Integraal van $$$\frac{\pi x^{2} z \tan{\left(2 \right)} \sec{\left(4 \right)}}{40}$$$ met betrekking tot $$$x$$$
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Uw invoer
Bepaal $$$\int \frac{\pi x^{2} z \tan{\left(2 \right)} \sec{\left(4 \right)}}{40}\, dx$$$.
Oplossing
Pas de constante-veelvoudregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ toe met $$$c=\frac{\pi z \tan{\left(2 \right)} \sec{\left(4 \right)}}{40}$$$ en $$$f{\left(x \right)} = x^{2}$$$:
$${\color{red}{\int{\frac{\pi x^{2} z \tan{\left(2 \right)} \sec{\left(4 \right)}}{40} d x}}} = {\color{red}{\left(\frac{\pi z \tan{\left(2 \right)} \sec{\left(4 \right)} \int{x^{2} d x}}{40}\right)}}$$
Pas de machtsregel $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ toe met $$$n=2$$$:
$$\frac{\pi z \tan{\left(2 \right)} \sec{\left(4 \right)} {\color{red}{\int{x^{2} d x}}}}{40}=\frac{\pi z \tan{\left(2 \right)} \sec{\left(4 \right)} {\color{red}{\frac{x^{1 + 2}}{1 + 2}}}}{40}=\frac{\pi z \tan{\left(2 \right)} \sec{\left(4 \right)} {\color{red}{\left(\frac{x^{3}}{3}\right)}}}{40}$$
Dus,
$$\int{\frac{\pi x^{2} z \tan{\left(2 \right)} \sec{\left(4 \right)}}{40} d x} = \frac{\pi x^{3} z \tan{\left(2 \right)} \sec{\left(4 \right)}}{120}$$
Voeg de integratieconstante toe:
$$\int{\frac{\pi x^{2} z \tan{\left(2 \right)} \sec{\left(4 \right)}}{40} d x} = \frac{\pi x^{3} z \tan{\left(2 \right)} \sec{\left(4 \right)}}{120}+C$$
Antwoord
$$$\int \frac{\pi x^{2} z \tan{\left(2 \right)} \sec{\left(4 \right)}}{40}\, dx = \frac{\pi x^{3} z \tan{\left(2 \right)} \sec{\left(4 \right)}}{120} + C$$$A