Integraal van $$$\frac{b^{2} \sin{\left(x \right)} \sin{\left(2 x \right)} \cos{\left(x \right)}}{a^{2}}$$$ met betrekking tot $$$x$$$

De rekenmachine zal de integraal/primitieve van $$$\frac{b^{2} \sin{\left(x \right)} \sin{\left(2 x \right)} \cos{\left(x \right)}}{a^{2}}$$$ met betrekking tot $$$x$$$ bepalen, waarbij de stappen worden getoond.

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Uw invoer

Bepaal $$$\int \frac{b^{2} \sin{\left(x \right)} \sin{\left(2 x \right)} \cos{\left(x \right)}}{a^{2}}\, dx$$$.

Oplossing

Herschrijf de integraand:

$${\color{red}{\int{\frac{b^{2} \sin{\left(x \right)} \sin{\left(2 x \right)} \cos{\left(x \right)}}{a^{2}} d x}}} = {\color{red}{\int{\frac{2 b^{2} \sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)}}{a^{2}} d x}}}$$

Pas de constante-veelvoudregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ toe met $$$c=\frac{2 b^{2}}{a^{2}}$$$ en $$$f{\left(x \right)} = \sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)}$$$:

$${\color{red}{\int{\frac{2 b^{2} \sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)}}{a^{2}} d x}}} = {\color{red}{\left(\frac{2 b^{2} \int{\sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)} d x}}{a^{2}}\right)}}$$

Herschrijf de integraand met behulp van de dubbelhoeksformule $$$\sin\left(x \right)\cos\left(x \right)=\frac{1}{2}\sin\left( 2 x \right)$$$:

$$\frac{2 b^{2} {\color{red}{\int{\sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)} d x}}}}{a^{2}} = \frac{2 b^{2} {\color{red}{\int{\frac{\sin^{2}{\left(2 x \right)}}{4} d x}}}}{a^{2}}$$

Pas de constante-veelvoudregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ toe met $$$c=\frac{1}{4}$$$ en $$$f{\left(x \right)} = \sin^{2}{\left(2 x \right)}$$$:

$$\frac{2 b^{2} {\color{red}{\int{\frac{\sin^{2}{\left(2 x \right)}}{4} d x}}}}{a^{2}} = \frac{2 b^{2} {\color{red}{\left(\frac{\int{\sin^{2}{\left(2 x \right)} d x}}{4}\right)}}}{a^{2}}$$

Pas de machtsreductieformule $$$\sin^{2}{\left(\alpha \right)} = \frac{1}{2} - \frac{\cos{\left(2 \alpha \right)}}{2}$$$ toe met $$$\alpha=2 x$$$:

$$\frac{b^{2} {\color{red}{\int{\sin^{2}{\left(2 x \right)} d x}}}}{2 a^{2}} = \frac{b^{2} {\color{red}{\int{\left(\frac{1}{2} - \frac{\cos{\left(4 x \right)}}{2}\right)d x}}}}{2 a^{2}}$$

Pas de constante-veelvoudregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ toe met $$$c=\frac{1}{2}$$$ en $$$f{\left(x \right)} = 1 - \cos{\left(4 x \right)}$$$:

$$\frac{b^{2} {\color{red}{\int{\left(\frac{1}{2} - \frac{\cos{\left(4 x \right)}}{2}\right)d x}}}}{2 a^{2}} = \frac{b^{2} {\color{red}{\left(\frac{\int{\left(1 - \cos{\left(4 x \right)}\right)d x}}{2}\right)}}}{2 a^{2}}$$

Integreer termgewijs:

$$\frac{b^{2} {\color{red}{\int{\left(1 - \cos{\left(4 x \right)}\right)d x}}}}{4 a^{2}} = \frac{b^{2} {\color{red}{\left(\int{1 d x} - \int{\cos{\left(4 x \right)} d x}\right)}}}{4 a^{2}}$$

Pas de constantenregel $$$\int c\, dx = c x$$$ toe met $$$c=1$$$:

$$\frac{b^{2} \left(- \int{\cos{\left(4 x \right)} d x} + {\color{red}{\int{1 d x}}}\right)}{4 a^{2}} = \frac{b^{2} \left(- \int{\cos{\left(4 x \right)} d x} + {\color{red}{x}}\right)}{4 a^{2}}$$

Zij $$$u=4 x$$$.

Dan $$$du=\left(4 x\right)^{\prime }dx = 4 dx$$$ (de stappen zijn te zien »), en dan geldt dat $$$dx = \frac{du}{4}$$$.

De integraal kan worden herschreven als

$$\frac{b^{2} \left(x - {\color{red}{\int{\cos{\left(4 x \right)} d x}}}\right)}{4 a^{2}} = \frac{b^{2} \left(x - {\color{red}{\int{\frac{\cos{\left(u \right)}}{4} d u}}}\right)}{4 a^{2}}$$

Pas de constante-veelvoudregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ toe met $$$c=\frac{1}{4}$$$ en $$$f{\left(u \right)} = \cos{\left(u \right)}$$$:

$$\frac{b^{2} \left(x - {\color{red}{\int{\frac{\cos{\left(u \right)}}{4} d u}}}\right)}{4 a^{2}} = \frac{b^{2} \left(x - {\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{4}\right)}}\right)}{4 a^{2}}$$

De integraal van de cosinus is $$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$:

$$\frac{b^{2} \left(x - \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{4}\right)}{4 a^{2}} = \frac{b^{2} \left(x - \frac{{\color{red}{\sin{\left(u \right)}}}}{4}\right)}{4 a^{2}}$$

We herinneren eraan dat $$$u=4 x$$$:

$$\frac{b^{2} \left(x - \frac{\sin{\left({\color{red}{u}} \right)}}{4}\right)}{4 a^{2}} = \frac{b^{2} \left(x - \frac{\sin{\left({\color{red}{\left(4 x\right)}} \right)}}{4}\right)}{4 a^{2}}$$

Dus,

$$\int{\frac{b^{2} \sin{\left(x \right)} \sin{\left(2 x \right)} \cos{\left(x \right)}}{a^{2}} d x} = \frac{b^{2} \left(x - \frac{\sin{\left(4 x \right)}}{4}\right)}{4 a^{2}}$$

Vereenvoudig:

$$\int{\frac{b^{2} \sin{\left(x \right)} \sin{\left(2 x \right)} \cos{\left(x \right)}}{a^{2}} d x} = \frac{b^{2} \left(4 x - \sin{\left(4 x \right)}\right)}{16 a^{2}}$$

Voeg de integratieconstante toe:

$$\int{\frac{b^{2} \sin{\left(x \right)} \sin{\left(2 x \right)} \cos{\left(x \right)}}{a^{2}} d x} = \frac{b^{2} \left(4 x - \sin{\left(4 x \right)}\right)}{16 a^{2}}+C$$

Antwoord

$$$\int \frac{b^{2} \sin{\left(x \right)} \sin{\left(2 x \right)} \cos{\left(x \right)}}{a^{2}}\, dx = \frac{b^{2} \left(4 x - \sin{\left(4 x \right)}\right)}{16 a^{2}} + C$$$A