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Math Help - 2011 MIT Integration Bee Qualifying Test

  1. #16
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    7. \displaystyle \int{\sec^4{x}\tan^2{x}\,dx} = \int{\sec^2{x}\sec^2{x}\tan^2{x}\,dx}.


    Now using Integration by Parts with \displaystyle u = \sec^2{x} \implies du = 2\tan{x}\sec^2{x}\,dx and \displaystyle dv = \sec^2{x}\tan^2{x}\,dx \implies v = \frac{\tan^3{x}}{3} and the integral becomes

    \displaystyle \frac{\sec^2{x}\tan^3{x}}{3} - \int{\frac{2\tan^4{x}\sec^2{x}}{3}\,dx}

    \displaystyle = \frac{\sec^2{x}\tan^3{x}}{3} - \frac{2}{3}\int{\tan^4{x}\sec^2{x}\,dx}.

    Now make the substitution \displaystyle u = \tan{x} \implies du = \sec^2{x}\,dx and the integral becomes

    \displaystyle \frac{\sec^2{x}\tan^3{x}}{3} - \frac{2}{3}\int{u^4\,du}

    \displaystyle = \frac{\sec^2{x}\tan^3{x}}{3} - \frac{2}{3}\left(\frac{u^5}{5}\right) + C

    \displaystyle = \frac{\sec^2{x}\tan^3{x}}{3} - \frac{2\tan^5{x}}{15} + C.
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  2. #17
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    8. \displaystyle \int{\sqrt{\csc{x} - \sin{x}}\,dx}= \int{\sqrt{\frac{1}{\sin{x}} - \sin{x}}\,dx}

    \displaystyle = \int{\sqrt{\frac{1}{\sin{x}} - \frac{\sin^2{x}}{\sin{x}}}\,dx}

    \displaystyle = \int{\sqrt{\frac{1 - \sin^2{x}}{\sin{x}}}\,dx}

    \displaystyle = \int{\sqrt{\frac{\cos^2{x}}{\sin{x}}}\,dx}

    \displaystyle = \int{\frac{\cos{x}}{\sqrt{\sin{x}}}\,dx}

    \displaystyle = \int{\cos{x}(\sin{x})^{-\frac{1}{2}}\,dx}.


    Now make the substitution \displaystyle u = \sin{x} \implies du = \cos{x}\,dx and the integral becomes

    \displaystyle \int{u^{-\frac{1}{2}}\,du}

    \displaystyle = 2u^{\frac{1}{2}} + C

    \displaystyle = 2\sqrt{\sin{x}} + C.
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  3. #18
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    9. \displaystyle \int{\cos^6{x}\,dx} = \int{\left(\cos^2{x}\right)^3\,dx}

    \displaystyle = \int{\left(\frac{1}{2} + \frac{1}{2}\cos{2x}\right)^3\,dx}

    \displaystyle = \int{\frac{1}{8}\cos^3{2x} + \frac{3}{8}\cos^2{2x} + \frac{3}{8}\cos{2x} + \frac{1}{8}\,dx}

    \displaystyle = \int{-\frac{1}{16} [-2\cos{2x}(1 - \sin^2{2x})] + \frac{3}{16} + \frac{3}{16}\cos{4x} + \frac{3}{8}\cos{2x} + \frac{1}{8}\,dx}

    \displaystyle = \int{-2\cos{2x}\left(\frac{1}{16}\sin^2{2x} - \frac{1}{16}\right) + \frac{3}{16}\cos{4x} + \frac{3}{8}\cos{2x} + \frac{5}{16}\,dx}

    \displaystyle = \int{\frac{1}{16}u^2 -\frac{1}{16}\,du} + \frac{3}{64}\sin{4x} + \frac{3}{16}\sin{2x} + \frac{5}{16}x + C after making the substitution \displaystyle u = \sin{2x}

    \displaystyle = \frac{1}{48}u^3 -\frac{1}{16}u + \frac{3}{64}\sin{4x} + \frac{3}{16}\sin{2x} + \frac{5}{16}x + C

    \displaystyle = \frac{1}{48}\sin^3{2x} -\frac{1}{16}\sin{2x} + \frac{3}{64}\sin{4x} + \frac{3}{16}\sin{2x} + \frac{5}{16}x + C

    which I'm sure scrubs up much nicer than does here...
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  4. #19
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    12. \displaystyle \int{\frac{dx}{\cos{x}}} = \int{\frac{\cos{x}\,dx}{\cos^2{x}}}

    \displaystyle = \int{\frac{\cos{x}\,dx}{1 - \sin^2{x}}}

    \displaystyle = \int{\frac{\cos{x}\,dx}{(1 - \sin{x})(1 + \sin{x})}}.


