Příklad 4.64
[!example] Vypočtěte $\(\Large \int_{0}^{\pi} 2x^2\cdot\cos(2x)\;dx\)$
\[\large
\begin{align}
&= \int_{0}^{\pi} x^2\cdot\cos(2x)\;2dx \\
\end{align}
\]
[!tip]+ Substituce
\[\large\begin{aligned} u &= 2x \\ du &= 2dx \\ \end{aligned}\]
\[\large
\begin{align}
&= \int_{0}^{2\pi} \left(\frac{u}{2}\right)^2\cdot\cos(u)\;du \\
&= \int_{0}^{2\pi} \frac{1}{4}u^2\cdot\cos(u)\;du \\
&= \frac{1}{4}\int_{0}^{2\pi} u^2\cdot\cos(u)\;du \\
\end{align}
\]
[!tip]+ Aplikace [[Per Partes#Metoda DI|DI metody]] Sestavíme si DI tabulku:
D I + \(u^2\) \(\cos(u)\) - \(2u\) \(\sin(u)\) + \(2\) \(-\cos(u)\) - \(0\) \(-\sin(u)\) Zapíšeme výsledek: \(u^2\sin(u)+2u\cos(u)-2\sin(u)\)
\[\large
\begin{align}
&= \frac{1}{4}\left[u^2\sin(u)+2u\cos(u)-2\sin(u)\right]_{0}^{2\pi} \\
&= \frac{1}{4}\left[4\pi\right] \\
&= \boxed{\pi}
\end{align}
\]