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    Problem 413*

    Find

    \displaystyle \int_0^\frac{\pi}{2} \frac{1}{1+\tan^\alpha(x)}dx.
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    (Original post by henpen)
    Problem 413*

    Find

    \displaystyle \int_0^\frac{\pi}{2} \frac{1}{1+\tan^\alpha(x)}dx.
    pi/4
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    (Original post by henpen)
    Problem 413*

    Find

    \displaystyle  \int_0^\frac{\pi}{2} \frac{1}{1+\tan^\alpha(x)}dx.
    Solution 413*

    Spoiler:
    Show
    \text{Let } I = \displaystyle \int_{0}^{\frac{\pi}{2}} \frac{dx}{1+\tan^{\alpha}x}

    \begin{aligned} x \mapsto \left( \frac{\pi}{2} + 0 \right) - x \implies I = \displaystyle \int_{0}^{\frac{\pi}{2}} \frac{dx}{1+\tan^{\alpha}x} & = \displaystyle \int_{0}^{\frac{\pi}{2}} \frac{dx}{1+\cot^{\alpha}x}

    \displaystyle \begin{aligned} \therefore 2I & = \int_{0}^{\frac{\pi}{2}} \left( \frac{dx}{1+\tan^{\alpha}x} + \frac{1}{1+\cot^{\alpha}x} \right) dx = \int_{0}^{\frac{\pi}{2}} dx = \Big[ x \Big]_{0}^{\pi/2} = \frac{\pi}{2} \implies I = \frac{\pi}{4} \end{aligned}
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    (Original post by henpen)
    Problem 413*

    Find

    \displaystyle \int_0^\frac{\pi}{2} \frac{1}{1+\tan^\alpha(x)}dx.
    ​Solution 413
    I=\displaystyle \int_0^\frac{\pi}{2} \frac{1}{1+\tan^\alpha(x)}dx \\ \\ I=\displaystyle \int_0^\frac{\pi}{2} \frac{\cos^\alpha(x)}{ \cos^\alpha(x)+\sin^\alpha(x)}dx  = \displaystyle \int_0^\frac{\pi}{2} \frac{\sin^\alpha(x)}{ \cos^\alpha(x)+\sin^\alpha(x)}dx \\ \\ 2I=\displaystyle \int_0^\frac{\pi}{2}dx= \dfrac{\pi}{2} \\ \\ I=\dfrac{\pi}{4}
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    (Original post by Khallil)
    .
    Beat me to it aha
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    (Original post by Flauta)
    Beat me to it aha
    It was a 1* question. How could I not jump at the chance?!
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    (Original post by Khallil)
    It was a 1* question. How could I not jump at the chance?!
    Indeed. I'm hoping that someone else will post an unnecessarily complicated solution now
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    Hurrah, I've reignited this thread!

    Problem 214**

    Find

    \displaystyle \int_0^1 \frac{x^2-1}{\log(x)}dx.
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    (Original post by henpen)
    Hurrah, I've reignited this thread!

    Problem 214**

    Find

    \displaystyle \int_0^1 \frac{x^2-1}{\log(x)}dx.
    log3
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    (Original post by henpen)
    Hurrah, I've reignited this thread!

    Problem 214**

    Find

    \displaystyle \int_0^1 \frac{x^2-1}{\log(x)}dx.

    (Original post by Tarquin Digby)
    log3
    Any hints?
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    (Original post by CD315)
    Any hints?
    Differentiate \displaystyle \int_0^1 \frac{x^a-1}{\log{x}}\,\mathrm{d}x w.r.t a.
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    Problem 216**

    Find

     \displaystyle \int_0^\infty \frac{dx}{a\cos^2(x)+b\sin^2(x)}  .

    Problem 217**

    Find

     \displaystyle \int_0^\pi \log(1-2\alpha \cos(x)+\alpha^2)dx, \alpha \in \mathbb{R}, |\alpha|\ne 1.
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    Problem 218*

    Evaluate
    \displaystyle \int_{-\pi}^{\pi} \frac{sin(nx)}{(1+2^x)sinx}dx.
    For natural n.

    A word of advice: * doesn't mean it's easy...
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    (Original post by henpen)
    Problem 216**

    Find

     \displaystyle \int_0^\infty \frac{dx}{a\cos^2(x)+b\sin^2(x)}  .

    Problem 217**

    Find

     \displaystyle \int_0^\pi \log(1-2\alpha \cos(x)+\alpha^2)dx, \alpha \in \mathbb{R}, |\alpha|\ne 1.
    216....

    (ab)^-0.5*arctan(pi/2*sqrt(b/a))

    ?
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    (Original post by Llewellyn)
    Problem 218*

    Evaluate
    \displaystyle \int_{-\pi}^{\pi} \frac{sin(nx)}{(1+2^x)sinx}dx.
    For natural n.

