Integrarea prin schimbarea de variabila

Prezentam o teorema care ne ajuta sa calculam integralele cu ajutorul metodei de schimbare de variabila .
Teorema. Fie \phi:J\rightarrow I o functie derivabila cu derivata continua si functia f:I\rightarrow R o functie continua si \alpha, \beta \in J. Atunci:
\int^{\beta}_{\alpha}f\left(\varphi\left(t\right)\right)\varphi^{'}\left(t\right)dt=\int^{\varphi\left(\beta\right)}_{\varphi\left(\alpha\right)}f\left(x\right)dx
astfel se fac schimbarile  de variabila si de simbol
\varphi\left(t\right)=x si \varphi\left(t\right)^{'}dt=dx, t\in j, x\in I.
 Exercitii :
\int^{1}_{0}\frac{\sqrt{x}}{x+3}dx=
Observam ca rezolvam integrala prin metoda schimbarii de variabila, astfel notam :
\sqrt{x}=t\Rightarrow x=t^{2}
Acum derivam in functie de x si in functie de t.
x^{'}dx=\left(t^{2}\right)^{'}\Rightarrow 1\cdot dx=2t dt
Iar acum ne ocupam de capetele intervalului, astfel pentru
x=1\Rightarrow \sqrt{1}=t\Rightarrow t=1
Pentru x=0\Rightarrow x=0\Rightarrow \sqrt{0}=t\Rightarrow t=0
Astfel integrala devine :
\int^{1}_{0}\frac{t}{t^{2}+3}\cdot 2tdt=\int^{1}_{0}\frac{2t^{2}}{t^{2}+3}dt=2\int^{1}_{0}\frac{t^{2}}{t^{2}+3}dt=2\int^{1}_{0}\frac{t^{2}+3-3}{t^{2}+3}dt=2\int^{1}_{0}\frac{t^{2}+3}{t^{2}+3}dt-2\int^{1}_{0}\frac{3}{t^{2}+3}dt=2\int^{1}_{0}dt-2\cdot 3\int^{1}_{0}\frac{dt}{t^{2}+3}=2t|^{1}_{0}-
6\frac{1}{\sqrt{3}}arctan\frac{x}{\sqrt{3}}|^{1}_{0}=2\cdot 1-2\cdot 0-\frac{6}{\sqrt{3}}arctan\frac{1}{\sqrt{3}}+\frac{6}{\sqrt{3}}arctan\frac{0}{1}=2-\frac{6\sqrt{3}}{3}\frac{\pi}{3}-0=2-2\sqrt{3}\cdot\frac{\pi}{6}=2-\sqrt{3}\frac{\pi}{3}.

b) \int^{e}_{1}\frac{\ln x}{\sqrt{x}}dx=
Rezolvam integrala tot cu ajutorul metodei schimbarii de variabila.
Astfel avem functia continua
\left(x\right)=\frac{\ln x}{\sqrt{x}}, \forall x\in \left(0,+\infty\right)

Astfel notam :
\sqrt{x}=t\Rightarrow x=t^{2}, \forall x\in \left(0,+\infty\right)
functia
\varphi:\left(0,+\infty\right)\rightarrow \left(0,+\infty\right)
este derivabila cu derivata continua
Derivam acum in functie de x si de t
x^{'} dx=\left(t^{2}\right)^{'}dt\Rightarrow dx=2t dt
Acum ne ocupam si de capetele intervalului, astfel pentru
x=e\Rightarrow e=t^{2}\Rightarrow t=\sqrt{e}
si

x=1\Rightarrow 1=t^{2}\Rightarrow t=1
Astfel obtinem integrala :

\int^{\sqrt{e}}_{1}\frac{\ln t^{2}}{t}\cdot 2t dt=2\int^{\sqrt{e}}_{1}\ln t^{2}dt=2\int^{\sqrt{e}}_{1} 2\ln t dt=4\int^{\sqrt{e}}_{1} \ln t dt=4\int^{\sqrt{e}}_{1}t^{'}\cdot \ln t dt=4 t\cdot \ln t|^{\sqrt{e}}_{1}-4\int^{\sqrt{e}}_{1} t\cdot \left(\ln t\right)^{'}dt=4\sqrt{e}\ln\sqrt{e}-4\cdot 1\cdot \ln 1-4\int^{\sqrt{e}}_{1}t\cdot\frac{1}{t} dt=4\sqrt{e}\ln e^{\frac{1}{2}}-4\cdot 1\cdot 0-4\int^{\sqrt{e}}_{1}dt=  4\sqrt{e}\frac{1}{2}\ln e-0-4\cdot t|^{\sqrt{e}}_{1}=2\sqrt{e}\cdot 1-0-4\cdot\sqrt{e}+4\cdot 1=  2\sqrt{e}-4\sqrt{e}+4=-2\sqrt{e}+4=2\left(-\sqrt{e}+2\right).
Dupa ce am folosit schimbarea de variabila am folosit si metoda integrarii prin parti.
c) \int^{\frac{\pi}{2}}_{\frac{\pi}{6}}\frac{\sin 2x}{\sqrt{\sin x}} dx=  \int^{\frac{\pi}{2}}_{\frac{\pi}{6}}\frac{2\sin x\cdot\cos x}{\sqrt{\sin x}}
Notam :
\sqrt{\sin x}=t\Rightarrow \sin x=t^{2}
Acum calculam derivata in functie de x si in functie de t.
\left(\sin x\right)^{'}dx=\left(t^{2}\right)^{'}\Rightarrow \cos x dx =2t dt
Acum aflam capetele intevalului :
x=\frac{\pi}{2}\Rightarrow \sqrt{\sin \frac{\pi}{2}}=t\Rightarrow t=1
x=\frac{\pi}{6}\Rightarrow \sqrt{\sin \frac{\pi}{6}}=t\Rightarrow t=\sqrt{\frac{1}{2}}=\frac{\sqrt{2}}{2}.
Astfel obtinem :
\int^{1}_{\frac{\sqrt{2}}{2}}\frac{2\cdot t^{2}}{t}2t dt=\int^{1}_{\frac{\sqrt{2}}{2}}\frac{4\cdot t^{3}}{t} dt=4\int^{1}_{\frac{\sqrt{2}}{2}}t^{2} dt=4\cdot \frac{t^{3}}{3}|^{1}_{\frac{\sqrt{2}}{2}}=
4\frac{1^{3}}{3}-4\left(\frac{\sqrt{2}}{2}\right)^{3}=
4\frac{1}{3}-4\frac{1}{3}\cdot \frac{\sqrt{8}}{8}=4\frac{1}{3}-4\frac{1}{3}\cdot \frac{2\sqrt{2}}{8}=
\frac{4}{3}-\frac{4}{3}\cdot\frac{\sqrt{2}}{4}=\frac{4}{3}-\frac{4\sqrt{2}}{12}=\frac{4}{3}-\frac{\sqrt{2}}{3}=\frac{4-\sqrt{2}}{3}

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