    Now make the substitution \displaystyle u = \sin{x} \implies du = \cos{x}\,dx and the integral becomes

    \displaystyle \int{\frac{du}{(1 - u)(1 + u)}}.


    Now using Partial Fractions:

    \displaystyle \frac{A}{1 - u} + \frac{B}{1 + u} = \frac{1}{(1 - u)(1 + u)}

    \displaystyle \frac{A(1 + u) + B(1 - u)}{(1 - u)(1 + u)} = \frac{1}{(1 - u)(1 + u)}

    \displaystyle A(1 + u) + B(1 -  u) = 1

    \displaystyle A + Au + B - Bu = 1

    \displaystyle (A - B)u + A + B = 0u + 1

    \displaystyle A-B = 0 and \displaystyle A + B = 1

    \displaystyle A = B \implies A = \frac{1}{2}, B = \frac{1}{2}.


    So \displaystyle \int{\frac{du}{(1 - u)(1 + u)}}= \int{\frac{1}{2(1 - u)} + \frac{1}{2(1 + u)}\,du}

    \displaystyle = -\frac{1}{2}\ln{|1 - u|} + \frac{1}{2}\ln{|1 + u|} + C

    \displaystyle = \frac{1}{2}\ln{\left|\frac{1+u}{1 - u}\right|} + C

    \displaystyle = \frac{1}{2}\ln{\left|\frac{1 + \sin{x}}{1 - \sin{x}}\right|} + C

    \displaystyle = \frac{1}{2}\ln{\left|\frac{(1 + \sin{x})^2}{(1 - \sin{x})(1 + \sin{x})}\right|} + C

    \displaystyle = \frac{1}{2}\ln{\left|\frac{(1 + \sin{x})^2}{\cos^2{x}}\right|} + C

    \displaystyle = \ln{\left|\left(\frac{1 + \sin{x}}{\cos{x}}\right)^2\right|^{\frac{1}{2}}} + C

    \displaystyle = \ln{\left|\frac{1 + \sin{x}}{\cos{x}}\right|} + C

    \displaystyle = \ln{|\sec{x} + \tan{x}|} + C.
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  5. #20
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    17. \displaystyle \int{\frac{dx}{2 + e^x}} = \int{\frac{e^x\,dx}{2e^x + e^{2x}}}.

    Now let \displaystyle u = e^x \implies du = e^x\,dx and the integral becomes

    \displaystyle \int{\frac{du}{2u + u^2}}

    \displaystyle = \int{\frac{du}{u(2 + u)}}.


    Now using Partial Fractions:

    \displaystyle \frac{A}{u} + \frac{B}{2 + u} = \frac{1}{u(2 + u)}

    \displaystyle \frac{A(2 +u) + Bu}{u(2 + u)} = \frac{1}{u(2 + u)}

    \displaystyle A(2 + u) + Bu = 1

    \displaystyle 2A + Au + Bu = 1

    \displaystyle (A + B)u + 2A = 0u + 1

    \displaystyle A + B = 0 and \displaystyle 2A = 1

    \displaystyle A = \frac{1}{2} and \displaystyle B = -\frac{1}{2}.


    So \displaystyle \int{\frac{du}{u(2+u)}} = \int{\frac{1}{2u} - \frac{1}{2(2 + u)}\,du}

    \displaystyle = \frac{1}{2}\ln{|u|} - \frac{1}{2}\ln{|2+u|} + C

    \displaystyle = \frac{1}{2}\ln{\left|\frac{u}{2 + u}\right|} + C

    \displaystyle = \frac{1}{2}\ln{\left|\frac{e^x}{2 + e^x}\right|} + C
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  6. #21
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    19. \displaystyle \int{\frac{4x\,dx}{1 - x^4}} = 2\int{\frac{2x\,dx}{1 - (x^2)^2}}.