    A word of advice: * doesn't mean it's easy...
    u=-x, and im not suggesting its an odd function

    in thi case * did mean easy
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    (Original post by Tarquin Digby)
    I thought you're not allowed to integrate over a discontinuity? (denominator is 0 when x=0)
    u can in certain cases, you have to see what it comes out as and then youll know
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    (Original post by henpen)
    Problem 413*

    Find

    \displaystyle \int_0^\frac{\pi}{2} \frac{1}{1+\tan^\alpha(x)}dx.
    funny how everyone was astounded when LOTF got this right at the beginning of the thread (a specific case so slightly easier) but now everyones calling it an easy question

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    (Original post by Llewellyn)
    Problem 218*

    Evaluate
    \displaystyle \int_{-\pi}^{\pi} \frac{sin(nx)}{(1+2^x)sinx}dx.
    For natural n.

    A word of advice: * doesn't mean it's easy...
    Solution 218:

    Spoiler:
    Show


    Let  y = -x

    So,  \displaystyle \int_{-\pi}^{\pi} \frac{sin(nx)}{(1+2^x)sinx}dx = -\displaystyle \int_{\pi}^{-\pi} \frac{sin(nx)}{(1+2^{-x})sinx}dx = \displaystyle \int_{-\pi}^{\pi} 2^x \frac{sin(nx)}{(1+2^x)sinx}dx

    So,  2I = \displaystyle \int_{-\pi}^{\pi} \frac{sin(nx)}{sinx} \ dx

     I =  \displaystyle \int_{0}^{\pi} \frac{sin(nx)}{sinx} \ dx

     sin(nx) = cosxsin(n-1)x+sinxcos(n-1)x

     sin(n-2)x = cosxsin(n-1)x-sinxcos(n-1)x

    So,  I_n = \displaystyle \int_{0}^{\pi} cos(n-1)x + \displaystyle \int_{0}^{\pi} \frac{cosxsin(n-1)x}{sinx} \ dx

     I_{n-2} =  - \displaystyle \int_{0}^{\pi} cos(n-1)x + \displaystyle \int_{0}^{\pi} \frac{cosxsin(n-1)x}{sinx} \ dx

     I_n = I_{n-2}

    From this we can easily deduce that:

     I_{2n-1} = \pi

     I_{2n} = 0



    (Original post by Tarquin Digby)
    I thought you're not allowed to integrate over a discontinuity? (denominator is 0 when x=0)
    In this case you can because the function is well behaved as you approach that particular discontinuity.
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    (Original post by henpen)
    Problem 413*

    Find

    \displaystyle \int_0^\frac{\pi}{2} \frac{1}{1+\tan^\alpha(x)}dx.
    funny how everyone was astounded when LOTF got this right at the beginning of the thread (a specific case so slightly easier) but now everyones calling it an easy question

    (Original post by Llewellyn)
    Problem 218*

    Evaluate
    \displaystyle \int_{-\pi}^{\pi} \frac{sin(nx)}{(1+2^x)sinx}dx.
    For natural n.

    A word of advice: * doesn't mean it's easy...
    pi/0 for o/e
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    (Original post by DJMayes)
    Solution 218:

    Spoiler:
    Show


    Let  y = -x

    So,  \displaystyle \int_{-\pi}^{\pi} \frac{sin(nx)}{(1+2^x)sinx}dx = -\displaystyle \int_{\pi}^{-\pi} \frac{sin(nx)}{(1+2^{-x})sinx}dx = \displaystyle \int_{-\pi}^{\pi} 2^x \frac{sin(nx)}{(1+2^x)sinx}dx

    So,  2I = \displaystyle \int_{-\pi}^{\pi} \frac{sin(nx)}{sinx} \ dx

     I =  \displaystyle \int_{0}^{\pi} \frac{sin(nx)}{sinx} \ dx

     sin(nx) = cosxsin(n-1)x+sinxcos(n-1)x

     sin(n-2)x = cosxsin(n-1)x-sinxcos(n-1)x

    So,  I_n = \displaystyle \int_{0}^{\pi} cos(n-1)x + \displaystyle \int_{0}^{\pi} \frac{cosxsin(n-1)x}{sinx} \ dx

     I_{n-2} =  - \displaystyle \int_{0}^{\pi} cos(n-1)x + \displaystyle \int_{0}^{\pi} \frac{cosxsin(n-1)x}{sinx} \ dx

     I_n = I_{n-2}

    From this we can easily deduce that:

     I_{2n-1} = \pi

     I_{2n} = 0





    In this case you can because the function is well behaved as you approach that particular discontinuity.
    you used my hint well young padawan...maybe you should aim to apply to warwick
 
 
 
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