Integral - malo neobican

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Re: Integral - malo neobican 11.05.2011. u 14:27 · pre 188 meseci #21

A sad jedan specijalan slucaj prethodnog:
$\begin{array}{l} T3. \to \int {{x^{ - 1}}} dx = \ln x + C \\ {\rm{Dokaz}}\,{\rm{(Proof):}} \\ \int {{x^{ - 1}}} dx = \left| {{x^{ - 1}} = {e^{ - \ln x}}} \right| = \int {{e^{ - \ln x}}dx} = \left| \begin{array}{l} - \ln x = t \\ - \frac{{dx}}{x} = dt \\ dx = - xdt = - {e^{ - t}}dt \\ \ln x = - t \to x = {e^{ - t}} \\ \end{array} \right| = \int {{e^t}( - {e^{ - t}})dt} = - \int {dt} = - t = - ( - \ln x) = \ln x + C \\ \end{array}$

Re: Integral - malo neobican 11.05.2011. u 14:30 · pre 188 meseci #22

Kad znamo T3 resavamo i sledece:
$\begin{array}{l} T9. \to \int {\frac{{dx}}{{{{\sin }^2}x}}} = - {\mathop{\rm ctg}\nolimits} x + C \\ {\rm{Dokaz}}\,{\rm{(Proof):}} \\ \int {\frac{{dx}}{{{{\sin }^2}x}}} = \left| \begin{array}{l} {\mathop{\rm tg}\nolimits} x = t \\ x = {\mathop{\rm arctg}\nolimits} t \\ dx = \frac{1}{{1 + {t^2}}}dt \\ \sin x = \frac{t}{{\sqrt {1 + {t^2}} }} \\ \end{array} \right| = \int {\frac{{\frac{1}{{1 + {t^2}}}}}{{{{\left( {\frac{t}{{\sqrt {1 + {t^2}} }}} \right)}^2}}}dt} = \int {\frac{{\frac{1}{{1 + {t^2}}}}}{{\frac{{{t^2}}}{{1 + {t^2}}}}}dt} = \int {\frac{{dt}}{{{t^2}}}} = \frac{{{t^{ - 1}}}}{{ - 1}} = - \frac{1}{t} = - \frac{1}{{{\mathop{\rm tg}\nolimits} x}} = - {\mathop{\rm ctg}\nolimits} x \\ \end{array}$

$\begin{array}{l} T15. \to \int {\frac{{dx}}{{{{\sinh }^2}x}}} = - {\mathop{\rm ctgh}\nolimits} x + C \\ {\rm{Dokaz}}\,{\rm{(Proof):}} \\ \int {\frac{{dx}}{{{{\sinh }^2}x}}} = \left| \begin{array}{l} {\mathop{\rm tgh}\nolimits} x = t \\ x = {\mathop{\rm arctgh}\nolimits} t \\ dx = \frac{1}{{1 - {t^2}}}dt \\ \sinh x = \frac{t}{{\sqrt {1 - {t^2}} }} \\ \end{array} \right| = \int {\frac{{\frac{1}{{1 - {t^2}}}}}{{{{\left( {\frac{t}{{\sqrt {1 - {t^2}} }}} \right)}^2}}}dt} = \int {\frac{{\frac{1}{{1 - {t^2}}}}}{{\frac{{{t^2}}}{{1 - {t^2}}}}}dt} = \int {\frac{{dt}}{{{t^2}}}} = \frac{{{t^{ - 1}}}}{{ - 1}} = - \frac{1}{t} = - \frac{1}{{{\mathop{\rm tgh}\nolimits} x}} = - {\mathop{\rm ctgh}\nolimits} x \\ \end{array}$

