integrale assurdo!

[Noisemaker]

Ho trovato questo integrale ""assurdo"" ... potrebbe essere corretto?


[tex]\displaystyle\int_{2}^{5}\,\, \frac{\cos\left(\frac{1}{x+2}\right)\ln\left[\sin^2\left(\frac{1}{x+2}\right)+4\right]}{\left(x+2 \right)^2\left[1+\sin\left(\frac{1}{x+2}\right)\right]^3 } \,\,dx[/tex]

La funzione integarnda è definita per [tex]x\not=-2 ,[/tex] e dunque l'integrale nell'intervallo [tex][2;5],[/tex] non è improrio; calcolimao l'insieme delle primitive: posto

[tex]\displaystyle\frac{1}{x+2}=t, \to dx=-\frac{1}{t^2}\,\,dt[/tex]


[tex]\displaystyle\int \frac{\cos\left(\frac{1}{x+2}\right)\ln\left(\sin^2\left(\frac{1}{x+2}\right)+4\right)}{\left(x+2 \right)^2\left(1+\sin\left(\frac{1}{x+2}\right)\right)^3 } \,\,dx[/tex] [tex]\displaystyle=-\int\frac{\cos t\ln\left(\sin^2t+4\right)}{\frac{1}{t^2}\left(1+\sin t\right) ^3 }\cdot\frac{1}{t^2}\,\,dt=[/tex] [tex]\displaystyle-\int\frac{\ln\left(\sin^2t+4\right)}{ \left(1+\sin t\right) ^3 } \,\,d(\sin t)[/tex]

posto [tex]\sin t=y[/tex]

[tex]\displaystyle-\int\frac{\ln\left(y^2 +4\right)}{ \left(1+y\right) ^3 } \,\,dy, \quad \text{ed essendo}\quad \int\frac{1}{ \left(1+y\right) ^3 } \,\,dy=[/tex] [tex]\displaystyle-\frac{1}{2(1+y)^2}\quad \text{si ha integrando per parti:}[/tex]


[tex]\displaystyle=-\int \ln\left(y^2 +4\right) \,\,d\left(-\frac{1}{2(1+y)^2}\right)=[/tex] [tex]\displaystyle\frac{1}{2} \int \ln\left(y^2 +4\right) \,\,d\left( \frac{1}{ (1+y)^2}\right) \stackrel{\bf(P)}{=}[/tex] [tex]\displaystyle\frac{1}{2}\left(\frac{\ln\left(y^2 +4\right)}{ (1+y)^2} -\int \frac{1}{ (1+y)^2} \,\,d\left(\ln\left(y^2 +4\right)\right)\right)[/tex]

[tex]\displaystyle=\frac{1}{2}\left(\frac{\ln\left(y^2 +4\right)}{ (1+y)^2} -\int \frac{2y}{(y^2 +4) (1+y)^2} \,\,dy\right)[/tex]

[tex]\quad\mbox{considerando l ultimo integrale, si ha:}[/tex]

[tex]\displaystyle\int \frac{2y}{(y^2 +4) (1+y)^2} \,\,dy= \int \frac{A}{ 1+y } + \frac{B}{ (1+y)^2}+\frac{Cy+D}{(y^2 +4) } \,\,dy=[/tex] [tex]\displaystyle A\ln|1+y|- \frac{B}{ 1+y }+\int \frac{Cy+D}{(y^2 +4) } \,\,dy[/tex]

[tex]\displaystyle\text{dove}\quad A = \frac{6}{25}, B= -\frac{2}{5}, C= -\frac{6}{25}, D = \frac{16}{25},\quad \text{e dunque:}[/tex]

[tex]\displaystyle \frac{6}{25}\ln|1+y|+ \frac{ 2 }{ 5(1+y)}-\int \frac{ \frac{6}{25}x-\frac{16}{25}}{(y^2 +4) } \,\,dy=[/tex] [tex]\displaystyle\frac{6}{25}\ln|1+y|+ \frac{ 2 }{ 5(1+y)}-\frac{2}{25}\int \frac{ 3y-8}{(y^2 +4) } \,\,dy[/tex]

[tex]\text{considerando l ultimo integrale, si ha:}[/tex]

[tex]\displaystyle\int \frac{ 3y-8}{(y^2 +4) } \,\,dy=[/tex] [tex]\displaystyle\int \frac{ 3y }{(y^2 +4) } \,\,dy-\int \frac{8}{(y^2 +4) } \,\,dy=[/tex] [tex]\displaystyle\frac{3}{2}\int \frac{ 2y }{(y^2 +4) } \,\,dy-8\int \frac{1}{4(\frac{y^2}{4} +1) } \,\,dy[/tex]

[tex]\displaystyle=\frac{3}{2}\ln |y^2 +4|-2\int \frac{1}{ \left(\frac{y}{2}\right)^2 +1 } \,\,dy=[/tex] [tex]\displaystyle\frac{3}{2}\ln |y^2 +4|-4\int \frac{\frac{1}{2}}{ \left(\frac{y}{2}\right)^2 +1 } \,\,dy=[/tex] [tex]\displaystyle\frac{3}{2}\ln |y^2 +4|-4\arctan\frac{y}{2}[/tex]


[tex]\quad\mbox{ si ha:}[/tex]

[tex]\displaystyle \frac{\ln\left(y^2 +4\right)}{ 2(1+y)^2} - \frac{6}{50}\ln|1+y|-[/tex] [tex]\displaystyle\frac{ 2 }{ 10(1+y)}+\frac{2}{50}\left(\frac{3}{2}\ln |y^2 +4|-4\arctan\frac{y}{2}\right)[/tex]

[tex]\quad \text{essendo}[/tex] [tex]\sin t=y[/tex]

[tex]\displaystyle \frac{\ln\left(\sin^2 t +4\right)}{ 2(1+\sin t)^2} - \frac{6}{50}\ln|1+\sin t|-[/tex] [tex]\displaystyle\frac{ 2 }{ 10(1+\sin t)}+\frac{2}{50}\left(\frac{3}{2}\ln |\sin^2 t +4|-4\arctan\frac{\sin t}{2}\right)[/tex]

[tex]\text{essendo per}[/tex] [tex]\displaystyle x=2 \to t=\frac{1}{4},\quad x=5 \to t=\frac{1}{7}[/tex]

[tex]\displaystyle \frac{\ln\left(\sin^2 t +4\right)}{ 2(1+\sin t)^2} - \frac{6}{50}\ln|1+\sin t|-[/tex] [tex]\displaystyle\frac{ 2 }{ 10(1+\sin t)}+\frac{2}{50}\left(\frac{3}{2}\ln |\sin^2 t +4|-4\arctan\frac{\sin t}{2}\right)\right|_{\frac{1}{4}}^{\frac{1}{7}}\sim 0.09[/tex]

[Massimo Gobbino]

Beh, alla fine non è poi così assurdo. I conti non li ho controllati (non sarebbe male che qualcuno lo facesse indipendentemente per vedere se viene uguale ...), ma i cambi di variabile sono quelli giusti.

[Noisemaker]