352. Some definite integrals.

by Virgil Nicula, Jul 30, 2012, 3:08 PM

PP1. Calculate $I=\int_0^{n\pi}\frac{1}{\cos^2{x} + 4\sin{2x} + 4}\ \mathrm{dx}$ , where $n\in\mathbb N^*$ .

Proof. Let $f(x)=\frac{1}{\cos^2{x} + 4\sin{2x} + 4}$ , where $x\in\mathbb R$ . Thus, $(\forall )\ x\in\mathbb R\ ,\ f(x+\pi )=f(x)$ , i.e. $f$ has the period $T=\pi$ . Thus, $I=\int_0^{n\pi}f(x)\ \mathrm{dx}= $ $ n\cdot \int_0^{\pi}f(x)\ \mathrm{dx}= $

$n\cdot \int_{-\frac{\pi}{2}}^{\frac {\pi}{2}}f\left(\frac {\pi}{2}-x\right)\ \mathrm{dx}= $ $n\cdot \int_{0}^{\frac {\pi}{2}}\left[f\left(\frac {\pi}{2}-x\right)+f\left(\frac {\pi}{2}+x\right)\right]\ \mathrm{dx}= $ $n\cdot \int_{0}^{\frac {\pi}{2}}\left(\frac {1}{\sin^2x+4\sin 2x+4}+\frac {1}{\sin^2x-4\sin 2x+4}\right) \mathrm{dx}=$

$\int_0^{\frac {\pi}2}\left[\frac {\sin^2x+\cos^2x}{\sin^2x+8\sin x\cos x+4\left(\sin^2x+\cos^2x\right)}+\frac {\sin^2x+\cos^2x}{\sin^2x-8\sin x\cos x+4\left(\sin^2x+\cos^2x\right)}\right]\ \mathrm{dx}=$

$\int_0^{\frac {\pi}2}\left[\frac {\tan^2x+1}{\tan^2x+8\tan x+4\left(\tan^2x+1\right)}+\frac {\tan^2x+1}{\tan^2x-8\tan x+4\left(\tan^2x+1\right)}\right]\ \mathrm{dx}=$

$\int_0^{\frac{\pi}2}\left[\frac 1{\tan^2x+8\tan x+4\left(\tan^2x+1\right)}+\frac 1{\tan^2x-8\tan x+4\left(\tan^2x+1\right)}\right]$ $\cdot \left(\tan x\right)'\ \mathrm{dx}\ \stackrel{(t=\tan x)}{=}$ $\int_0^{\infty}$ $\left[\frac 1{5t^2+8t+4}+\frac 1{5t^2-8t+4}\right]\ \mathrm{dt}=$

$\int_0^{\infty}$ $\left[\frac {(5t+4)'}{(5t+4)^2+4}+\frac {(5t-4)'}{(5t-4)^2+4}\right]\ \mathrm{dt}=$ $\left|\frac 12\cdot \left(\arctan\frac {5t+4}2+\arctan\frac {5t-4}2\right)\right|_0^{\infty}=\frac {\pi}2$ . In conclusion, $\boxed{\ I=\frac {n\pi}2\ }$ .


PP2. Prove that $I\equiv \int_0^1\frac {\ln (x+1)}{x^2+1}\ \mathrm{dx}=\frac {\pi\ln2}8$ and $\frac {\ln 2}2-\frac 12+\frac {\pi}8\ <\ I\ <\ \frac {\ln 2}2$ .

Proof. $\boxed{I=\int^1_0 \frac{\ln (x+1)}{x^2+1}\ \mathrm{dx}=\frac {\pi\ln2}{8}}$ . Indeed, $I=\int^1_0 \ln(1+x)(\arctan x)'\ \mathrm{dx}=$ $\int^{\frac{\pi}{4}}_0 \ln (1+\tan x)\ \mathrm{dx}=$ $\int^{\frac{\pi}{4}}_0 \ln \left[1+\tan \left(\frac{\pi}{4} -x\right)\right] \ \mathrm{dx}=$ $\int^{\frac{\pi}{4}}_0\ln \frac{2}{1+\tan x}\ \mathrm{dx}=$

$\frac{\pi}{4}\ln 2 -I$ $\Longrightarrow I=\frac{\pi\ln 2}{8}$ . Prove easily that $x-\frac {x^2}2\le \ln (x+1)\le x\ .$ Therefore, $\int_0^1\frac {2x-x^2}{2\left(x^2+1\right)}\ \mathrm{dx}=$ $\frac 12\cdot\int_0^1\left(\frac {2x}{x^2+1}-1+\frac 1{x^2+1}\right)\ \mathrm{dx}\le $

$\int_0^1\frac {\ln (x+1)}{x^2+1}\le$ $ \int_0^1\frac x{x^2+1}\ \mathrm{dx}\implies$ $\frac {\ln 2}2-\frac 12+\frac {\pi}8\ <\ I\ <\ \frac {\ln 2}2$ . See here
This post has been edited 17 times. Last edited by Virgil Nicula, Nov 17, 2015, 8:35 AM

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  • wow... i am lost.

    369302 views!

    -piphi

    by piphi, Jun 10, 2020, 11:44 PM

  • That was a lot! But, really good solutions and format! Nice blog!!!! :)

    by CSPAL, May 27, 2020, 4:17 PM

  • impressive :D
    awesome. 358,000 visits?????

    by OlympusHero, May 14, 2020, 8:43 PM

72 shouts
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  • Joined: Jun 22, 2005
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  • Total entries: 456
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