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Thread: more integrals

  1. #1
    Super Member Random Variable's Avatar
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    more integrals

    $\displaystyle \int \sin (101x) \sin^{99} x \ dx $

    $\displaystyle \int_{0}^{1} \frac{\tan^{-1} x}{1+x} \ dx $
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  2. #2
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    If $\displaystyle y=\sin(100x)\sin^{100}x$ then $\displaystyle y'=100\cos(100x)\sin^{100}x+100\sin(100x)\sin^{99} x\,\cos x$ $\displaystyle =100[\sin(100x)\cos x+\cos(100x)\sin x]\sin^{99}x=100\sin(101x)\sin^{99}x$, which quickly leads to the answer.

    Consider the integral $\displaystyle I=\int_0^1\frac{\ln(1+x)}{1+x^2}\mathrm dx$. Substitute $\displaystyle x=\frac{1-u}{1+u}$ so that $\displaystyle \mathrm dx=-\frac2{(1+u)^2}\mathrm du$. Also $\displaystyle 1+x=\frac2{1+u}$ and $\displaystyle 1+x^2=\frac{2(1+u^2)}{(1+u)^2}$.

    Then $\displaystyle I=\int_0^1\frac{\ln\bigl(\frac2{1+u}\bigr)}{1+u^2} \mathrm du=\int_0^1\frac{\ln 2-\ln(1+u)}{1+u^2}\mathrm du=\int_0^1\frac{\ln 2}{1+u^2}\mathrm du-I=\frac\pi4\ln 2-I$. Thus $\displaystyle I=\frac\pi8\ln 2$.

    Thus $\displaystyle \int_0^1\frac{\tan^{-1}x}{1+x}\mathrm dx=\Bigl[\tan^{-1}x\ln(1+x)\Bigr]_0^1-\int_0^1\frac{\ln(1+x)}{1+x^2}\mathrm dx=\frac\pi4\ln2-I=\frac\pi8\ln2$.
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  3. #3
    Super Member Random Variable's Avatar
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    If $\displaystyle y=\sin(100x)\sin^{100}x$ then $\displaystyle y'=100\cos(100x)\sin^{100}x+100\sin(100x)\sin^{99} x\,\cos x$ $\displaystyle =100[\sin(100x)\cos x+\cos(100x)\sin x]\sin^{99}x=100\sin(101x)\sin^{99}x$, which quickly leads to the answer.
    How did you know that an antiderivative would have that general form (i.e. $\displaystyle A \sin (100x) \sin^{100}x + C$)?
    Last edited by Random Variable; Sep 3rd 2009 at 06:57 PM.
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  4. #4
    Super Member Random Variable's Avatar
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    This is what I would have done for the first integral:

    $\displaystyle \int \sin (101x) \sin^{99}x \ dx = \text{Im} \int e^{i101x}\Big(\frac{e^{ix}-e^{-ix}}{2i}\Big)^{99} dx $

    $\displaystyle \int e^{i101x}\Big(\frac{e^{ix}-e^{-ix}}{2i}\Big)^{99} dx = \frac{i}{2^{99}} \int e^{i101x}(e^{ix}-e^{-ix})^{99} \ dx = $ $\displaystyle \frac{i}{2^{99}} \int e^{2ix}(e^{2ix}-1)^{99} \ dx $

    let $\displaystyle u = e^{2ix}-1 $

    then $\displaystyle du = 2ie^{2ix} $

    $\displaystyle = \frac{1}{2^{100}} \int u ^{99} \ du = \frac{1}{2^{100}} \frac{u^{100}}{100} + C = \frac{1}{2^{100}}\frac{(e^{2ix}-1)^{100}}{100} + C$

    $\displaystyle = \frac{1}{100} \Big(\frac{e^{2ix}-1}{2}\Big)^{100} + C = \frac{1}{100} \ e^{i100x} \Big(\frac{e^{ix}-e^{-ix}}{2}\Big)^{100} + C$

    $\displaystyle = \frac{1}{100} \ e^{i100x} \Big(\frac{e^{ix}-e^{-ix}}{2i}\Big)^{100} + C = \frac{e^{i100x} \sin^{100} x}{100} + C$


    $\displaystyle \text{Im} \Big(\frac{e^{i100x} \sin^{100} x}{100}\Big) = \frac{\sin (100x) \sin^{100} x}{100}$

    so $\displaystyle \int \sin (101x) \sin^{99}x \ dx = \frac{\sin (100x) \sin^{100} x}{100} + C$
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