    Let \displaystyle u = x^2 \implies du = 2x\,dx and the integral becomes

    \displaystyle 2\int{\frac{du}{1 - u^2}}

    \displaystyle = 2\int{\frac{du}{(1-u)(1+u)}}

    \displaystyle = 2\int{\frac{1}{2(1 + u)} + \frac{1}{2(1 - u)}\,du} after using Partial Fractions

    \displaystyle = 2\left(\frac{1}{2}\ln{|1 + u|} - \frac{1}{2}\ln{|1-u|}\right) + C

    \displaystyle = \ln{|1 + u|} - \ln{|1 - u|} + C

    \displaystyle = \ln{\left|\frac{1+u}{1 - u}\right|} + C

    \displaystyle = \ln{\left|\frac{1 + x^2}{1 - x^2}\right|} + C.
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  7. #22
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    Prove It:
    for #17, If you multiply and devide by e^{-x} it will be solved in 1 second.
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  8. #23
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    21. \displaystyle \int_0^6{\sqrt{6x - x^2}\,dx} = \int_0^6{\sqrt{-(x^2 - 6x)}\,dx}

    \displaystyle = \int_0^6{\sqrt{-\left[x^2 - 6x + (-3)^2 - (-3)^2\right]}\,dx}

    \displaystyle = \int_0^6{\sqrt{-\left[(x - 3)^2 - 9\right]}\,dx}

    \displaystyle = \int_0^6{\sqrt{9 - (x - 3)^2}\,dx}.


    Note that \displaystyle y = \sqrt{9 - (x - 3)^2} is a semicircle centred at \displaystyle (3,0) of radius \displaystyle 3.

    So this integral is the area of that semicircle.

    \displaystyle \int_0^6{\sqrt{9-(x-3)^2}\,dx} = \frac{1}{2}\cdot \pi \cdot 3^2

    \displaystyle = \frac{9\pi}{4}.
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  9. #24
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    \displaystyle \int_0^1{\frac{x^3 - 3x^2 + 3x - 1}{x^4 + 4x^3 + 6x^2 + 4x + 1}\,dx} = \int_0^1{\frac{(x - 1)^3}{(x + 1)^4}}.

    Now let \displaystyle u = x+1 \implies x - 1 = u - 2 and \displaystyle du = dx.

    When \displaystyle x = 0, u = 1 and when \displaystyle x = 1, u = 2.

    The integral becomes

    \displaystyle \int_1^2{\frac{(u-2)^3}{u^4}\,du}

    \displaystyle = \int_1^2{\frac{u^3 - 6u^2 + 12u - 8}{u^4}\,du}

    \displaystyle = \int_1^2{u^{-1} - 6u^{-2} + 12u^{-3} - 8u^{-4}\,du}

    \displaystyle = \left[\ln{|u|} + 6u^{-1} - 6u^{-2} + \frac{8u^{-3}}{3}\right]_1^2

    \displaystyle = \left(\ln{|2|} + 3 - \frac{3}{2} + \frac{1}{3}\right) - \left(\ln{|1|} + 6 - 6 + \frac{8}{3}\right)

    \displaystyle = \ln{(2)} - \frac{5}{6}.
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  10. #25
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    25. \displaystyle \int{\sqrt{\frac{1-x}{1+x}}\,dx} = \int{\sqrt{\frac{(1 - x)(1-x)}{(1 + x)(1 - x)}}\,dx}

    \displaystyle = \int{\sqrt{\frac{(1 - x)^2}{1 - x^2}}\,dx}

    \displaystyle = \int{\frac{1 - x}{\sqrt{1 - x^2}}\,dx}.


