Integralen av $$$\sin{\left(x \right)} \cos^{2}{\left(\cos{\left(x \right)} \right)}$$$

Bestäm $$$\int \sin{\left(x \right)} \cos^{2}{\left(\cos{\left(x \right)} \right)}\, dx$$$.

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

Då $$$du=\left(\cos{\left(x \right)}\right)^{\prime }dx = - \sin{\left(x \right)} dx$$$ (stegen kan ses »), och vi har att $$$\sin{\left(x \right)} dx = - du$$$.

Alltså,

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

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

$${\color{red}{\int{\left(- \cos^{2}{\left(u \right)}\right)d u}}} = {\color{red}{\left(- \int{\cos^{2}{\left(u \right)} d u}\right)}}$$

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

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

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

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

Integrera termvis:

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

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

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

Låt $$$v=2 u$$$ vara.

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

Alltså,

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

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

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

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

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

Kom ihåg att $$$v=2 u$$$:

$$- \frac{u}{2} - \frac{\sin{\left({\color{red}{v}} \right)}}{4} = - \frac{u}{2} - \frac{\sin{\left({\color{red}{\left(2 u\right)}} \right)}}{4}$$

Kom ihåg att $$$u=\cos{\left(x \right)}$$$:

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

Alltså,

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

Lägg till integrationskonstanten:

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

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