/6 sec t tan t

    I have a feeling this is not the right approach because the result is an overly complicated integral once you apply the substitution . . .

    146 \int \frac {11 sec t tan t}{6x^2 \sqrt{36((11/6)sect)^2}}

    Is this correct?
    x^2=\frac{121}{36}sec^2t

    You also forgot the -121 under the square root
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  3. #3
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    Quote Originally Posted by Archduke01 View Post
    \int \frac {146}{x^2 \sqrt{36x^2 - 121}} dx

    I draw the 146 outside the integral and set the substitution as;

    6x = 11 sec t
    x = 11/6 sec t
    dx = 11/6 sec t tan t

    I have a feeling this is not the right approach because the result is an overly complicated integral once you apply the substitution . . .

    146 \int \frac {11 sec t tan t}{6x^2 \sqrt{36((11/6)sect)^2}}

    Is this correct?
    6x = 11\sec{t}

    36x^2 = 121\sec^2{t}

    x^2 = \frac{121}{36}\sec^2{t}

    dx = \frac{11}{6}\sec{t}\tan{t} \, dt<br />

    146 \int \frac{\frac{11}{6}\sec{t}\tan{t}}{\frac{121}{36}\s  ec^2{t}\sqrt{121\sec^2{t} - 121}} \, dt

    146 \int  \frac{\frac{1}{6}\sec{t}\tan{t}}{\frac{121}{36}\se  c^2{t}\sqrt{\sec^2{t}  - 1}} \, dt

    146 \cdot \frac{1}{6} \cdot \frac{36}{121} \int   \frac{\sec{t}\tan{t}}{\sec^2{t}\sqrt{\sec^2{t}   - 1}} \, dt

    \frac{876}{121} \int \cos{t} \, dt<br />

    finish?
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    Quote Originally Posted by Archduke01 View Post
    \int \frac {146}{x^2 \sqrt{36x^2 - 121}} dx

    I draw the 146 outside the integral and set the substitution as;

    6x = 11 sec t
    x = 11/6 sec t
    dx = 11/6 sec t tan t

    I have a feeling this is not the right approach because the result is an overly complicated integral once you apply the substitution . . .

    146 \int \frac {11 sec t tan t}{6x^2 \sqrt{36((11/6)sect)^2}}

    Is this correct?
    Hyperbolic substitution works well in this case.

    \int{\frac{146}{x^2\sqrt{36x^2 - 121}}\,dx} = \int{\frac{146}{6x^2\sqrt{x^2 - \frac{121}{36}}}\,dx}

     = \int{\frac{73}{3x^2\sqrt{x^2 - \frac{121}{36}}}\,dx}

    Now make the substitution x = \frac{11}{6}\cosh{t} so that dx = \frac{11}{6}\sinh{t}\,dt.

    The integral becomes:

    \int{\frac{73}{3\left(\frac{11}{6}\cosh{t}\right)^  2\sqrt{\left(\frac{11}{6}\cosh{t}\right)^2 - \frac{121}{36}}}\,\frac{11}{6}\sinh{t}\,dt}

     = \int{\frac{803\sinh{t}}{18\left(\frac{121}{36}\cos  h^2{t}\right)\sqrt{\frac{121}{36}\cosh^2{t} - \frac{121}{36}}}\,dt}

     = \int{\frac{803\sinh{t}}{\frac{121}{2}\cosh^2{t}\cd  ot \frac{11}{6}\sqrt{\cosh^2{t} - 1}}\,dt}

     = \int{\frac{803\sinh{t}}{\frac{1331}{12}\cosh^2{t}\  sqrt{\sinh^2{t}}}\,dt}

     = \frac{876}{121}\int{\frac{\sinh{t}}{\cosh^2{t}\sin  h{t}}\,dt}

     = \frac{876}{121}\int{\frac{1}{\cosh^2{t}}\,dt}

     = \frac{876}{121}\int{\textrm{sech}^2{t}\,dt}

     = \frac{876}{121}\tanh{t} + C


    Now remembering that we made the substitution x = \frac{11}{6}\cosh{t}

    \cosh{t} = \frac{6x}{11}.


    Also, since \tanh{t} = \frac{\sqrt{\cosh^2{t} - 1}}{\cosh{t}}

    \tanh{t} = \frac{\sqrt{\left(\frac{6x}{11}\right)^2 - 1}}{\frac{6x}{11}}

     = \frac{11\sqrt{\frac{36x^2 - 121}{121}}}{6x}

     = \frac{\sqrt{36x^2 - 121}}{6x}


    Therefore, our integral, which we got to

    \frac{876}{121}\tanh{t} + C

    becomes

    \frac{876}{121}\left(\frac{\sqrt{36x - 121)}}{6x}\right) + C

     = \frac{146\sqrt{36x - 121}}{121x} + C.
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