Integralen av $$$12 int_{0}^{6} \sqrt{36 - x^{2}}$$$ med avseende på $$$x$$$

Bestäm $$$\int 12 int_{0}^{6} \sqrt{36 - x^{2}}\, dx$$$.

Tillämpa konstantfaktorregeln $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ med $$$c=12 int_{0}^{6}$$$ och $$$f{\left(x \right)} = \sqrt{36 - x^{2}}$$$:

$${\color{red}{\int{12 int_{0}^{6} \sqrt{36 - x^{2}} d x}}} = {\color{red}{\left(12 int_{0}^{6} \int{\sqrt{36 - x^{2}} d x}\right)}}$$

Låt $$$x=6 \sin{\left(u \right)}$$$ vara.

Då $$$dx=\left(6 \sin{\left(u \right)}\right)^{\prime }du = 6 \cos{\left(u \right)} du$$$ (stegen kan ses »).

Det följer också att $$$u=\operatorname{asin}{\left(\frac{x}{6} \right)}$$$.

Alltså,

$$$\sqrt{36 - x^{2}} = \sqrt{36 - 36 \sin^{2}{\left( u \right)}}$$$

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

$$$\sqrt{36 - 36 \sin^{2}{\left( u \right)}}=6 \sqrt{1 - \sin^{2}{\left( u \right)}}=6 \sqrt{\cos^{2}{\left( u \right)}}$$$

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

$$$6 \sqrt{\cos^{2}{\left( u \right)}} = 6 \cos{\left( u \right)}$$$

Alltså,

$$12 int_{0}^{6} {\color{red}{\int{\sqrt{36 - x^{2}} d x}}} = 12 int_{0}^{6} {\color{red}{\int{36 \cos^{2}{\left(u \right)} d u}}}$$

Tillämpa konstantfaktorregeln $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ med $$$c=36$$$ och $$$f{\left(u \right)} = \cos^{2}{\left(u \right)}$$$:

$$12 int_{0}^{6} {\color{red}{\int{36 \cos^{2}{\left(u \right)} d u}}} = 12 int_{0}^{6} {\color{red}{\left(36 \int{\cos^{2}{\left(u \right)} d u}\right)}}$$

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

$$432 int_{0}^{6} {\color{red}{\int{\cos^{2}{\left(u \right)} d u}}} = 432 int_{0}^{6} {\color{red}{\int{\left(\frac{\cos{\left(2 u \right)}}{2} + \frac{1}{2}\right)d u}}}$$

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

$$432 int_{0}^{6} {\color{red}{\int{\left(\frac{\cos{\left(2 u \right)}}{2} + \frac{1}{2}\right)d u}}} = 432 int_{0}^{6} {\color{red}{\left(\frac{\int{\left(\cos{\left(2 u \right)} + 1\right)d u}}{2}\right)}}$$

Integrera termvis:

$$216 int_{0}^{6} {\color{red}{\int{\left(\cos{\left(2 u \right)} + 1\right)d u}}} = 216 int_{0}^{6} {\color{red}{\left(\int{1 d u} + \int{\cos{\left(2 u \right)} d u}\right)}}$$

Tillämpa konstantregeln $$$\int c\, du = c u$$$ med $$$c=1$$$:

$$216 int_{0}^{6} \left(\int{\cos{\left(2 u \right)} d u} + {\color{red}{\int{1 d u}}}\right) = 216 int_{0}^{6} \left(\int{\cos{\left(2 u \right)} d u} + {\color{red}{u}}\right)$$

Låt $$$v=2 u$$$ vara.

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

Alltså,

$$216 int_{0}^{6} \left(u + {\color{red}{\int{\cos{\left(2 u \right)} d u}}}\right) = 216 int_{0}^{6} \left(u + {\color{red}{\int{\frac{\cos{\left(v \right)}}{2} d v}}}\right)$$

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

$$216 int_{0}^{6} \left(u + {\color{red}{\int{\frac{\cos{\left(v \right)}}{2} d v}}}\right) = 216 int_{0}^{6} \left(u + {\color{red}{\left(\frac{\int{\cos{\left(v \right)} d v}}{2}\right)}}\right)$$

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

$$216 int_{0}^{6} \left(u + \frac{{\color{red}{\int{\cos{\left(v \right)} d v}}}}{2}\right) = 216 int_{0}^{6} \left(u + \frac{{\color{red}{\sin{\left(v \right)}}}}{2}\right)$$

Kom ihåg att $$$v=2 u$$$:

$$216 int_{0}^{6} \left(u + \frac{\sin{\left({\color{red}{v}} \right)}}{2}\right) = 216 int_{0}^{6} \left(u + \frac{\sin{\left({\color{red}{\left(2 u\right)}} \right)}}{2}\right)$$

Kom ihåg att $$$u=\operatorname{asin}{\left(\frac{x}{6} \right)}$$$:

$$216 int_{0}^{6} \left(\frac{\sin{\left(2 {\color{red}{u}} \right)}}{2} + {\color{red}{u}}\right) = 216 int_{0}^{6} \left(\frac{\sin{\left(2 {\color{red}{\operatorname{asin}{\left(\frac{x}{6} \right)}}} \right)}}{2} + {\color{red}{\operatorname{asin}{\left(\frac{x}{6} \right)}}}\right)$$

Alltså,

$$\int{12 int_{0}^{6} \sqrt{36 - x^{2}} d x} = 216 int_{0}^{6} \left(\frac{\sin{\left(2 \operatorname{asin}{\left(\frac{x}{6} \right)} \right)}}{2} + \operatorname{asin}{\left(\frac{x}{6} \right)}\right)$$

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:

$$\int{12 int_{0}^{6} \sqrt{36 - x^{2}} d x} = 216 int_{0}^{6} \left(\frac{x \sqrt{1 - \frac{x^{2}}{36}}}{6} + \operatorname{asin}{\left(\frac{x}{6} \right)}\right)$$

Förenkla ytterligare:

$$\int{12 int_{0}^{6} \sqrt{36 - x^{2}} d x} = 6 int_{0}^{6} \left(x \sqrt{36 - x^{2}} + 36 \operatorname{asin}{\left(\frac{x}{6} \right)}\right)$$

Lägg till integrationskonstanten:

$$\int{12 int_{0}^{6} \sqrt{36 - x^{2}} d x} = 6 int_{0}^{6} \left(x \sqrt{36 - x^{2}} + 36 \operatorname{asin}{\left(\frac{x}{6} \right)}\right)+C$$

$$$\int 12 int_{0}^{6} \sqrt{36 - x^{2}}\, dx = 6 int_{0}^{6} \left(x \sqrt{36 - x^{2}} + 36 \operatorname{asin}{\left(\frac{x}{6} \right)}\right) + C$$$A