Some interesting integrals I saw online

Posted: August 2, 2012 in Mathematics

(1) Evaluate \displaystyle\int\sin(101x)\sin^{99}x\;dx

(2) Evaluate \displaystyle\int_0^1\frac{\arctan{x}}{x+1}\;dx

(3) For n>1, prove that \displaystyle\int_0^\infty\frac{dx}{(x+\sqrt{1+x^2})^n}=\frac{n}{n^2-1}

(4) If f is a bounded non-negative function, then show that
\displaystyle\int_0^\infty f\left(x+\frac{1}{x}\right)\frac{\log{x}}{x}dx=0

(5) Evaluate \displaystyle\int_0^1\log(\sqrt{1-x}+\sqrt{1+x})\;dx

(6) Evaluate \displaystyle\int_{-\pi/2}^{\pi/2}\frac{1}{2007^x+1}\cdot\frac{\sin^{2008}x}{\sin^{2008}x+\cos^{2008}x}\;dx

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Comments
  1. mathgarage says:

    (2) solution
    \displaystyle\int_0^1\frac{\arctan{x}}{x+1}dx
    integration by part here let u=\frac{1}{x+1} and dv=\arctan x\;dx
    we obtain,
    =\displaystyle\arctan{x}\log(x+1)\bigg|_0^1-\int_0^1\frac{\log(x+1)}{x^2+1}dx
    here sub x=\tan\theta
    =\displaystyle\frac{\pi}{4}\log{2}-\int_0^{\pi/4}\log(\tan\theta+1)\;d\theta
    =\displaystyle\frac{\pi}{4}\log{2}-\int_0^{\pi/4}\log\left(\tan(\frac{\pi}{4}-\theta)+1\right)\;d\theta
    apply the trig identity \displaystyle\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A \tan B}
    =\displaystyle\frac{\pi}{4}\log{2}-\int_0^{\pi/4}\log\left(\frac{1-\tan\theta}{1+\tan\theta}+1\right)\;d\theta
    =\displaystyle\frac{\pi}{4}\log{2}-\int_0^{\pi/4}\log\frac{2}{1+\tan\theta}\;d\theta
    =\displaystyle\frac{\pi}{4}\log{2}-\int_0^{\pi/4}(\log 2 - \log(1+\tan\theta))\;d\theta
    =\displaystyle\frac{\pi}{4}\log{2}-\frac{1}{2}\log{2}\left(\frac{\pi}{4}\right)
    =\displaystyle\frac{\pi}{8}\log{2}

  2. mathgarage says:

    (3) solution
    \displaystyle\int_0^\infty\frac{dx}{(x+\sqrt{1+x^2})^n}
    let x=\sinh\theta
    =\displaystyle\int_0^\infty\frac{\cosh\theta \;d\theta}{(\sinh\theta+\cosh\theta)^n}
    =\displaystyle\frac{1}{2}\int_0^\infty\frac{e^\theta+e^{-\theta}}{e^{n\theta}}d\theta
    =\displaystyle\frac{1}{2}\int_0^\infty \left(e^{(1-n)\theta}+e^{-(n+1)\theta}\right)d\theta
    =\displaystyle\frac{1}{2} \left(\frac{e^{(1-n)\theta}}{1-n}+\frac{e^{-(n+1)\theta}}{-(n+1)}\right)\bigg|_{\theta=0}^\infty
    since n>1
    =\displaystyle-\frac{1}{2} \left(\frac{1}{1-n}-\frac{1}{n+1}\right)
    =\displaystyle\frac{1}{2} \left(\frac{1}{n-1}+\frac{1}{n+1}\right)
    =\displaystyle\frac{n}{n^2-1}

  3. mathgarage says:

    (4) solution
    \displaystyle\int_0^\infty f\left(x+\frac{1}{x}\right) \frac{\log x}{x}\;dx
    Let w=\log x, then we have
    =\displaystyle\int_{-\infty}^\infty f(e^w+e^{-w})w\;dw=0
    Since \displaystyle f(e^w+e^{-w})w is odd

  4. mathgarage says:

    (5) solution
    integration by parts,
    let \displaystyle u=\log(\sqrt{1-x}+\sqrt{1+x})
    \displaystyle du=\frac{1}{\sqrt{1-x}+\sqrt{1+x}}\left(\frac{-1}{2\sqrt{1-x}}+\frac{1}{2\sqrt{1+x}}\right)\;dx
    \displaystyle du=\frac{1}{2}\cdot\frac{\sqrt{1-x}-\sqrt{1+x}}{-2x}\cdot\frac{\sqrt{1-x}-\sqrt{1+x}}{\sqrt{1-x^2}}\;dx
    \displaystyle du=-\frac{2-2\sqrt{1-x^2}}{4x\sqrt{1-x^2}}dx
    \displaystyle du=\frac{\sqrt{1-x^2}-1}{2x\sqrt{1-x^2}}dx
    then
    \displaystyle\int_0^1\log(\sqrt{1-x}+\sqrt{1+x})\;dx
    =\displaystyle x\log(\sqrt{1-x}+\sqrt{1+x})\bigg|_0^1-\frac{1}{2}\int_0^1\frac{\sqrt{1-x^2}-1}{\sqrt{1-x^2}}\;dx
    =\displaystyle\frac{1}{2}\log{2}-\frac{1}{2}+\frac{1}{2}\int_0^1\frac{1}{\sqrt{1-x^2}}\;dx
    let x=\sin\theta
    =\displaystyle\frac{1}{2}\log{2}-\frac{1}{2}+\frac{1}{2}\int_0^{\pi/2}\;d\theta
    =\displaystyle\frac{1}{2}\log{2}-\frac{1}{2}+\frac{\pi}{4}

  5. sigmundlok says:

    Q.1 的做法太神了…係由答案做返轉頭 定係你睇到d咩令你知道要金做…而且話時話 Q.3 所let的sinh x 係咩?? 都係sin,cos之類的野??

  6. mathgarage says:

    We use sin and cos to parametrize circle and ellipses. For hyperbola, we have sinh and cosh.
    For details, you can read http://en.wikipedia.org/wiki/Hyperbolic_function

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