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Math Help - 2 Integral questions

  1. #1
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    2 Integral questions

    1.  \int \frac {1} {t^2 \sqrt {t^2 + 1}} dt

    2.  \int^{\pi}_{\pi/2} \cos(x)^2 \sin(x) dx
    Is this one's result -  \frac {1} {2} ?
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  2. #2
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    Quote Originally Posted by nirva

    2.  \int^{\pi}_{\pi/2} \cos(x)^2 \sin(x) dx
    Is this one's result -  \frac {1} {2} ?
    Let,
    u=\cos x then, \frac{du}{dx}=-\sin x Thus, you have,
    \int^{\pi}_{\pi/2} \cos^2 x  \sin x dx becomes,
    -\int^{-1}_{0} u^2 \frac{du}{dx} dx
    Thus,
    \left \int_{-1}^0 u^2 du=\frac{1}{3}u^3 \right| ^0_{-1}
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  3. #3
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    Quote Originally Posted by nirva
    1.  \int \frac {1} {t^2 \sqrt {t^2 + 1}} dt
    This one is more complicated.
    This involves trigonomteric substitution.

    You have,
    \int \frac{1}{t^2\sqrt{1+t^2}}dt
    Rationalize,
    \int \frac{\sqrt{1+t^2}}{t^2(1+t^2)}dt
    Let,
    u=\tan^{-1}t
    Then,
    \frac{du}{dt}=\frac{1}{1+t^2}
    \csc u=\frac{\sqrt{1+t^2}}{t}
    \cot u=\frac{1}{t}
    Thus, express integrand as,
    \int \frac{\sqrt{1+t^2}}{t}\cdot \frac{1}{t} \cdot \frac{1}{1+t^2} dt
    Thus, substitute the previous equations,
    \int \csc u \cot u \frac{du}{dx} dx
    Thus,
    \int \csc u\cot u du
    Now, just determine this new integral.

    Use trigonometric identities thus,
    \int \frac{1}{\sin u}\cdot \frac{\cos u}{\sin u} du
    Thus,
    \int \frac{\cos u}{\sin^2u}du
    Let, x=\sin u
    Then, \frac{dx}{du}=\cos u
    Express integrand as,
    \int \frac{1}{\sin^2u} \cdot \cos u du
    Make the above substitutions,
    \int \frac{1}{x^2} \frac{dx}{du} du
    Thus,
    \int x^{-2} dx=-\frac{1}{x}+C
    Substitute for x to get,
    -\frac{1}{\sin u}+C
    Thus,
    -\csc u+C
    Substitute for u to get,
    -\csc (\tan^{-1}(t))+C
    Thus,
    -\frac{\sqrt{1+t^2}}{t}+C
    Last edited by ThePerfectHacker; May 30th 2006 at 02:39 PM.
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  4. #4
    TD!
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    I find the same answer but I don't see why you do the rationalizing and rewriting if the substitution t = tan(u) works immediately.

    <br />
\int {\frac{{dt}}{{t^2 \sqrt {1 + t^2 } }}} \mathop  \to \limits_{dt = \sec ^2 u du}^{t = \tan u} \int {\frac{{\sec ^2 udu}}{{\tan ^2 u\sqrt {1 + \tan ^2 u} }}}  = \int {\frac{{\cos u}}{{\sin ^2 u}}du} <br />


    <br />
\int {\frac{1}{{\sin ^2 u}}d\left( {\sin u} \right) =  - \frac{1}{{\sin u}} + C \to }  - \frac{1}{{\sin \left( {\arctan t} \right)}} + C<br />

    Which is the same as your last step, yeilding:

    <br />
-\frac{\sqrt{1+t^2}}{t}+C<br />
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  5. #5
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    Quote Originally Posted by TD!
    I find the same answer but I don't see why you do the rationalizing and rewriting if the substitution t = tan(u) works immediately.

    <br />
\int {\frac{{dt}}{{t^2 \sqrt {1 + t^2 } }}} \mathop  \to \limits_{dt = \sec ^2 u du}^{t = \tan u} \int {\frac{{\sec ^2 udu}}{{\tan ^2 u\sqrt {1 + \tan ^2 u} }}}  = \int {\frac{{\cos u}}{{\sin ^2 u}}du} <br />


    <br />
\int {\frac{1}{{\sin ^2 u}}d\left( {\sin u} \right) =  - \frac{1}{{\sin u}} + C \to }  - \frac{1}{{\sin \left( {\arctan t} \right)}} + C<br />

    Which is the same as your last step, yeilding:

    <br />
-\frac{\sqrt{1+t^2}}{t}+C<br />
    Indeed I did see that! You method lacks mathematical beauty mine does not.
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  6. #6
    TD!
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    My beaty lies in compactness
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    Quote Originally Posted by TD!
    My beaty lies in compactness
    (Sigh!) I wish I was compact. Then I might be beautiful.

    -Dan
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  8. #8
    TD!
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    And not only beautiful, also closed and bounded!
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