formulario sugli integrali
INTEGRALI
PRIMITIVA DI UNA FUNZIONE
E’ una funzione $F\left( x \right)$tale che $F’\left( x \right)=f\left( x \right)$
INTEGRALI INDEFINITI
$\int{f\left( x \right)dx}=F\left( x \right)+c$
INDEGRALI DEFINITI
$\int\limits_{a}^{b}{f\left( x \right)dx}=F\left( b \right)-F\left( a \right)$
INTEGRALI IMMEDIATI DELLE FUNZIONI FONDAMENTALI
$\int{{{x}^{\alpha }}dx}=\frac{{{x}^{\alpha +1}}}{\alpha +1}\,\,+c\,\,con\,\,\,\alpha \ne -1$
$\int{\frac{1}{x}dx}=\ln \left| x \right|+c$
$\int{{{e}^{x}}dx}={{e}^{x}}+c$
$\int{\sin x\,dx\,}=-\cos x\,+\,c$
$\int{\cos x\,dx\,}=\sin x\,+\,c$
$\int{\frac{1}{{{\cos }^{2}}x}\,dx\,}=\int{\left( 1+{{\tan }^{2}}x \right)dx\,}=\tan x\,+\,c$
$\int{\frac{1}{{{\sin }^{2}}x}\,dx\,}=\int{\left( 1+{{\cot }^{2}}x \right)\,dx\,}=-\cot x\,+\,c$
$\int{\frac{1}{1+{{x}^{2}}}dx}=\arctan x+c$
$\int{\frac{1}{\sqrt{1-{{x}^{2}}}}dx}=\arcsin x+c$
INTEGRALI LA CUI PRIMITIVA E’ UNA FUNZIONE COMPOSTA
$\int{{{\left[ f\left( x \right) \right]}^{\alpha }}{f}’\left( x \right)dx}=\frac{{{\left[ f\left( x \right) \right]}^{\alpha +1}}}{\alpha +1}\,\,+c\,\,con\,\,\,\alpha \ne -1$
$\int{\frac{{f}’\left( x \right)}{f\left( x \right)}dx}=\ln \left| f\left( x \right) \right|+c$
$\int{{{e}^{f\left( x \right)}}{f}’\left( x \right)dx}={{e}^{f\left( x \right)}}+c$
$\int{\sin \left[ f\left( x \right) \right]\,{f}’\left( x \right)dx\,}=-\cos x\,+\,c$
$\int{\cos \left[ f\left( x \right) \right]\,{f}’\left( x \right)dx\,}=\sin x\,+\,c$
$\int{\frac{{f}’\left( x \right)}{{{\cos }^{2}}f\left( x \right)}\,dx\,}=\int{\left( 1+{{\tan }^{2}}\left[ f\left( x \right) \right] \right){f}’\left( x \right)dx\,}=\tan f\left( x \right)\,+\,c$
$\int{\frac{{f}’\left( x \right)}{{{\sin }^{2}}\left[ f\left( x \right) \right]}\,dx\,}=\int{\left( 1+{{\cot }^{2}}\left[ f\left( x \right) \right] \right)\,{f}’\left( x \right)dx\,}=-\cot f\left( x \right)\,+\,c$
$\int{\frac{f\left( x \right)}{1+f{{\left( x \right)}^{2}}}dx}=\arctan f\left( x \right)+c$
$\int{\frac{{f}’\left( x \right)}{\sqrt{1-{{\left[ f\left( x \right) \right]}^{2}}}}dx}=\arcsin x+c$
INTEGRAZIONE PER PARTI
CASO INDEFINITO
$\int{{f}'(x)g(x)dx=f\left( x \right)g\left( x \right)-\int{f\left( x \right)g’\left( x \right)dx}}$
CASO DEFINITO
$\int\limits_{a}^{b}{{f}'(x)g(x)dx}=\left. \left[ f\left( x \right)g\left( x \right) \right] \right|_{a}^{b}-\int\limits_{a}^{b}{f\left( x \right)g’\left( x \right)dx}$
INTEGRAZIONE PER SOSTITUZIONE
$\int\limits_{a}^{b}{f\left( x \right)dx}$
Sostituzione: $t=g(x)$
PASSO 1
Ci sono due opzioni:
1. Invertire e poi derivare
$x={{g}^{-1}}\left( t \right)\,\,\,\,\,\,\,\Rightarrow \,\,\,\,\,\,dx=D\left[ {{g}^{-1}}\left( t \right) \right]\,\,\,dt$
2. Derivare direttamente:
$t=g\left( x \right)\,\,\,\,\,\,\Rightarrow \,\,\,\,\,dt=\,\,D\left[ g\left( t \right) \right]\,\,dt$
PASSO 2
Sostituire nella funzione
$t\rightleftarrows g\left( x \right)$ e le $x\rightleftarrows {{g}^{-1}}\left( t \right)$
PASSO 3
Se l’integrale è definito cambio gli estremi di integrazione
$\begin{cases} x=a\,\,\,\,\,\Rightarrow \,\,\,\,\,t=g(a) \\ x=b\,\,\,\,\,\Rightarrow \,\,\,\,\,t=g(b) \\ \end{cases}$
Quindi ottengo:
$\int\limits_{a}^{b}{f\left( x \right)dx}=\int\limits_{g(a)}^{g(b)}{f\left[ {{g}^{-1}}\left( t \right) \right]}\,\,\cdot \,\,D\left[ {{g}^{-1}}\left( t \right) \right]dt$
PRINCIPALI SOSTITUZIONI
Sia $R\left( t \right)$ , una qualsiasi funzione razionale (rapporto di polinomi).
