Integralen av $$$4 x^{3} - \frac{1}{\cos{\left(2 x \right)}}$$$

Bestäm $$$\int \left(4 x^{3} - \frac{1}{\cos{\left(2 x \right)}}\right)\, dx$$$.

Integrera termvis:

$${\color{red}{\int{\left(4 x^{3} - \frac{1}{\cos{\left(2 x \right)}}\right)d x}}} = {\color{red}{\left(\int{4 x^{3} d x} - \int{\frac{1}{\cos{\left(2 x \right)}} d x}\right)}}$$

Låt $$$u=2 x$$$ vara.

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

Alltså,

$$\int{4 x^{3} d x} - {\color{red}{\int{\frac{1}{\cos{\left(2 x \right)}} d x}}} = \int{4 x^{3} d x} - {\color{red}{\int{\frac{1}{2 \cos{\left(u \right)}} d u}}}$$

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

$$\int{4 x^{3} d x} - {\color{red}{\int{\frac{1}{2 \cos{\left(u \right)}} d u}}} = \int{4 x^{3} d x} - {\color{red}{\left(\frac{\int{\frac{1}{\cos{\left(u \right)}} d u}}{2}\right)}}$$

Skriv om cosinus i termer av sinus med hjälp av formeln $$$\cos\left( u \right)=\sin\left( u + \frac{\pi}{2}\right)$$$ och skriv sedan om sinus med dubbelvinkelformeln $$$\sin\left( u \right)=2\sin\left(\frac{ u }{2}\right)\cos\left(\frac{ u }{2}\right)$$$:

$$\int{4 x^{3} d x} - \frac{{\color{red}{\int{\frac{1}{\cos{\left(u \right)}} d u}}}}{2} = \int{4 x^{3} d x} - \frac{{\color{red}{\int{\frac{1}{2 \sin{\left(\frac{u}{2} + \frac{\pi}{4} \right)} \cos{\left(\frac{u}{2} + \frac{\pi}{4} \right)}} d u}}}}{2}$$

Multiplicera täljare och nämnare med $$$\sec^2\left(\frac{ u }{2} + \frac{\pi}{4} \right)$$$:

$$\int{4 x^{3} d x} - \frac{{\color{red}{\int{\frac{1}{2 \sin{\left(\frac{u}{2} + \frac{\pi}{4} \right)} \cos{\left(\frac{u}{2} + \frac{\pi}{4} \right)}} d u}}}}{2} = \int{4 x^{3} d x} - \frac{{\color{red}{\int{\frac{\sec^{2}{\left(\frac{u}{2} + \frac{\pi}{4} \right)}}{2 \tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}} d u}}}}{2}$$

Låt $$$v=\tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}$$$ vara.

Då $$$dv=\left(\tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}\right)^{\prime }du = \frac{\sec^{2}{\left(\frac{u}{2} + \frac{\pi}{4} \right)}}{2} du$$$ (stegen kan ses »), och vi har att $$$\sec^{2}{\left(\frac{u}{2} + \frac{\pi}{4} \right)} du = 2 dv$$$.

Alltså,

$$\int{4 x^{3} d x} - \frac{{\color{red}{\int{\frac{\sec^{2}{\left(\frac{u}{2} + \frac{\pi}{4} \right)}}{2 \tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}} d u}}}}{2} = \int{4 x^{3} d x} - \frac{{\color{red}{\int{\frac{1}{v} d v}}}}{2}$$

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

$$\int{4 x^{3} d x} - \frac{{\color{red}{\int{\frac{1}{v} d v}}}}{2} = \int{4 x^{3} d x} - \frac{{\color{red}{\ln{\left(\left|{v}\right| \right)}}}}{2}$$

Kom ihåg att $$$v=\tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}$$$:

$$- \frac{\ln{\left(\left|{{\color{red}{v}}}\right| \right)}}{2} + \int{4 x^{3} d x} = - \frac{\ln{\left(\left|{{\color{red}{\tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}}}}\right| \right)}}{2} + \int{4 x^{3} d x}$$

Kom ihåg att $$$u=2 x$$$:

$$- \frac{\ln{\left(\left|{\tan{\left(\frac{\pi}{4} + \frac{{\color{red}{u}}}{2} \right)}}\right| \right)}}{2} + \int{4 x^{3} d x} = - \frac{\ln{\left(\left|{\tan{\left(\frac{\pi}{4} + \frac{{\color{red}{\left(2 x\right)}}}{2} \right)}}\right| \right)}}{2} + \int{4 x^{3} d x}$$

Tillämpa konstantfaktorregeln $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ med $$$c=4$$$ och $$$f{\left(x \right)} = x^{3}$$$:

$$- \frac{\ln{\left(\left|{\tan{\left(x + \frac{\pi}{4} \right)}}\right| \right)}}{2} + {\color{red}{\int{4 x^{3} d x}}} = - \frac{\ln{\left(\left|{\tan{\left(x + \frac{\pi}{4} \right)}}\right| \right)}}{2} + {\color{red}{\left(4 \int{x^{3} d x}\right)}}$$

Tillämpa potensregeln $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ med $$$n=3$$$:

$$- \frac{\ln{\left(\left|{\tan{\left(x + \frac{\pi}{4} \right)}}\right| \right)}}{2} + 4 {\color{red}{\int{x^{3} d x}}}=- \frac{\ln{\left(\left|{\tan{\left(x + \frac{\pi}{4} \right)}}\right| \right)}}{2} + 4 {\color{red}{\frac{x^{1 + 3}}{1 + 3}}}=- \frac{\ln{\left(\left|{\tan{\left(x + \frac{\pi}{4} \right)}}\right| \right)}}{2} + 4 {\color{red}{\left(\frac{x^{4}}{4}\right)}}$$

Alltså,

$$\int{\left(4 x^{3} - \frac{1}{\cos{\left(2 x \right)}}\right)d x} = x^{4} - \frac{\ln{\left(\left|{\tan{\left(x + \frac{\pi}{4} \right)}}\right| \right)}}{2}$$

Lägg till integrationskonstanten:

$$\int{\left(4 x^{3} - \frac{1}{\cos{\left(2 x \right)}}\right)d x} = x^{4} - \frac{\ln{\left(\left|{\tan{\left(x + \frac{\pi}{4} \right)}}\right| \right)}}{2}+C$$

$$$\int \left(4 x^{3} - \frac{1}{\cos{\left(2 x \right)}}\right)\, dx = \left(x^{4} - \frac{\ln\left(\left|{\tan{\left(x + \frac{\pi}{4} \right)}}\right|\right)}{2}\right) + C$$$A