Integraal van $$$\sqrt{x} \sqrt{1 - x}$$$

Bepaal $$$\int \sqrt{x} \sqrt{1 - x}\, dx$$$.

De invoer is herschreven: $$$\int{\sqrt{x} \sqrt{1 - x} d x}=\int{\sqrt{- x^{2} + x} d x}$$$.

Voltooi het kwadraat (stappen zijn te zien »): $$$- x^{2} + x = \frac{1}{4} - \left(x - \frac{1}{2}\right)^{2}$$$:

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

Zij $$$u=x - \frac{1}{2}$$$.

Dan $$$du=\left(x - \frac{1}{2}\right)^{\prime }dx = 1 dx$$$ (de stappen zijn te zien »), en dan geldt dat $$$dx = du$$$.

Dus,

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

Zij $$$u=\frac{\sin{\left(v \right)}}{2}$$$.

Dan $$$du=\left(\frac{\sin{\left(v \right)}}{2}\right)^{\prime }dv = \frac{\cos{\left(v \right)}}{2} dv$$$ (zie » voor de stappen).

Bovendien volgt dat $$$v=\operatorname{asin}{\left(2 u \right)}$$$.

De integraand wordt

$$$\sqrt{\frac{1}{4} - u ^{2}} = \sqrt{\frac{1}{4} - \frac{\sin^{2}{\left( v \right)}}{4}}$$$

Gebruik de identiteit $$$1 - \sin^{2}{\left( v \right)} = \cos^{2}{\left( v \right)}$$$:

$$$\sqrt{\frac{1}{4} - \frac{\sin^{2}{\left( v \right)}}{4}}=\frac{\sqrt{1 - \sin^{2}{\left( v \right)}}}{2}=\frac{\sqrt{\cos^{2}{\left( v \right)}}}{2}$$$

Aangenomen dat $$$\cos{\left( v \right)} \ge 0$$$, verkrijgen we het volgende:

$$$\frac{\sqrt{\cos^{2}{\left( v \right)}}}{2} = \frac{\cos{\left( v \right)}}{2}$$$

Dus,

$${\color{red}{\int{\sqrt{\frac{1}{4} - u^{2}} d u}}} = {\color{red}{\int{\frac{\cos^{2}{\left(v \right)}}{4} d v}}}$$

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

$${\color{red}{\int{\frac{\cos^{2}{\left(v \right)}}{4} d v}}} = {\color{red}{\left(\frac{\int{\cos^{2}{\left(v \right)} d v}}{4}\right)}}$$

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

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

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

$$\frac{{\color{red}{\int{\left(\frac{\cos{\left(2 v \right)}}{2} + \frac{1}{2}\right)d v}}}}{4} = \frac{{\color{red}{\left(\frac{\int{\left(\cos{\left(2 v \right)} + 1\right)d v}}{2}\right)}}}{4}$$

Integreer termgewijs:

$$\frac{{\color{red}{\int{\left(\cos{\left(2 v \right)} + 1\right)d v}}}}{8} = \frac{{\color{red}{\left(\int{1 d v} + \int{\cos{\left(2 v \right)} d v}\right)}}}{8}$$

Pas de constantenregel $$$\int c\, dv = c v$$$ toe met $$$c=1$$$:

$$\frac{\int{\cos{\left(2 v \right)} d v}}{8} + \frac{{\color{red}{\int{1 d v}}}}{8} = \frac{\int{\cos{\left(2 v \right)} d v}}{8} + \frac{{\color{red}{v}}}{8}$$

Zij $$$w=2 v$$$.

Dan $$$dw=\left(2 v\right)^{\prime }dv = 2 dv$$$ (de stappen zijn te zien »), en dan geldt dat $$$dv = \frac{dw}{2}$$$.