$\int{R\left( f\left( x \right) \right){f}’\left( x \right)dx}$
$f\left( x \right)=t\,\,\,\,\Rightarrow \,\,\,\,{f}’\left( x \right)dx=dt$
Sostituendo: $\int{R\left( t \right)dt}$
ESPONENZIALI
$\int{R\left( {{e}^{\alpha x}} \right)dx}$
${{e}^{\alpha x}}=t\,\,\,\,\,\Rightarrow \,\,\,\,\alpha x=\ln t\,\,\,\,\,\Rightarrow \,\,\,dx=\frac{1}{\alpha t}dt$
Sostituendo:$\int{R\left( t \right)\,\,\frac{1}{\alpha t}}\,dt$
LOGARITMI
$\int{R\left( \ln x \right)\frac{1}{x}dx}$
$\ln x=t\,\,\,\Rightarrow \,\,\,\frac{1}{x}dx=dt$
Sostituendo: $\int{R\left( t \right)\,}\,dt$
POTENZE
$\int{R\left( {{x}^{\alpha }} \right){{x}^{\alpha -1}}dx}$
${{x}^{\alpha }}=t\,\,\,\Rightarrow \,\,\,\alpha {{x}^{\alpha -1}}dx=dt\,\,\Rightarrow \,\,{{x}^{\alpha -1}}dx=\frac{1}{\alpha }dt$
Sostituendo: $\frac{1}{\alpha }\int{R\left( t \right)\,}\,dt$
RADICI DI POLINOMI DI PRIMO GRADO
$\int{R\left( \sqrt{ax+b},\,x \right)dx}$
$\sqrt{ax+b}=t\,\,\,\,\,\Rightarrow \,ax+b={{t}^{2}}\,\Rightarrow \,\,x\,=\frac{{{t}^{2}}-b}{a}\,\Rightarrow \,\,\,dx=\frac{2t}{a}dt$
Sostituendo: $\int{R\left( t,\,\frac{{{t}^{2}}-b}{a} \right)\,\,\,\frac{2t}{a}dt}$
INTEGRALI IN SENO, COSENO E TANGENTE
SOSTITUZIONE 1
$\int{R\left( \sin x \right)}\cos x\,dx$
Porre: $\sin x=t\,\,\Rightarrow \,\,\cos xdx=dt$
Sostituendo: $\int{R\left( t \right)}\,dt$
$\int{R\left( \cos x \right)}\sin x\,dx$
Porre: $\cos x=t\,\,\Rightarrow \,\,-\sin xdx=dt$
Sostituendo: $-\int{R\left( t \right)}\,dt$
SOSTITUZIONE 2
$\int{R\left( \sin x,\cos x \right)dx}$
Parametrizzazione di seno e coseno:
$\sin x=\frac{2t}{{{t}^{2}}+1}$ $\cos x=\frac{1-{{t}^{2}}}{1+{{t}^{2}}}$ $\tan \left( \frac{x}{2} \right)=t\,\,\,\,\,\,\Rightarrow $ $dx=\frac{2}{1+{{t}^{2}}}dt$
Sostituendo: $\int{R\left( \frac{2t}{{{t}^{2}}+1},\frac{1-{{t}^{2}}}{1+{{t}^{2}}} \right)\frac{2}{1+{{t}^{2}}}dt}$
SOSTITUZIONE 3
$\int{R\left( \tan x,{{\sin }^{2}}x,{{\cos }^{2}}x \right)dx}$
$\tan x=t\,\,\,\Rightarrow \,\,x=\arctan t\,\,\,\Rightarrow \,\,dx=\frac{1}{1+{{t}^{2}}}$