$\begin{array}{l} T16. \to \int {\frac{{dx}}{{1 - {x^2}}}} = \frac{1}{2}\ln \left| {\frac{{1 + x}}{{1 - x}}} \right| + C = {\mathop{\rm arctgh}\nolimits} x + C \\ {\rm{Dokaz}}\,{\rm{(Proof):}} \\ \int {\frac{{dx}}{{1 - {x^2}}}} = \int {\frac{1}{{(1 - x)(1 + x)}}dx} = \frac{1}{2}\int {\frac{{2 + x - x}}{{(1 - x)(1 + x)}}dx} = \frac{1}{2}\int {\frac{{(1 + x) + (1 - x)}}{{(1 - x)(1 + x)}}dx = } \\ = \frac{1}{2}\int {\frac{{(1 + x)}}{{(1 - x)(1 + x)}}dx} + \frac{1}{2}\int {\frac{{(1 - x)}}{{(1 - x)(1 + x)}}dx = } \frac{1}{2}\left( {\int {\frac{1}{{1 - x}}dx} + \int {\frac{1}{{1 + x}}dx} } \right) = \left| {\begin{array}{*{20}{c}} \begin{array}{l} 1 - x = t \\ - dx = dt \\ \end{array} & \begin{array}{l} 1 + x = u \\ dx = du \\ \end{array} \\ \end{array}} \right| = \\ = \frac{1}{2}\left( {\int {\frac{1}{t}( - dt)} + \int {\frac{1}{u}du} } \right) = \frac{1}{2}\left( { - \ln t + \ln u} \right) = \frac{1}{2}\left( { - \ln \left| {1 - x} \right| + \ln \left| {1 + x} \right|} \right) = \frac{1}{2}\left( {\ln \left| {1 + x} \right| - \ln \left| {1 - x} \right|} \right) = \\ = \frac{1}{2}\ln \left| {\frac{{1 + x}}{{1 - x}}} \right| \\ \end{array}$

$\begin{array}{l} T17. \to \int {\frac{{dx}}{{{x^2} - 1}}} = \frac{1}{2}\ln \left| {\frac{{x - 1}}{{x + 1}}} \right| + C = - {\mathop{\rm arcctgh}\nolimits} x + C \\ {\rm{Dokaz}}\,{\rm{(Proof):}} \\ \int {\frac{{dx}}{{{x^2} - 1}}} = \frac{1}{2}\int {\frac{{(x + 1) - (x - 1)}}{{(x - 1)(x + 1)}}dx} = \frac{1}{2}\int {\frac{{dx}}{{x - 1}}} - \frac{1}{2}\int {\frac{{dx}}{{x + 1}}} = \left| {\begin{array}{*{20}{c}} \begin{array}{l} x - 1 = t \\ dx = dt \\ \end{array} & \begin{array}{l} x + 1 = u \\ dx = du \\ \end{array} \\ \end{array}} \right| = \frac{1}{2}\left( {\int {\frac{{dt}}{t}} - \int {\frac{{du}}{u}} } \right) = \\ = \frac{1}{2}\left( {\ln t - \ln u} \right) = \frac{1}{2}\left( {\ln \left| {x - 1} \right| - \ln \left| {x + 1} \right|} \right) = \frac{1}{2}\ln \left| {\frac{{x - 1}}{{x + 1}}} \right| \\ \end{array}$

Re: Integral - malo neobican 11.05.2011. u 14:33 · pre 188 meseci #23

Za T18 i T19 moramo takodje upotrebiti malo zonglerskog znanja ali ovog puta sa hiberbolicnim funkcijama

