Integralen av $$$\frac{1}{1 - \sin{\left(2 x \right)}}$$$

Bestäm $$$\int \frac{1}{1 - \sin{\left(2 x \right)}}\, dx$$$.

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,

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

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

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

Skriv om $$$1$$$ som $$$\sin^2\left(\frac{ u }{2}\right)+\cos^2\left(\frac{ u }{2}\right)$$$ och tillämpa dubbelvinkelformeln för sinus $$$\sin\left( u \right)=2\sin\left(\frac{ u }{2}\right)\cos\left(\frac{ u }{2}\right)$$$:

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

Kvadratkomplettera (stegen kan ses »):

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

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

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

Låt $$$v=\tan{\left(\frac{u}{2} \right)} - 1$$$ vara.

Då $$$dv=\left(\tan{\left(\frac{u}{2} \right)} - 1\right)^{\prime }du = \frac{\sec^{2}{\left(\frac{u}{2} \right)}}{2} du$$$ (stegen kan ses »), och vi har att $$$\sec^{2}{\left(\frac{u}{2} \right)} du = 2 dv$$$.

Integralen kan omskrivas som

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

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

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

Tillämpa potensregeln $$$\int v^{n}\, dv = \frac{v^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ med $$$n=-2$$$:

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

Kom ihåg att $$$v=\tan{\left(\frac{u}{2} \right)} - 1$$$:

$$- {\color{red}{v}}^{-1} = - {\color{red}{\left(\tan{\left(\frac{u}{2} \right)} - 1\right)}}^{-1}$$

Kom ihåg att $$$u=2 x$$$:

$$- \left(-1 + \tan{\left(\frac{{\color{red}{u}}}{2} \right)}\right)^{-1} = - \left(-1 + \tan{\left(\frac{{\color{red}{\left(2 x\right)}}}{2} \right)}\right)^{-1}$$

Alltså,

$$\int{\frac{1}{1 - \sin{\left(2 x \right)}} d x} = - \frac{1}{\tan{\left(x \right)} - 1}$$

Lägg till integrationskonstanten:

$$\int{\frac{1}{1 - \sin{\left(2 x \right)}} d x} = - \frac{1}{\tan{\left(x \right)} - 1}+C$$

$$$\int \frac{1}{1 - \sin{\left(2 x \right)}}\, dx = - \frac{1}{\tan{\left(x \right)} - 1} + C$$$A