and 
Then 
#2 16. 12. 2016 02:13 — Editoval Pavel (16. 12. 2016 02:16)
Re: Definite Integrals
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\mathsf I_1&=\pi\ln 2\\
\mathsf I_2&=\int_0^{\pi}\ln\left(\sin\frac x2\right)\,\mathrm dx
=\left[\text{subst. }y=\frac x2\right]=2\int_0^{\frac{\pi}2}\ln\left(\sin y\right)\,\mathrm dy\\
\mathsf I_3&=\int_0^{\pi}\ln\left(\cos\frac x2\right)\,\mathrm dx
=\left[\text{subst. }y=\frac x2\right]=2\int_0^{\frac{\pi}2}\ln\left(\cos y\right)\,\mathrm dy
=2\int_0^{\frac{\pi}2}\ln\left(\sin\left(y+\frac{\pi}2\right)\right)\,\mathrm dy\\
&=\left[\text{subst. }t=y+\frac{\pi}2\right]=2\int_{\frac{\pi}2}^{\pi}\ln\left(\sin t\right)\,\mathrm dt
$](/mathtex/1a/1adb5b1929eec24be715062d8bffeeed.gif "kopírovat do textarea")
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\mathsf I_4&=\frac 12\pi^2\ln 2\\
\mathsf I_5&=\int_0^{\pi}x\ln\left(\sin\frac x2\right)\,\mathrm dx
=\left[\text{subst. }y=\frac x2\right]=4\int_0^{\frac{\pi}2}y\ln\left(\sin y\right)\,\mathrm dy\\
\mathsf I_6&=\int_0^{\pi}x\ln\left(\cos\frac x2\right)\,\mathrm dx
=\left[\text{subst. }y=\frac x2\right]=4\int_0^{\frac{\pi}2}y\ln\left(\cos y\right)\,\mathrm dy
=4\int_0^{\frac{\pi}2}y\ln\left(\sin\left(y+\frac{\pi}2\right)\right)\,\mathrm dy\\
&=\left[\text{subst. }t=y+\frac{\pi}2\right]
=4\int_{\frac{\pi}2}^{\pi}\left(t-\frac{\pi}2\right)\ln\left(\sin t\right)\,\mathrm dt
=4\int_{\frac{\pi}2}^{\pi}t\ln\left(\sin t\right)\,\mathrm dt-2\pi\int_{\frac{\pi}2}^{\pi}\ln\left(\sin t\right)\,\mathrm dt\\
&=\left[\text{subst. }u=\pi-t\right]
=4\int_{\frac{\pi}2}^{\pi}t\ln\left(\sin t\right)\,\mathrm dt-\pi\left(\int_{\frac{\pi}2}^{\pi}\ln\left(\sin t\right)\,\mathrm dt-\int_{\frac{\pi}2}^0\ln
\left(\sin(\pi-u)\right)\,\mathrm du\right)\\
&=4\int_{\frac{\pi}2}^{\pi}t\ln\left(\sin t\right)\,\mathrm dt-\pi\left(\int_{\frac{\pi}2}^{\pi}\ln\left(\sin t\right)\,\mathrm dt+\int_0^{\frac{\pi}2}\ln
\left(\sin u\right)\,\mathrm du\right)\\
&=4\int_{\frac{\pi}2}^{\pi}t\ln\left(\sin t\right)\,\mathrm dt-\pi\int_0^{\pi}\ln\left(\sin t\right)\,\mathrm dt
\stackrel{{\color{blue}\spadesuit}}{=}4\int_{\frac{\pi}2}^{\pi}t\ln\left(\sin t\right)\,\mathrm dt+\pi^2\ln 2
$](/mathtex/be/be008da2bd4285f2458c3229d69c5ca0.gif "kopírovat do textarea")
:
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\mathsf I_7&=\frac 12\pi^3\ln 2\\
\mathsf I_8&=\int_0^{\pi}x^2\ln\left(\sin\frac x2\right)\,\mathrm dx
=\left[\text{subst. }y=\frac x2\right]=8\int_0^{\frac{\pi}2}y^2\ln\left(\sin y\right)\,\mathrm dy\\
\mathsf I_9&=\int_0^{\pi}x^2\ln\left(\cos\frac x2\right)\,\mathrm dx
