Неодређени интеграли – парцијална интеграција 2

Текст задатака објашњених у видео лекцији:

Пр.4)   $\int {x\cos xdx} $

Пр.5)   $\int {\left( {4{x^2} - 3} \right)} \cos \left( {3 - x} \right)dx$

Пр.4)  Први начин:

$\int {x\cos xdx}= $

\[\begin{array}{*{20}{c}}
\begin{gathered}
u = x \hfill \\
du = dx \hfill \\
\end{gathered} &{}&\begin{gathered}
dv = \cos x \hfill \\
v = \int {\cos xdx = \sin x + C} \hfill \\
\end{gathered}
\end{array}\]

$ = x\sin x - \int {\sin xdx = x\sin x + } \cos x + C$

Други начин:

$\int {x\cos xdx}  = $

\[\begin{array}{*{20}{c}}
\begin{gathered}
\cos x = u \hfill \\
- \sin xdx = du \hfill \\
\end{gathered} &{}&\begin{gathered}
dv = xdx \hfill \\
v = \int {xdx = \frac{{{x^2}}}{2} + C} \hfill \\
\end{gathered}
\end{array}\]

$ = \frac{{{x^2}}}{2}\cos x - \int {\frac{{{x^2}}}{2}\left( { - \sin x} \right)} dx = \frac{{{x^2}}}{2}\cos x + \frac{1}{2}\int {{x^2}\sin x} dx$

Пр.5)   $\int {\left( {4{x^2} - 3} \right)} \cos \left( {3 - x} \right)dx = $

\[\begin{array}{*{20}{c}}
\begin{gathered}
u = 4{x^2} - 3 \hfill \\
du = 8xdx \hfill \\
\hfill \\
\hfill \\
\hfill \\
\hfill \\
\hfill \\
\hfill \\
\end{gathered} &{}&\begin{gathered}
dv = \cos \left( {3 - x} \right) \hfill \\
v = \int {\cos \left( {3 - x} \right)dx = } \hfill \\
3 - x = t \hfill \\
- dx = dt \hfill \\
dx = - dt \hfill \\
= - \int {\cos tdt = - } \sin t + C \hfill \\
v = - \sin \left( {3 - x} \right) + C \hfill \\
\end{gathered}
\end{array}\]

$ = \left( {4{x^2} - 3} \right)\left( { - \sin \left( {3 - x} \right)} \right) + \int {\sin \left( {3 - x} \right)8xdx = } $

$ = \left( {3 - 4{x^2}} \right)\left( {\sin \left( {3 - x} \right)} \right) + 8\int {\sin \left( {3 - x} \right)xdx = } $

\[\begin{array}{*{20}{c}}
\begin{gathered}
u = x \hfill \\
du = dx \hfill \\
\hfill \\
\hfill \\
\hfill \\
\hfill \\
\hfill \\
\end{gathered} &{}&\begin{gathered}
dv = \sin \left( {3 - x} \right)dx \hfill \\
v = \int {\sin \left( {3 - x} \right)dx = } \hfill \\
3 - x = t \hfill \\
- dx = dt \hfill \\
dx = - dt \hfill \\
v = \int {\sin tdt = \cos t + C = \cos \left( {3 - x} \right) + C} \hfill \\
\end{gathered}
\end{array}\]

$ = \left( {3 - 4{x^2}} \right)\left( {\sin \left( {3 - x} \right)} \right) + 8\left( {x\cos \left( {3 - x} \right) - \int {\cos \left( {3 - x} \right)dx} } \right) = $

$ = \left( {3 - 4{x^2}} \right)\left( {\sin \left( {3 - x} \right)} \right) + 8\left( {x\cos \left( {3 - x} \right) - \sin \left( {3 - x} \right)} \right) + C = $

$ = \left( {11 - 4{x^2}} \right)\left( {\sin \left( {3 - x} \right)} \right) + 8x\cos \left( {3 - x} \right) + C$