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Math Help - integrate-really difficult

  1. #1
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    integrate-really difficult




    Guys!!! please help me with this. it is quite unsolvable.
    Attached Thumbnails Attached Thumbnails integrate-really difficult-new-acdsee6-bmp-image.bmp  
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  2. #2
    Member Abu-Khalil's Avatar
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    \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
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  3. #3
    Member Nacho's Avatar
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    <br />
\begin{gathered}<br />
  \int {\ln \left( {x\tan x} \right)dx}  = \int {\ln \left( x \right)dx + \int {\ln \left( {\tan x} \right)dx} }  \hfill \\<br />
  {\text{Integrating by part in both integrals}}{\text{, tanking }}u{\text{ the logarythm}} \hfill \\<br />
  x\ln \left| x \right| - x + x\ln \left| x \right| + \underbrace {\int { - x \cdot \frac{1}<br />
{{\tan x}} \cdot \sec ^2 xdx} }_I \hfill \\ <br />
\end{gathered} <br />

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

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

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

    Restoring and adding you have the answer
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  4. #4
    Math Engineering Student
    Krizalid's Avatar
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    Quote Originally Posted by Abu-Khalil View Post

    \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.

    Quote Originally Posted by Nacho View Post

    <br />
  \pi  - \arccos \left( { - u} \right) = \arccos \left( u \right)<br />
    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.)
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  5. #5
    Junior Member
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    solution

    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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