${{\cos }^{2}}x=\frac{1}{1+{{t}^{2}}}$
${{\sin }^{2}}x=\frac{{{t}^{2}}}{1+{{t}^{2}}}$
Sostituendo: $\int{R\left( \operatorname{t}\,,\,\,\frac{{{t}^{2}}}{1+{{t}^{2}}},\,\,\frac{1}{1+{{t}^{2}}} \right)\,\,\frac{1}{1+{{t}^{2}}}dt}$
ALCUNI INTEGRALI CHE E’ BENE RICORDARE O TENERE SOTTO MANO
$\int{{{\sin }^{2}}x\,dx}=\frac{x}{2}-\frac{1}{4}\sin 2x$
$\int{{{\cos }^{2}}x\,dx}=\frac{x}{2}+\frac{1}{4}\sin 2x$
$\int{\tan x}=-\ln \left| \cos x \right|+c$
$\int{\cot x}=\ln \left| \sin x \right|+c$
$\int{\frac{1}{\sin x}dx}=\ln \left| \tan \frac{x}{2} \right|+c$
$\int{\frac{1}{\cos x}dx}=\ln \left| \tan \left( \frac{x}{2}+\frac{\pi }{4} \right) \right|+c$
$\int{\sqrt{{{x}^{2}}\pm {{a}^{2}}}}dx=\frac{1}{2}\left( x\sqrt{{{a}^{2}}\pm {{x}^{2}}}\pm {{a}^{2}}\ln \left( \sqrt{{{x}^{2}}\pm {{a}^{2}}+}x \right) \right)+c$
$\int{\frac{1}{\sqrt{{{x}^{2}}\pm {{a}^{2}}}}}dx=\ln \left( \sqrt{{{x}^{2}}\pm {{a}^{2}}+}x \right)+c$
$\int{\frac{1}{\sqrt{{{a}^{2}}-{{x}^{2}}}}}dx=\arctan \left( \frac{x}{\sqrt{{{a}^{2}}-{{x}^{2}}}} \right)+c$
$\int{\frac{1}{a+{{x}^{2}}}}dx=\frac{1}{\sqrt{a}}\arctan \frac{x}{\sqrt{a}}+c$
$\int{{{a}_{n}}{{x}^{n}}+{{a}_{n-1}}{{x}^{n-1}}+…+{{a}_{1}}x}+{{a}_{0}}\,\,dx={{a}_{n}}\frac{{{x}^{n+1}}}{n+1}+{{a}_{n-1}}\frac{{{x}^{n}}}{n}+..+{{a}_{1}}\frac{{{x}^{2}}}{2}+{{a}_{0}}x$
$\int{\frac{A}{ax+b}dx}=\frac{A\ln \left| ax+b \right|}{a}$
$\int{\frac{Ax+B}{a{{x}^{2}}+bx+c}dx}=\frac{ A}{2a}\,\ln \left| a{{x}^{2}}+bx+c \right|+\frac{2aB-bA}{a\,\sqrt{-\Delta }}\arctan \left( \frac{2ax+b}{\sqrt{-\Delta }} \right)+k\,\,\,\,\,\,\,\,\,\,\,\,\,se\,\,\,\,\Delta <0$
$\int{\frac{B}{a{{x}^{2}}+bx+c}dx}=\frac{2aB}{a\,\sqrt{-\Delta }}\arctan \left( \frac{2ax+b}{\sqrt{-\Delta }} \right)+k$
$\int{\frac{Ax+B}{a{{x}^{2}}+c}dx}=\frac{ A}{2a}\,\ln \left| a{{x}^{2}}+c \right|+\frac{B}{\,\sqrt{ac}}\arctan \left( \sqrt{\frac{a}{c}}\,\,x \right)+k$
$\int{\frac{1}{{{\left( ax+b \right)}^{n}}}dx}=-\frac{1}{n-1}\,\,\frac{1}{a{{\left( ax+b \right)}^{n-1}}}$