integrate-really difficult

**sammy nqh** · February 27th 2009, 06:28 AM

Guys!!! please help me with this. it is quite unsolvable.

**Abu-Khalil** · February 27th 2009, 07:35 AM

$\int_\frac{\pi}{6}^\frac{\pi}{3}\log(x\tan x)dx=\left.x\log(x\tan x)\right|^\frac{\pi}{3}_\frac{\pi}{6}-\underbrace{\int_\frac{\pi}{6}^\frac{\pi}{3}\frac{ x}{x\tan x}(\tan x+x\sec^2 x)dx}_{I_1}$ and $I_1=\int_\frac{\pi}{6}^\frac{\pi}{3}1+x\sec^2 x\cot x dx$ .

Now let $u=\tan x \Rightarrow du=\sec^2 x dx$ then $\int_\frac{\pi}{6}^\frac{\pi}{3}x\sec^2 x\cot x dx=\int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}}\frac{\arct an u}{u}du=\left.\log u \arctan u\right|^{\sqrt{3}}_{\frac{1}{\sqrt{3}}}=\underbra ce{\int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}}\frac{\log u}{1+u^2}du}_{=0}.$

For the last integral, see here http://www.mathhelpforum.com/math-he...lp-needed.html

**Nacho** · February 27th 2009, 07:40 AM

$ \begin{gathered} \int {\ln \left( {x\tan x} \right)dx} = \int {\ln \left( x \right)dx + \int {\ln \left( {\tan x} \right)dx} } \hfill \\ {\text{Integrating by part in both integrals}}{\text{, tanking }}u{\text{ the logarythm}} \hfill \\ x\ln \left| x \right| - x + x\ln \left| x \right| + \underbrace {\int { - x \cdot \frac{1} {{\tan x}} \cdot \sec ^2 xdx} }_I \hfill \\ \end{gathered} $

$ \begin{gathered} I = \int {\frac{{ - x}} {{\cos x\sin x}}dx\overbrace = ^{u = \cos x}\int {\frac{{\arccos \left( u \right)}} {{u\left( {1 - u^2 } \right)}}} } du \hfill \\ \pi - \arccos \left( { - u} \right) = \arccos \left( u \right) \hfill \\ \end{gathered} $

$ \begin{gathered} I = \int {\frac{{\pi - \arccos \left( { - u} \right)}} {{u\left( {1 - u^2 } \right)}}} du\overbrace = ^{ - u = z}\int {\frac{{\pi - \arccos \left( z \right)}} {{z\left( {1 - z^2 } \right)}}dz} \hfill \\ I = \int {\frac{\pi } {{z\left( {1 - z^2 } \right)}}dz} - \underbrace {\int {\frac{{\arccos \left( z \right)}} {{z\left( {1 - z^2 } \right)}}dz} }_I \hfill \\ \end{gathered} $

$ \begin{gathered} 2I = \pi \int {\frac{1} {z} + \frac{z} {{1 - z^2 }}dz = \pi \ln \left| z \right| - \frac{\pi } {2}\int {\frac{{ - 2z}} {{1 - z^2 }}dz} } \hfill \\ I = \frac{\pi } {2}\left\{ {\ln \left| z \right| - \ln \left| {1 - z^2 } \right|} \right\} + C \hfill \\ \end{gathered} $

Restoring and adding you have the answer

**Krizalid** · February 27th 2009, 08:19 AM

Originally Posted by Abu-Khalil

$\int_\frac{\pi}{6}^\frac{\pi}{3}x\sec^2 x\cot x dx=\int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}}\frac{\arct an u}{u}du$

I'll take it from here: in fact consider $\int_{\frac{1}{n}}^{n}{\frac{\arctan x}{x}\,dx},$ whereas by putting $x\to\frac1x$ this leads to $\int_{\frac{1}{n}}^{n}{\frac{1}{x}\arctan \left( \frac{1}{x} \right)\,dx},$ whereat by adding these (provided that for each $x\in(-\infty,0)\cup(0,\infty)$ is $\arctan x+\arctan \frac{1}{x}=\frac{\pi }{2}$ ) we get $\frac{\pi }{2}\int_{\frac{1}{n}}^{n}{\frac{dx}{x}}=\pi \ln n$ and the integral equals $\frac{\pi }{2}\ln n$ which holds for all $n>0.$

Originally Posted by Nacho

$ \pi - \arccos \left( { - u} \right) = \arccos \left( u \right) $

Alas, this statement does not hold for all $u,$ besides, by differentiating your result, we don't get the integrand. (Actually, there's no primitive in terms of elementary functions.)

**sbcd90** · February 27th 2009, 10:02 AM

I=integrate{ln(xtanx) dx}[lim pi/6 to pi/3]

I=integrate{ln((pi/2-x)cotx) dx}[lim pi/6 to pi/3]

so,adding,2I=integrate{ln(x*(pi/2-x)) dx)[lim pi/6 to pi/3]

2I=integrate {lnx}[lim pi/6 to pi/3] + integrate {ln(pi/2-x)}[lim pi/6 to pi/3]

the rest is easy.

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