=\left[\text{subst. }y=\frac x2\right]=8\int_0^{\frac{\pi}2}y^2\ln\left(\cos y\right)\,\mathrm dy
=8\int_0^{\frac{\pi}2}y^2\ln\left(\sin\left(y+\frac{\pi}2\right)\right)\,\mathrm dy\\
&=\left[\text{subst. }t=y+\frac{\pi}2\right]
=8\int_{\frac{\pi}2}^{\pi}\left(t-\frac{\pi}2\right)^2\ln\left(\sin t\right)\,\mathrm dt\\
&=8\int_{\frac{\pi}2}^{\pi}t^2\ln\left(\sin t\right)\,\mathrm dt-8\pi\int_{\frac{\pi}2}^{\pi}t\ln\left(\sin t\right)\,\mathrm dt+2\pi^2\int_{\frac{\pi}2}^
{\pi}\ln\left(\sin t\right)\,\mathrm dt\\
&=8\int_{\frac{\pi}2}^{\pi}t^2\ln\left(\sin t\right)\,\mathrm dt-8\pi\int_{\frac{\pi}2}^{\pi}t\ln\left(\sin t\right)\,\mathrm dt+\pi^2\int_0^{\pi}\ln\left
(\sin t\right)\,\mathrm dt\\
&\stackrel{{\color{blue}\spadesuit}}{=}8\int_{\frac{\pi}2}^{\pi}t^2\ln\left(\sin t\right)\,\mathrm dt-8\pi\int_{\frac{\pi}2}^{\pi}t\ln\left(\sin t\right)\,\mathrm dt-\pi^3\ln 2
$](/mathtex/2a/2a780382cc67149db499a48c0bd578f5.gif "kopírovat do textarea")
:
![$
\mathsf I_{10}&=\frac 14\pi^4\ln 2\\
\mathsf I_{11}&=\int_0^{\pi}x^3\ln\left(\sin\frac x2\right)\,\mathrm dx
=\left[\text{subst. }y=\frac x2\right]=16\int_0^{\frac{\pi}2}y^3\ln\left(\sin y\right)\,\mathrm dy\\
\mathsf I_{12}&=\int_0^{\pi}x^3\ln\left(\cos\frac x2\right)\,\mathrm dx
=\left[\text{subst. }y=\frac x2\right]=16\int_0^{\frac{\pi}2}y^3\ln\left(\cos y\right)\,\mathrm dy
=16\int_0^{\frac{\pi}2}y^3\ln\left(\sin\left(y+\frac{\pi}2\right)\right)\,\mathrm dy\\
&=\left[\text{subst. }t=y+\frac{\pi}2\right]
=16\int_{\frac{\pi}2}^{\pi}\left(t-\frac{\pi}2\right)^3\ln\left(\sin t\right)\,\mathrm dt\\
&=16\int_{\frac{\pi}2}^{\pi}t^3\ln\left(\sin t\right)\,\mathrm dt-24\pi\int_{\frac{\pi}2}^{\pi}t^2\ln\left(\sin t\right)\,\mathrm dt+12\pi^2\int_{\frac{\pi}2}^{\pi}t\ln\left(\sin t\right)\,\mathrm dt-2\pi^3\int_{\frac{\pi}2}^{\pi}\ln\left(\sin t\right)\,\mathrm dt\\
&=16\int_{\frac{\pi}2}^{\pi}t^3\ln\left(\sin t\right)\,\mathrm dt-24\pi\int_{\frac{\pi}2}^{\pi}t^2\ln\left(\sin t\right)\,\mathrm dt+12\pi^2\int_{\frac{\pi}2}^{\pi}t\ln\left(\sin t\right)\,\mathrm dt-\pi^3\int_0^{\pi}\ln\left(\sin t\right)\,\mathrm dt\\
&\stackrel{{\color{blue}\spadesuit}}{=}16\int_{\frac{\pi}2}^{\pi}t^3\ln\left(\sin t\right)\,\mathrm dt-24\pi\underbrace{\int_{\frac{\pi}2}^{\pi}t^2\ln\left(\sin t\right)\,\mathrm dt}_{\mathsf I_{13}}+12\pi^2\int_{\frac{\pi}2}^{\pi}t\ln\left(\sin t\right)\,\mathrm dt+\pi^4\ln 2\\
\mathsf I_{13}&=\frac 12\left(\int_{\frac{\pi}2}^{\pi}t^2\ln\left(\sin t\right)\,\mathrm dt+\int_{\frac{\pi}2}^{\pi}t^2\ln\left(\sin t\right)\,\mathrm dt\right)
=\left[\text{subst. }u=\pi-t\right]\\