$\begin{array}{l} T18. \to \int {\frac{{dx}}{{\sqrt {{x^2} + 1} }}} = \ln \left( {x + \sqrt {{x^2} + 1} } \right) + C = {\mathop{\rm arcsinh}\nolimits} x + C \\ {\rm{Dokaz}}\,{\rm{(Proof):}} \\ \int {\frac{{dx}}{{\sqrt {{x^2} + 1} }}} = \left| \begin{array}{l} x = \sinh t \\ dx = \cosh t \\ {\mathop{\rm arcsinh}\nolimits} x = t \\ \end{array} \right| = \int {\frac{{\cosh t}}{{\sqrt {{{\sinh }^2}t + 1} }}dx} = \left| {{{\cosh }^2}t - {{\sinh }^2}t = 1} \right| = \int {\frac{{\cosh t}}{{\cosh t}}dx} = \int {dt} = t = {\mathop{\rm arcsinh}\nolimits} x \\ \sinh x = \frac{{{e^x} - {e^{ - x}}}}{2} \to \left| {x = \sinh t} \right| \to x = \frac{{{e^t} - {e^{ - t}}}}{2} \\ 2x = {e^t} - {e^{ - t}}/ \cdot {e^t} \\ 2x{e^t} = {e^{2t}} - 1 \\ {({e^t})^2} - 2x{e^t} - 1 = 0 \to \left| \begin{array}{l} a{x^2} + bx + c = 0 \\ {x_{1/2}} = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \\ \end{array} \right| \to e_{1/2}^t = \frac{{2x \pm \sqrt {4{x^2} + 4} }}{2} = x \pm \sqrt {{x^2} + 1} \\ {e^t} = x + \sqrt {{x^2} + 1} \,\,\,/\ln \to \ln {e^t} = \ln (x + \sqrt {{x^2} + 1} ) \\ t = \ln (x + \sqrt {{x^2} + 1} ) \to \int {\frac{{dx}}{{\sqrt {{x^2} + 1} }}} = t = \ln \left( {x + \sqrt {{x^2} + 1} } \right) \\ \end{array}$

$\begin{array}{l} T19. \to \int {\frac{{dx}}{{\sqrt {{x^2} - 1} }}} = \ln \left( {x + \sqrt {{x^2} - 1} } \right) + C = {\mathop{\rm arccosh}\nolimits} x + C \\ {\rm{Dokaz}}\,{\rm{(Proof):}} \\ \int {\frac{{dx}}{{\sqrt {{x^2} - 1} }}} = \left| \begin{array}{l} x = \cosh t \\ dx = \sinh t \\ {\mathop{\rm arccosh}\nolimits} x = t \\ \end{array} \right| = \int {\frac{{\sinh t}}{{\sqrt {{{\cosh }^2}t - 1} }}dx} = \left| {{{\cosh }^2}t - {{\sinh }^2}t = 1} \right| = \int {\frac{{\sinh t}}{{\sinh t}}dx} = \int {dt} = t = {\mathop{\rm arccosh}\nolimits} x \\ \cosh x = \frac{{{e^x} + {e^{ - x}}}}{2} \to \left| {x = \cosh t} \right| \to x = \frac{{{e^t} + {e^{ - t}}}}{2} \\ 2x = {e^t} + {e^{ - t}}/ \cdot {e^t} \\ 2x{e^t} = {e^{2t}} + 1 \\ {({e^t})^2} - 2x{e^t} + 1 = 0 \to \left| \begin{array}{l} a{x^2} + bx + c = 0 \\ {x_{1/2}} = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \\ \end{array} \right| \to e_{1/2}^t = \frac{{2x \pm \sqrt {4{x^2} - 4} }}{2} = x \pm \sqrt {{x^2} - 1} \\ {e^t} = x + \sqrt {{x^2} - 1} \,\,\,/\ln \to \ln {e^t} = \ln (x + \sqrt {{x^2} - 1} ) \\ t = \ln (x + \sqrt {{x^2} - 1} ) \to \int {\frac{{dx}}{{\sqrt {{x^2} - 1} }}} = t = \ln \left( {x + \sqrt {{x^2} - 1} } \right) \\ \end{array}$

Re: Integral - malo neobican 11.05.2011. u 14:56 · pre 188 meseci #24

$\begin{array}{l} T5. \to \int {{a^x}} dx = \frac{{{a^x}}}{{\ln a}} + C \\ {\rm{Dokaz}}\,{\rm{(Proof):}} \\ \int {{a^x}} dx = \left| \begin{array}{l} {a^x} = t \\ {a^x}\ln adx = dt \\ dx = \frac{{dt}}{{{a^x}\ln a}} = \frac{{dt}}{{t\ln a}} \\ \end{array} \right| = \int {t\frac{{dt}}{{t\ln a}}} = \frac{1}{{\ln a}}\int {dt} = \frac{1}{{\ln a}} \cdot t = \frac{{{a^x}}}{{\ln a}} \\ \end{array}$

