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

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

Tillämpa konstantfaktorregeln $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ med $$$c=\frac{1}{2}$$$ och $$$f{\left(x \right)} = \sec^{2}{\left(\frac{x}{2} \right)}$$$:

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

Låt $$$u=\frac{x}{2}$$$ vara.

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

Integralen kan omskrivas som

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

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

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

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 $$$u=\frac{x}{2}$$$:

$$\tan{\left({\color{red}{u}} \right)} = \tan{\left({\color{red}{\left(\frac{x}{2}\right)}} \right)}$$

Alltså,

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

Lägg till integrationskonstanten:

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

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