Integralen av $$$\sqrt{1 - 7 x^{2}}$$$

Bestäm $$$\int \sqrt{1 - 7 x^{2}}\, dx$$$.

Låt $$$x=\frac{\sqrt{7} \sin{\left(u \right)}}{7}$$$ vara.

Då $$$dx=\left(\frac{\sqrt{7} \sin{\left(u \right)}}{7}\right)^{\prime }du = \frac{\sqrt{7} \cos{\left(u \right)}}{7} du$$$ (stegen kan ses »).

Det följer också att $$$u=\operatorname{asin}{\left(\sqrt{7} x \right)}$$$.

Alltså,

$$$\sqrt{1 - 7 x^{2}} = \sqrt{1 - \sin^{2}{\left( u \right)}}$$$

Använd identiteten $$$1 - \sin^{2}{\left( u \right)} = \cos^{2}{\left( u \right)}$$$:

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

Om vi antar att $$$\cos{\left( u \right)} \ge 0$$$, erhåller vi följande:

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

Integralen kan skrivas om som

$${\color{red}{\int{\sqrt{1 - 7 x^{2}} d x}}} = {\color{red}{\int{\frac{\sqrt{7} \cos^{2}{\left(u \right)}}{7} d u}}}$$

Använd potensreduceringsformeln $$$\cos^{2}{\left(\alpha \right)} = \frac{\cos{\left(2 \alpha \right)}}{2} + \frac{1}{2}$$$ med $$$\alpha= u $$$:

$${\color{red}{\int{\frac{\sqrt{7} \cos^{2}{\left(u \right)}}{7} d u}}} = {\color{red}{\int{\frac{\sqrt{7} \left(\cos{\left(2 u \right)} + 1\right)}{14} d u}}}$$

Tillämpa konstantfaktorregeln $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ med $$$c=\frac{1}{2}$$$ och $$$f{\left(u \right)} = \frac{\sqrt{7} \left(\cos{\left(2 u \right)} + 1\right)}{7}$$$:

$${\color{red}{\int{\frac{\sqrt{7} \left(\cos{\left(2 u \right)} + 1\right)}{14} d u}}} = {\color{red}{\left(\frac{\int{\frac{\sqrt{7} \left(\cos{\left(2 u \right)} + 1\right)}{7} d u}}{2}\right)}}$$

Expand the expression:

$$\frac{{\color{red}{\int{\frac{\sqrt{7} \left(\cos{\left(2 u \right)} + 1\right)}{7} d u}}}}{2} = \frac{{\color{red}{\int{\left(\frac{\sqrt{7} \cos{\left(2 u \right)}}{7} + \frac{\sqrt{7}}{7}\right)d u}}}}{2}$$

Integrera termvis:

$$\frac{{\color{red}{\int{\left(\frac{\sqrt{7} \cos{\left(2 u \right)}}{7} + \frac{\sqrt{7}}{7}\right)d u}}}}{2} = \frac{{\color{red}{\left(\int{\frac{\sqrt{7}}{7} d u} + \int{\frac{\sqrt{7} \cos{\left(2 u \right)}}{7} d u}\right)}}}{2}$$

Tillämpa konstantregeln $$$\int c\, du = c u$$$ med $$$c=\frac{\sqrt{7}}{7}$$$:

$$\frac{\int{\frac{\sqrt{7} \cos{\left(2 u \right)}}{7} d u}}{2} + \frac{{\color{red}{\int{\frac{\sqrt{7}}{7} d u}}}}{2} = \frac{\int{\frac{\sqrt{7} \cos{\left(2 u \right)}}{7} d u}}{2} + \frac{{\color{red}{\left(\frac{\sqrt{7} u}{7}\right)}}}{2}$$

Tillämpa konstantfaktorregeln $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ med $$$c=\frac{\sqrt{7}}{7}$$$ och $$$f{\left(u \right)} = \cos{\left(2 u \right)}$$$:

$$\frac{\sqrt{7} u}{14} + \frac{{\color{red}{\int{\frac{\sqrt{7} \cos{\left(2 u \right)}}{7} d u}}}}{2} = \frac{\sqrt{7} u}{14} + \frac{{\color{red}{\left(\frac{\sqrt{7} \int{\cos{\left(2 u \right)} d u}}{7}\right)}}}{2}$$

Låt $$$v=2 u$$$ vara.

Då $$$dv=\left(2 u\right)^{\prime }du = 2 du$$$ (stegen kan ses »), och vi har att $$$du = \frac{dv}{2}$$$.

