Integralen av $$$4 x \sec^{2}{\left(x \right)}$$$

Bestäm $$$\int 4 x \sec^{2}{\left(x \right)}\, dx$$$.

Tillämpa konstantfaktorregeln $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ med $$$c=4$$$ och $$$f{\left(x \right)} = x \sec^{2}{\left(x \right)}$$$:

$${\color{red}{\int{4 x \sec^{2}{\left(x \right)} d x}}} = {\color{red}{\left(4 \int{x \sec^{2}{\left(x \right)} d x}\right)}}$$

För integralen $$$\int{x \sec^{2}{\left(x \right)} d x}$$$, använd partiell integration $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Låt $$$\operatorname{u}=x$$$ och $$$\operatorname{dv}=\sec^{2}{\left(x \right)} dx$$$.

Då gäller $$$\operatorname{du}=\left(x\right)^{\prime }dx=1 dx$$$ (stegen kan ses ») och $$$\operatorname{v}=\int{\sec^{2}{\left(x \right)} d x}=\tan{\left(x \right)}$$$ (stegen kan ses »).

Integralen blir

$$4 {\color{red}{\int{x \sec^{2}{\left(x \right)} d x}}}=4 {\color{red}{\left(x \cdot \tan{\left(x \right)}-\int{\tan{\left(x \right)} \cdot 1 d x}\right)}}=4 {\color{red}{\left(x \tan{\left(x \right)} - \int{\tan{\left(x \right)} d x}\right)}}$$

Skriv om tangenten som $$$\tan\left(x\right)=\frac{\sin\left(x\right)}{\cos\left(x\right)}$$$:

$$4 x \tan{\left(x \right)} - 4 {\color{red}{\int{\tan{\left(x \right)} d x}}} = 4 x \tan{\left(x \right)} - 4 {\color{red}{\int{\frac{\sin{\left(x \right)}}{\cos{\left(x \right)}} d x}}}$$

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,

$$4 x \tan{\left(x \right)} - 4 {\color{red}{\int{\frac{\sin{\left(x \right)}}{\cos{\left(x \right)}} d x}}} = 4 x \tan{\left(x \right)} - 4 {\color{red}{\int{\left(- \frac{1}{u}\right)d u}}}$$

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

$$4 x \tan{\left(x \right)} - 4 {\color{red}{\int{\left(- \frac{1}{u}\right)d u}}} = 4 x \tan{\left(x \right)} - 4 {\color{red}{\left(- \int{\frac{1}{u} d u}\right)}}$$

Integralen av $$$\frac{1}{u}$$$ är $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

$$4 x \tan{\left(x \right)} + 4 {\color{red}{\int{\frac{1}{u} d u}}} = 4 x \tan{\left(x \right)} + 4 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$

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

$$4 x \tan{\left(x \right)} + 4 \ln{\left(\left|{{\color{red}{u}}}\right| \right)} = 4 x \tan{\left(x \right)} + 4 \ln{\left(\left|{{\color{red}{\cos{\left(x \right)}}}}\right| \right)}$$

Alltså,

$$\int{4 x \sec^{2}{\left(x \right)} d x} = 4 x \tan{\left(x \right)} + 4 \ln{\left(\left|{\cos{\left(x \right)}}\right| \right)}$$

Förenkla:

$$\int{4 x \sec^{2}{\left(x \right)} d x} = 4 \left(x \tan{\left(x \right)} + \ln{\left(\left|{\cos{\left(x \right)}}\right| \right)}\right)$$

Lägg till integrationskonstanten:

$$\int{4 x \sec^{2}{\left(x \right)} d x} = 4 \left(x \tan{\left(x \right)} + \ln{\left(\left|{\cos{\left(x \right)}}\right| \right)}\right)+C$$

$$$\int 4 x \sec^{2}{\left(x \right)}\, dx = 4 \left(x \tan{\left(x \right)} + \ln\left(\left|{\cos{\left(x \right)}}\right|\right)\right) + C$$$A