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

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

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,

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

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

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

Integralen av $$$\sec^{2}{\left(u \right)}$$$ är $$$\int{\sec^{2}{\left(u \right)} d u} = \tan{\left(u \right)}$$$:

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

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

$$- \tan{\left({\color{red}{u}} \right)} = - \tan{\left({\color{red}{\cos{\left(x \right)}}} \right)}$$

Alltså,

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

Lägg till integrationskonstanten:

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

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