De integraal wordt

$$\frac{v}{8} + \frac{{\color{red}{\int{\cos{\left(2 v \right)} d v}}}}{8} = \frac{v}{8} + \frac{{\color{red}{\int{\frac{\cos{\left(w \right)}}{2} d w}}}}{8}$$

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

$$\frac{v}{8} + \frac{{\color{red}{\int{\frac{\cos{\left(w \right)}}{2} d w}}}}{8} = \frac{v}{8} + \frac{{\color{red}{\left(\frac{\int{\cos{\left(w \right)} d w}}{2}\right)}}}{8}$$

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

$$\frac{v}{8} + \frac{{\color{red}{\int{\cos{\left(w \right)} d w}}}}{16} = \frac{v}{8} + \frac{{\color{red}{\sin{\left(w \right)}}}}{16}$$

We herinneren eraan dat $$$w=2 v$$$:

$$\frac{v}{8} + \frac{\sin{\left({\color{red}{w}} \right)}}{16} = \frac{v}{8} + \frac{\sin{\left({\color{red}{\left(2 v\right)}} \right)}}{16}$$

We herinneren eraan dat $$$v=\operatorname{asin}{\left(2 u \right)}$$$:

$$\frac{\sin{\left(2 {\color{red}{v}} \right)}}{16} + \frac{{\color{red}{v}}}{8} = \frac{\sin{\left(2 {\color{red}{\operatorname{asin}{\left(2 u \right)}}} \right)}}{16} + \frac{{\color{red}{\operatorname{asin}{\left(2 u \right)}}}}{8}$$

We herinneren eraan dat $$$u=x - \frac{1}{2}$$$:

$$\frac{\sin{\left(2 \operatorname{asin}{\left(2 {\color{red}{u}} \right)} \right)}}{16} + \frac{\operatorname{asin}{\left(2 {\color{red}{u}} \right)}}{8} = \frac{\sin{\left(2 \operatorname{asin}{\left(2 {\color{red}{\left(x - \frac{1}{2}\right)}} \right)} \right)}}{16} + \frac{\operatorname{asin}{\left(2 {\color{red}{\left(x - \frac{1}{2}\right)}} \right)}}{8}$$

Dus,

$$\int{\sqrt{- x^{2} + x} d x} = \frac{\sin{\left(2 \operatorname{asin}{\left(2 x - 1 \right)} \right)}}{16} + \frac{\operatorname{asin}{\left(2 x - 1 \right)}}{8}$$

Gebruik de formules $$$\sin{\left(2 \operatorname{asin}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{1 - \alpha^{2}}$$$, $$$\sin{\left(2 \operatorname{acos}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{1 - \alpha^{2}}$$$, $$$\cos{\left(2 \operatorname{asin}{\left(\alpha \right)} \right)} = 1 - 2 \alpha^{2}$$$, $$$\cos{\left(2 \operatorname{acos}{\left(\alpha \right)} \right)} = 2 \alpha^{2} - 1$$$, $$$\sinh{\left(2 \operatorname{asinh}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{\alpha^{2} + 1}$$$, $$$\sinh{\left(2 \operatorname{acosh}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{\alpha - 1} \sqrt{\alpha + 1}$$$, $$$\cosh{\left(2 \operatorname{asinh}{\left(\alpha \right)} \right)} = 2 \alpha^{2} + 1$$$, $$$\cosh{\left(2 \operatorname{acosh}{\left(\alpha \right)} \right)} = 2 \alpha^{2} - 1$$$ om de uitdrukking te vereenvoudigen:

$$\int{\sqrt{- x^{2} + x} d x} = \frac{\sqrt{1 - \left(2 x - 1\right)^{2}} \left(2 x - 1\right)}{8} + \frac{\operatorname{asin}{\left(2 x - 1 \right)}}{8}$$

Voeg de integratieconstante toe:

$$\int{\sqrt{- x^{2} + x} d x} = \frac{\sqrt{1 - \left(2 x - 1\right)^{2}} \left(2 x - 1\right)}{8} + \frac{\operatorname{asin}{\left(2 x - 1 \right)}}{8}+C$$

$$$\int \sqrt{x} \sqrt{1 - x}\, dx = \left(\frac{\sqrt{1 - \left(2 x - 1\right)^{2}} \left(2 x - 1\right)}{8} + \frac{\operatorname{asin}{\left(2 x - 1 \right)}}{8}\right) + C$$$A