Integraal van $$$\frac{\pi \cos{\left(x \right)}}{2 \sqrt{\sin{\left(x \right)}}}$$$

Bepaal $$$\int \frac{\pi \cos{\left(x \right)}}{2 \sqrt{\sin{\left(x \right)}}}\, dx$$$.

Pas de constante-veelvoudregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ toe met $$$c=\frac{\pi}{2}$$$ en $$$f{\left(x \right)} = \frac{\cos{\left(x \right)}}{\sqrt{\sin{\left(x \right)}}}$$$:

$${\color{red}{\int{\frac{\pi \cos{\left(x \right)}}{2 \sqrt{\sin{\left(x \right)}}} d x}}} = {\color{red}{\left(\frac{\pi \int{\frac{\cos{\left(x \right)}}{\sqrt{\sin{\left(x \right)}}} d x}}{2}\right)}}$$

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

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

Dus,

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

Pas de machtsregel $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ toe met $$$n=- \frac{1}{2}$$$:

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

We herinneren eraan dat $$$u=\sin{\left(x \right)}$$$:

$$\pi \sqrt{{\color{red}{u}}} = \pi \sqrt{{\color{red}{\sin{\left(x \right)}}}}$$

Dus,

$$\int{\frac{\pi \cos{\left(x \right)}}{2 \sqrt{\sin{\left(x \right)}}} d x} = \pi \sqrt{\sin{\left(x \right)}}$$

Voeg de integratieconstante toe:

$$\int{\frac{\pi \cos{\left(x \right)}}{2 \sqrt{\sin{\left(x \right)}}} d x} = \pi \sqrt{\sin{\left(x \right)}}+C$$

$$$\int \frac{\pi \cos{\left(x \right)}}{2 \sqrt{\sin{\left(x \right)}}}\, dx = \pi \sqrt{\sin{\left(x \right)}} + C$$$A