Integralen av $$$\frac{\sqrt{\tan{\left(x \right)}}}{\sin{\left(x \right)} \cos{\left(x \right)}}$$$
Relaterad kalkylator: Kalkylator för bestämda och oegentliga integraler
Din inmatning
Bestäm $$$\int \frac{\sqrt{\tan{\left(x \right)}}}{\sin{\left(x \right)} \cos{\left(x \right)}}\, dx$$$.
Lösning
Låt $$$u=\tan{\left(x \right)}$$$ vara.
Då gäller $$$x=\operatorname{atan}{\left(u \right)}$$$ och $$$dx=\left(\operatorname{atan}{\left(u \right)}\right)^{\prime }du = \frac{du}{u^{2} + 1}$$$ (stegen kan ses »).
Alltså,
$${\color{red}{\int{\frac{\sqrt{\tan{\left(x \right)}}}{\sin{\left(x \right)} \cos{\left(x \right)}} d x}}} = {\color{red}{\int{\frac{1}{\sqrt{u}} d u}}}$$
Tillämpa potensregeln $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ med $$$n=- \frac{1}{2}$$$:
$${\color{red}{\int{\frac{1}{\sqrt{u}} d u}}}={\color{red}{\int{u^{- \frac{1}{2}} d u}}}={\color{red}{\frac{u^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1}}}={\color{red}{\left(2 u^{\frac{1}{2}}\right)}}={\color{red}{\left(2 \sqrt{u}\right)}}$$
Kom ihåg att $$$u=\tan{\left(x \right)}$$$:
$$2 \sqrt{{\color{red}{u}}} = 2 \sqrt{{\color{red}{\tan{\left(x \right)}}}}$$
Alltså,
$$\int{\frac{\sqrt{\tan{\left(x \right)}}}{\sin{\left(x \right)} \cos{\left(x \right)}} d x} = 2 \sqrt{\tan{\left(x \right)}}$$
Lägg till integrationskonstanten:
$$\int{\frac{\sqrt{\tan{\left(x \right)}}}{\sin{\left(x \right)} \cos{\left(x \right)}} d x} = 2 \sqrt{\tan{\left(x \right)}}+C$$
Svar
$$$\int \frac{\sqrt{\tan{\left(x \right)}}}{\sin{\left(x \right)} \cos{\left(x \right)}}\, dx = 2 \sqrt{\tan{\left(x \right)}} + C$$$A