Integraal van $$$\frac{e^{2 x}}{\sqrt{16 - e^{4 x}}}$$$

Bepaal $$$\int \frac{e^{2 x}}{\sqrt{16 - e^{4 x}}}\, dx$$$.

Zij $$$u=e^{2 x}$$$.

Dan $$$du=\left(e^{2 x}\right)^{\prime }dx = 2 e^{2 x} dx$$$ (de stappen zijn te zien »), en dan geldt dat $$$e^{2 x} dx = \frac{du}{2}$$$.

Dus,

$${\color{red}{\int{\frac{e^{2 x}}{\sqrt{16 - e^{4 x}}} d x}}} = {\color{red}{\int{\frac{1}{2 \sqrt{16 - u^{2}}} d u}}}$$

Pas de constante-veelvoudregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ toe met $$$c=\frac{1}{2}$$$ en $$$f{\left(u \right)} = \frac{1}{\sqrt{16 - u^{2}}}$$$:

$${\color{red}{\int{\frac{1}{2 \sqrt{16 - u^{2}}} d u}}} = {\color{red}{\left(\frac{\int{\frac{1}{\sqrt{16 - u^{2}}} d u}}{2}\right)}}$$

Zij $$$u=4 \sin{\left(v \right)}$$$.

Dan $$$du=\left(4 \sin{\left(v \right)}\right)^{\prime }dv = 4 \cos{\left(v \right)} dv$$$ (zie » voor de stappen).

Bovendien volgt dat $$$v=\operatorname{asin}{\left(\frac{u}{4} \right)}$$$.

Dus,

$$$\frac{1}{\sqrt{16 - u ^{2}}} = \frac{1}{\sqrt{16 - 16 \sin^{2}{\left( v \right)}}}$$$

Gebruik de identiteit $$$1 - \sin^{2}{\left( v \right)} = \cos^{2}{\left( v \right)}$$$:

$$$\frac{1}{\sqrt{16 - 16 \sin^{2}{\left( v \right)}}}=\frac{1}{4 \sqrt{1 - \sin^{2}{\left( v \right)}}}=\frac{1}{4 \sqrt{\cos^{2}{\left( v \right)}}}$$$

Aangenomen dat $$$\cos{\left( v \right)} \ge 0$$$, verkrijgen we het volgende:

$$$\frac{1}{4 \sqrt{\cos^{2}{\left( v \right)}}} = \frac{1}{4 \cos{\left( v \right)}}$$$

De integraal kan worden herschreven als

$$\frac{{\color{red}{\int{\frac{1}{\sqrt{16 - u^{2}}} d u}}}}{2} = \frac{{\color{red}{\int{1 d v}}}}{2}$$

Pas de constantenregel $$$\int c\, dv = c v$$$ toe met $$$c=1$$$:

$$\frac{{\color{red}{\int{1 d v}}}}{2} = \frac{{\color{red}{v}}}{2}$$

We herinneren eraan dat $$$v=\operatorname{asin}{\left(\frac{u}{4} \right)}$$$:

$$\frac{{\color{red}{v}}}{2} = \frac{{\color{red}{\operatorname{asin}{\left(\frac{u}{4} \right)}}}}{2}$$

We herinneren eraan dat $$$u=e^{2 x}$$$:

$$\frac{\operatorname{asin}{\left(\frac{{\color{red}{u}}}{4} \right)}}{2} = \frac{\operatorname{asin}{\left(\frac{{\color{red}{e^{2 x}}}}{4} \right)}}{2}$$

Dus,

$$\int{\frac{e^{2 x}}{\sqrt{16 - e^{4 x}}} d x} = \frac{\operatorname{asin}{\left(\frac{e^{2 x}}{4} \right)}}{2}$$

Voeg de integratieconstante toe:

$$\int{\frac{e^{2 x}}{\sqrt{16 - e^{4 x}}} d x} = \frac{\operatorname{asin}{\left(\frac{e^{2 x}}{4} \right)}}{2}+C$$

$$$\int \frac{e^{2 x}}{\sqrt{16 - e^{4 x}}}\, dx = \frac{\operatorname{asin}{\left(\frac{e^{2 x}}{4} \right)}}{2} + C$$$A