Re: Integral - malo neobican 11.05.2011. u 15:00 · pre 188 meseci #25

I malo je hvalilo da $\int {dx}$ resim preko njega samog

$\begin{array}{l} T1. \to \int {} dx = x + C\,\,\,\, \\ {\rm{Dokaz}}\,{\rm{(Proof):}} \\ \int {dx} = \int {\frac{{{x^n}}}{{{x^n}}}} dx = \int {{x^n}{x^{ - n}}} dx = \left| {\begin{array}{*{20}{c}} {dv = {x^n}dx} & {u = {x^{ - n}}} \\ {v = \frac{{{x^{n + 1}}}}{{n + 1}}} & {du = - n{x^{ - n - 1}}dx} \\ \end{array}} \right| = {x^{ - n}}\frac{{{x^{n + 1}}}}{{n + 1}} + \frac{n}{{n + 1}}\int {{x^{n + 1}}{x^{ - n - 1}}dx} = \frac{{{x^n}x}}{{{x^n}(n + 1)}} + \frac{n}{{n + 1}}\int {dx} \\ \int {dx} - \frac{n}{{n + 1}}\int {dx} = \frac{x}{{n + 1}} \\ \left( {1 - \frac{n}{{n + 1}}} \right)\int {dx} = \frac{x}{{n + 1}} \\ \left( {\frac{1}{{n + 1}}} \right)\int {dx} = \frac{x}{{n + 1}} \\ \int {dx} = x + C \\ \end{array}$

Re: Integral - malo neobican 11.05.2011. u 17:43 · pre 188 meseci #26

T18 je urađen onako kako sam i rekao. Uzeo si da je t = rešenju i onda ga izdiferenciraš.
To što ga ti zoveš arcsinh(x) a ne ln(x+koren(x*x + 1)) je presipanje iz šupljeg u prazno.
Nema tu ništa od žongliranja.
Žongliranje bi bila smena tg(x) = t.

Re: Integral - malo neobican 12.05.2011. u 02:09 · pre 188 meseci #27

Citat:

igorpet: I malo je hvalilo da $\int {dx}$ resim preko njega samog :)

$\begin{array}{l} T1. \to \int {} dx = x + C\,\,\,\, \\ {\rm{Dokaz}}\,{\rm{(Proof):}} \\ \int {dx} = \int {\frac{{{x^n}}}{{{x^n}}}} dx = \int {{x^n}{x^{ - n}}} dx = \left| {\begin{array}{*{20}{c}} {dv = {x^n}dx} & {u = {x^{ - n}}} \\ {v = \frac{{{x^{n + 1}}}}{{n + 1}}} & {du = - n{x^{ - n - 1}}dx} \\ \end{array}} \right| = {x^{ - n}}\frac{{{x^{n + 1}}}}{{n + 1}} + \frac{n}{{n + 1}}\int {{x^{n + 1}}{x^{ - n - 1}}dx} = \frac{{{x^n}x}}{{{x^n}(n + 1)}} + \frac{n}{{n + 1}}\int {dx} \\ \int {dx} - \frac{n}{{n + 1}}\int {dx} = \frac{x}{{n + 1}} \\ \left( {1 - \frac{n}{{n + 1}}} \right)\int {dx} = \frac{x}{{n + 1}} \\ \left( {\frac{1}{{n + 1}}} \right)\int {dx} = \frac{x}{{n + 1}} \\ \int {dx} = x + C \\ \end{array}$

Hm...
Ti si ovde koristio T2 koji se resio koristeci T1, a sada koristis T2 da dokazes za T1...
Ae probaj opet

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