&=\frac 12\left(\int_{\frac{\pi}2}^{\pi}t^2\ln\left(\sin t\right)\,\mathrm dt-\int_{\frac{\pi}2}^0(\pi-u)^2\ln\left(\sin(\pi-u)\right)\,\mathrm du\right)\\
&=\frac 12\left(\int_{\frac{\pi}2}^{\pi}t^2\ln\left(\sin t\right)\,\mathrm dt+\pi^2\int_0^{\frac{\pi}2}\ln\left(\sin u\right)\,\mathrm du-2\pi\int_0^{\frac{\pi}2}u\ln\left(\sin u\right)\,\mathrm du+\int_0^{\frac{\pi}2}u^2\ln\left(\sin u\right)\,\mathrm du\right)\\
&=\frac 12\left(\int_0^{\pi}x^2\ln\left(\sin x\right)\,\mathrm dx+\frac{\pi^2}2\int_0^{\pi}\ln\left(\sin x\right)\,\mathrm dx-2\pi\int_0^{\frac{\pi}2}x\ln\left(\sin x\right)\,\mathrm dx\right)\\
&\stackrel{{\color{blue}\spadesuit}}{=}
\frac 12\left(\int_0^{\pi}x^2\ln\left(\sqrt 2\sin x\right)\,\mathrm dx-\int_0^{\pi}x^2\ln\sqrt 2\,\mathrm dx-\frac{\pi^3}2\ln 2-2\pi\int_0^{\frac{\pi}2}x\ln\left(\sin x\right)\,\mathrm dx\right)\\
&\stackrel{{\color{blue}\clubsuit}}{=}
\frac 12\left(\frac{11}{42}\pi^3\ln 2+\frac 87\pi\int_{\frac{\pi}2}^{\pi}x\ln\left(\sin x\right)\,\mathrm dx-\frac{\pi^3}6\ln 2-\frac{\pi^3}2\ln 2-2\pi\int_0^{\frac{\pi}2}x\ln\left(\sin x\right)\,\mathrm dx\right)\\
&=\frac 12\left(-\frac{17}{42}\pi^3\ln 2+\frac{22}7\pi\int_{\frac{\pi}2}^{\pi}x\ln\left(\sin x\right)\,\mathrm dx-2\pi\int_0^{\pi}x\ln\left(\sin x\right)\,\mathrm dx\right)\\
&\stackrel{{\color{blue}\heartsuit}}{=}
-\frac{17}{84}\pi^3\ln 2+\frac{11}7\pi\int_{\frac{\pi}2}^{\pi}x\ln\left(\sin x\right)\,\mathrm dx+\frac{\pi^3}2\ln 2
=\frac{25}{84}\pi^3\ln 2+\frac{11}7\pi\int_{\frac{\pi}2}^{\pi}x\ln\left(\sin x\right)\,\mathrm dx\\
\mathsf I_{12}&=\dots=
16\int_{\frac{\pi}2}^{\pi}t^3\ln\left(\sin t\right)\,\mathrm dt-24\pi\mathsf I_{13}+12\pi^2\int_{\frac{\pi}2}^{\pi}t\ln\left(\sin t\right)\,\mathrm dt+\pi^4\ln 2\\
&=16\int_{\frac{\pi}2}^{\pi}t^3\ln\left(\sin t\right)\,\mathrm dt-24\pi\left(\frac{25}{84}\pi^3\ln 2+\frac{11}7\pi\int_{\frac{\pi}2}^{\pi}x\ln\left(\sin x\right)\,\mathrm dx\right)+12\pi^2\int_{\frac{\pi}2}^{\pi}t\ln\left(\sin t\right)\,\mathrm dt+\pi^4\ln 2\\
&=16\int_{\frac{\pi}2}^{\pi}t^3\ln\left(\sin t\right)\,\mathrm dt-\frac{43}{7}\pi^4\ln 2-\frac{180}7\pi^2\int_{\frac{\pi}2}^{\pi}x\ln\left(\sin x\right)\,\mathrm dx
$](/mathtex/d9/d98d5b295f52b785d897b861ba9617de.gif "kopírovat do textarea")
Backslash je v TeXu tak důležitý jako nekonečno při dělení nulou v tělesech charakteristiky 0.
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#3 08. 01. 2017 12:30
- stuart clark
- Příspěvky: 1015
- Reputace: 7
Re: Definite Integrals
↑ Pavel:Thanks Pavel for nice solution
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