Integralen blir

$$\frac{\sqrt{7} u}{14} + \frac{\sqrt{7} {\color{red}{\int{\cos{\left(2 u \right)} d u}}}}{14} = \frac{\sqrt{7} u}{14} + \frac{\sqrt{7} {\color{red}{\int{\frac{\cos{\left(v \right)}}{2} d v}}}}{14}$$

Tillämpa konstantfaktorregeln $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ med $$$c=\frac{1}{2}$$$ och $$$f{\left(v \right)} = \cos{\left(v \right)}$$$:

$$\frac{\sqrt{7} u}{14} + \frac{\sqrt{7} {\color{red}{\int{\frac{\cos{\left(v \right)}}{2} d v}}}}{14} = \frac{\sqrt{7} u}{14} + \frac{\sqrt{7} {\color{red}{\left(\frac{\int{\cos{\left(v \right)} d v}}{2}\right)}}}{14}$$

Integralen av cosinus är $$$\int{\cos{\left(v \right)} d v} = \sin{\left(v \right)}$$$:

$$\frac{\sqrt{7} u}{14} + \frac{\sqrt{7} {\color{red}{\int{\cos{\left(v \right)} d v}}}}{28} = \frac{\sqrt{7} u}{14} + \frac{\sqrt{7} {\color{red}{\sin{\left(v \right)}}}}{28}$$

Kom ihåg att $$$v=2 u$$$:

$$\frac{\sqrt{7} u}{14} + \frac{\sqrt{7} \sin{\left({\color{red}{v}} \right)}}{28} = \frac{\sqrt{7} u}{14} + \frac{\sqrt{7} \sin{\left({\color{red}{\left(2 u\right)}} \right)}}{28}$$

Kom ihåg att $$$u=\operatorname{asin}{\left(\sqrt{7} x \right)}$$$:

$$\frac{\sqrt{7} \sin{\left(2 {\color{red}{u}} \right)}}{28} + \frac{\sqrt{7} {\color{red}{u}}}{14} = \frac{\sqrt{7} \sin{\left(2 {\color{red}{\operatorname{asin}{\left(\sqrt{7} x \right)}}} \right)}}{28} + \frac{\sqrt{7} {\color{red}{\operatorname{asin}{\left(\sqrt{7} x \right)}}}}{14}$$

Alltså,

$$\int{\sqrt{1 - 7 x^{2}} d x} = \frac{\sqrt{7} \sin{\left(2 \operatorname{asin}{\left(\sqrt{7} x \right)} \right)}}{28} + \frac{\sqrt{7} \operatorname{asin}{\left(\sqrt{7} x \right)}}{14}$$

Använd formlerna $$$\sin{\left(2 \operatorname{asin}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{1 - \alpha^{2}}$$$, $$$\sin{\left(2 \operatorname{acos}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{1 - \alpha^{2}}$$$, $$$\cos{\left(2 \operatorname{asin}{\left(\alpha \right)} \right)} = 1 - 2 \alpha^{2}$$$, $$$\cos{\left(2 \operatorname{acos}{\left(\alpha \right)} \right)} = 2 \alpha^{2} - 1$$$, $$$\sinh{\left(2 \operatorname{asinh}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{\alpha^{2} + 1}$$$, $$$\sinh{\left(2 \operatorname{acosh}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{\alpha - 1} \sqrt{\alpha + 1}$$$, $$$\cosh{\left(2 \operatorname{asinh}{\left(\alpha \right)} \right)} = 2 \alpha^{2} + 1$$$, $$$\cosh{\left(2 \operatorname{acosh}{\left(\alpha \right)} \right)} = 2 \alpha^{2} - 1$$$ för att förenkla uttrycket:

$$\int{\sqrt{1 - 7 x^{2}} d x} = \frac{x \sqrt{1 - 7 x^{2}}}{2} + \frac{\sqrt{7} \operatorname{asin}{\left(\sqrt{7} x \right)}}{14}$$

Lägg till integrationskonstanten:

$$\int{\sqrt{1 - 7 x^{2}} d x} = \frac{x \sqrt{1 - 7 x^{2}}}{2} + \frac{\sqrt{7} \operatorname{asin}{\left(\sqrt{7} x \right)}}{14}+C$$

$$$\int \sqrt{1 - 7 x^{2}}\, dx = \left(\frac{x \sqrt{1 - 7 x^{2}}}{2} + \frac{\sqrt{7} \operatorname{asin}{\left(\sqrt{7} x \right)}}{14}\right) + C$$$A