    Now let \displaystyle x = \sin{\theta} \implies dx = \cos{\theta}\,d\theta and the integral becomes

    \displaystyle \int{\frac{1 - \sin{\theta}}{\sqrt{1 - \sin^2{\theta}}}\,\cos{\theta}\,d\theta}

    \displaystyle = \int{\frac{(1 - \sin{\theta})\cos{\theta}}{\cos{\theta}}\,d\theta}

    \displaystyle = \int{1 - \sin{\theta}\,d\theta}

    \displaystyle = \theta + \cos{\theta} + C

    \displaystyle = \theta+ \sqrt{1 - \sin^2{\theta}} + C

    \displaystyle = \arcsin{x}+ \sqrt{1 - x^2} + C.
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  11. #26
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    23. \displaystyle \int{x\,e^{e^{x^2} + x^2}\,dx} = \frac{1}{2}\int{e^{e^{x^2} + x^2}\cdot 2x\,dx}.

    Let \displaystyle u = x^2 \implies du = 2x\,dx and the integral becomes

    \displaystyle \frac{1}{2}\int{e^{e^u + u}\,du}

    \displaystyle = \frac{1}{2}\int{e^{e^u}\cdot e^u\,du}.


    Now let \displaystyle v = e^u \implies dv = e^u\,du and the integral becomes

    \displaystyle \frac{1}{2}\int{e^v\,dv}

    \displaystyle = \frac{1}{2}e^v + C

    \displaystyle = \frac{1}{2}e^{e^u} + C

    \displaystyle = \frac{1}{2}e^{e^{x^2}} + C.
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  12. #27
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    10. \displaystyle \int{\frac{dx}{1 + 2x^2 + x^4}} = \int{\frac{dx}{(1 + x^2)^2}}


    Let \displaystyle x = \tan{\theta} \implies dx = \sec^2{\theta}\,d\theta and the integral becomes

    \displaystyle \int{\frac{\sec^2{\theta}\,d\theta}{(1 + \tan^2{\theta})^2}}

    \displaystyle = \int{\frac{\sec^2{\theta}\,d\theta}{\sec^4{\theta}  }}

    \displaystyle = \int{\frac{d\theta}{\sec^2{\theta}}}

    \displaystyle = \int{\cos^2{\theta}\,d\theta}

    \displaystyle = \int{\frac{1}{2} + \frac{1}{2}\cos{2\theta}\,d\theta}

    \displaystyle =\frac{1}{2}\theta + \frac{1}{4}\sin{2\theta} + C

    \displaystyle =\frac{1}{2}\theta + \frac{1}{4}\left(\frac{2\tan{\theta}}{1 + \tan^2{\theta}}\right) + C

    \displaystyle = \frac{1}{2}\theta + \frac{\tan{\theta}}{2(1 + \tan^2{\theta})} + C

    \displaystyle = \frac{1}{2}\arctan{x} + \frac{x}{2(1 + x^2)} + C.
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  13. #28
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    Just out of interest, having never been in an Integral Bee before, how many do you have to get right in the time limit to qualify?
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  14. #29
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    Quote Originally Posted by Prove It View Post
    Just out of interest, having never been in an Integral Bee before, how many do you have to get right in the time limit to qualify?
    Don't know, I'm wondering that too. It might depend on the boundaries of that particular year. I think with sufficient training and wise choice of questions, most people can get more than ten right within the time limit. But you have to keep in mind that they are not doing them in the regular way -- with the explicit substitutions and all. For example, it's much more efficient to do # 10 as follows:

    \begin{aligned} & \int{\sqrt{\frac{1-x}{1+x}}\,dx} = \int{\frac{1 - x}{\sqrt{1 - x^2}}\,dx} \\& = \int \frac{1}{\sqrt{1-x^2}}\;{dx}-\int\frac{(x^2)'}{2\sqrt{1-x^2}}\;{dx} \\& \sin^{-1}{x}+\sqrt{1-x^2}+k. \end{aligned}
    Last edited by TheCoffeeMachine; January 27th 2011 at 05:36 AM.
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  15. #30
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    By the way, is a head-to-head video of the final round (not sure which year).
    It took the winner about 30 seconds to do \int \frac{x^4}{1-x^2}\;{dx}. What do you think?
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