Неодређени интеграли – парцијална интеграција 5

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Пр.11)   Решити $\int {\sqrt {4 - {x^2}} } dx$

Пр.11)    $\int {\sqrt {4 - {x^2}} } dx=$

$\sqrt {4 - {x^2}}  \cdot \frac{{\sqrt {4 - {x^2}} }}{{\sqrt {4 - {x^2}} }} = \frac{{4 - {x^2}}}{{\sqrt {4 - {x^2}} }} = \frac{4}{{\sqrt {4 - {x^2}} }} - \frac{{{x^2}}}{{\sqrt {4 - {x^2}} }}$

$ = \int {\frac{4}{{\sqrt {4 - {x^2}} }}} dx - \int {\frac{{{x^2}}}{{\sqrt {4 - {x^2}} }}} dx$

${I_1} = \int {\frac{4}{{\sqrt {4 - {x^2}} }}} dx = 4\int {\frac{1}{{\sqrt {4\left( {1 - {{\left( {\frac{x}{2}} \right)}^2}} \right)} }}} dx = $

$\frac{x}{2} = t$

$\frac{{dx}}{2} = dt$

$dx = 2dt$

$ = 2\int {\frac{{2dt}}{{\sqrt {1 - {t^2}} }}}  = 4\arcsin t + C = 4\arcsin \frac{x}{2} + C$

${I_2} = \int {\frac{{{x^2}}}{{\sqrt {4 - {x^2}} }}} dx = $

\[\begin{array}{*{20}{c}}
\begin{gathered}
u = x \hfill \\
du = dx \hfill \\
\hfill \\
\hfill \\
\hfill \\
\hfill \\
\hfill \\
\hfill \\
\hfill \\
\hfill \\
\hfill \\
\hfill \\
\end{gathered} &{}&\begin{gathered}
dv = \frac{x}{{\sqrt {4 - {x^2}} }}dx \hfill \\
v = \int {\frac{x}{{\sqrt {4 - {x^2}} }}dx = } \hfill \\
4 - {x^2} = s \hfill \\
- 2xdx = ds \hfill \\
xdx = \frac{{ds}}{{ - 2}} \hfill \\
\hfill \\
= \int {\frac{{\frac{{ds}}{{ - 2}}}}{{\sqrt s }}} = - \frac{1}{2}\int {{s^{ - \frac{1}{2}}}ds} \hfill \\
v = - \frac{1}{2} \cdot \frac{s}{{\frac{1}{2}}} + C = - \sqrt {4 - {x^2}} + C \hfill \\
\end{gathered}
\end{array}\]

$ =  - x\sqrt {4 - {x^2}}  + \int {\sqrt {4 - {x^2}} dx} $

$\int {\sqrt {4 - {x^2}} dx}  = {I_1} - {I_2}$

$\int {\sqrt {4 - {x^2}} dx}  = 4\arcsin \frac{x}{2} + x\sqrt {4 - {x^2}}  - \int {\sqrt {4 - {x^2}} dx} $

$2\int {\sqrt {4 - {x^2}} dx}  = 4\arcsin \frac{x}{2} + x\sqrt {4 - {x^2}} $

$\int {\sqrt {4 - {x^2}} dx}  = 2\arcsin \frac{x}{2} + \frac{x}{2}\sqrt {4 - {x^2}} $