## 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 $\displaystyle\int_{-\pi/2}^{\pi/2}\frac{1}{2007^x+1}\cdot\frac{\sin^{2008}x}{\sin^{2008}x+\cos^{2008}x}\;dx$

1. mathgarage says:

(1) solution
$\displaystyle\int\sin(101x)\sin^{99}x\;dx$
$=\displaystyle\int(\sin(100x)\cos(x)+\cos(100x)\sin(x))\sin^{99}x\;dx$
$=\displaystyle\int(\sin(100x)\cos(x)\sin^{99}x+\cos(100x)\sin^{100}x)\;dx$
$=\displaystyle\frac{\sin^{100}x\sin(100x)}{100}+C$

2. 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 $\displaystyle\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A \tan B}$
$=\displaystyle\frac{\pi}{4}\log{2}-\int_0^{\pi/4}\log\left(\frac{1-\tan\theta}{1+\tan\theta}+1\right)\;d\theta$
$=\displaystyle\frac{\pi}{4}\log{2}-\int_0^{\pi/4}\log\frac{2}{1+\tan\theta}\;d\theta$
$=\displaystyle\frac{\pi}{4}\log{2}-\int_0^{\pi/4}(\log 2 - \log(1+\tan\theta))\;d\theta$
$=\displaystyle\frac{\pi}{4}\log{2}-\frac{1}{2}\log{2}\left(\frac{\pi}{4}\right)$
$=\displaystyle\frac{\pi}{8}\log{2}$

3. 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 $n>1$
$=\displaystyle-\frac{1}{2} \left(\frac{1}{1-n}-\frac{1}{n+1}\right)$
$=\displaystyle\frac{1}{2} \left(\frac{1}{n-1}+\frac{1}{n+1}\right)$
$=\displaystyle\frac{n}{n^2-1}$

4. 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

5. 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 $x=\sin\theta$
$=\displaystyle\frac{1}{2}\log{2}-\frac{1}{2}+\frac{1}{2}\int_0^{\pi/2}\;d\theta$
$=\displaystyle\frac{1}{2}\log{2}-\frac{1}{2}+\frac{\pi}{4}$

6. mathgarage says:

(6) solution
$\displaystyle\int_{-\pi/2}^{\pi/2}\frac{1}{2007^x+1}\cdot\frac{\sin^{2008}x}{\sin^{2008}x+\cos^{2008}x}dx$
$=\displaystyle\int_0^{\pi/2}\frac{1}{2007^x+1}\cdot\frac{\sin^{2008}x}{\sin^{2008}x+\cos^{2008}x} + \frac{1}{2007^{-x}+1}\cdot\frac{\sin^{2008}(-x)}{\sin^{2008}(-x)+\cos^{2008}(-x)}dx$
$=\displaystyle\int_0^{\pi/2}\frac{1}{2007^x+1}\cdot\frac{\sin^{2008}x}{\sin^{2008}x+\cos^{2008}x} + \frac{2007^x}{2007^x+1}\cdot\frac{\sin^{2008}x}{\sin^{2008}x+\cos^{2008}x}dx$
$=\displaystyle\int_0^{\pi/2}\frac{\sin^{2008}x}{\sin^{2008}x+\cos^{2008}x} dx$
$=\displaystyle\int_0^{\pi/2}\frac{\sin^{2008}(\pi/2-x)}{\sin^{2008}(\pi/2-x)+\cos^{2008}(\pi/2-x)} dx$
$=\displaystyle\int_0^{\pi/2}\frac{\cos^{2008}x}{\sin^{2008}x+\cos^{2008}x} dx$
$=\displaystyle\frac{1}{2}\int_0^{\pi/2}dx$
$=\displaystyle\frac{\pi}{4}$

7. 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??

• mathgarage says:

for Q1. because of the power rule. its more of experience
sinh is hyperbolic sine. By definition
$\sinh x = \displaystyle\frac{e^x-e^{-x}}{2}$
$\cosh x = \displaystyle\frac{e^x+e^{-x}}{2}$
you can figure out the rest

• sigmundlok says:

sor~ 我想問多個問題~ why我們需要有sinh,cosh這些野的出現??

• sigmundlok says:

應該金講~ why我們會去研究佢呢?

8. 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