Integral de $$$\sqrt{\frac{1 - \sqrt{x}}{\sqrt{x}}}$$$

Halla $$$\int \sqrt{\frac{1 - \sqrt{x}}{\sqrt{x}}}\, dx$$$.

La entrada se reescribe: $$$\int{\sqrt{\frac{1 - \sqrt{x}}{\sqrt{x}}} d x}=\int{\frac{\sqrt{1 - \sqrt{x}}}{\sqrt[4]{x}} d x}$$$.

Sea $$$u=\sqrt[4]{x}$$$.

Entonces $$$du=\left(\sqrt[4]{x}\right)^{\prime }dx = \frac{1}{4 x^{\frac{3}{4}}} dx$$$ (los pasos pueden verse »), y obtenemos que $$$\frac{dx}{x^{\frac{3}{4}}} = 4 du$$$.

Por lo tanto,

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

Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=4$$$ y $$$f{\left(u \right)} = u^{2} \sqrt{1 - u^{2}}$$$:

$${\color{red}{\int{4 u^{2} \sqrt{1 - u^{2}} d u}}} = {\color{red}{\left(4 \int{u^{2} \sqrt{1 - u^{2}} d u}\right)}}$$

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

Entonces $$$du=\left(\sin{\left(v \right)}\right)^{\prime }dv = \cos{\left(v \right)} dv$$$ (los pasos pueden verse »).

Además, se sigue que $$$v=\operatorname{asin}{\left(u \right)}$$$.

Por lo tanto,

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

Utiliza la identidad $$$1 - \sin^{2}{\left( v \right)} = \cos^{2}{\left( v \right)}$$$:

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

Suponiendo que $$$\cos{\left( v \right)} \ge 0$$$, obtenemos lo siguiente:

$$$\sqrt{\cos^{2}{\left( v \right)}} \sin^{2}{\left( v \right)} = \sin^{2}{\left( v \right)} \cos{\left( v \right)}$$$

La integral puede reescribirse como

$$4 {\color{red}{\int{u^{2} \sqrt{1 - u^{2}} d u}}} = 4 {\color{red}{\int{\sin^{2}{\left(v \right)} \cos^{2}{\left(v \right)} d v}}}$$

Reescribe el integrando utilizando la fórmula del ángulo doble $$$\sin\left( v \right)\cos\left( v \right)=\frac{1}{2}\sin\left( 2 v \right)$$$:

$$4 {\color{red}{\int{\sin^{2}{\left(v \right)} \cos^{2}{\left(v \right)} d v}}} = 4 {\color{red}{\int{\frac{\sin^{2}{\left(2 v \right)}}{4} d v}}}$$

Aplica la regla del factor constante $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ con $$$c=\frac{1}{4}$$$ y $$$f{\left(v \right)} = \sin^{2}{\left(2 v \right)}$$$:

$$4 {\color{red}{\int{\frac{\sin^{2}{\left(2 v \right)}}{4} d v}}} = 4 {\color{red}{\left(\frac{\int{\sin^{2}{\left(2 v \right)} d v}}{4}\right)}}$$

Aplica la fórmula de reducción de potencia $$$\sin^{2}{\left(\alpha \right)} = \frac{1}{2} - \frac{\cos{\left(2 \alpha \right)}}{2}$$$ con $$$\alpha=2 v $$$:

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

Aplica la regla del factor constante $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ con $$$c=\frac{1}{2}$$$ y $$$f{\left(v \right)} = 1 - \cos{\left(4 v \right)}$$$:

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

Integra término a término:

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

Aplica la regla de la constante $$$\int c\, dv = c v$$$ con $$$c=1$$$:

$$- \frac{\int{\cos{\left(4 v \right)} d v}}{2} + \frac{{\color{red}{\int{1 d v}}}}{2} = - \frac{\int{\cos{\left(4 v \right)} d v}}{2} + \frac{{\color{red}{v}}}{2}$$

Sea $$$w=4 v$$$.

Entonces $$$dw=\left(4 v\right)^{\prime }dv = 4 dv$$$ (los pasos pueden verse »), y obtenemos que $$$dv = \frac{dw}{4}$$$.

La integral puede reescribirse como

$$\frac{v}{2} - \frac{{\color{red}{\int{\cos{\left(4 v \right)} d v}}}}{2} = \frac{v}{2} - \frac{{\color{red}{\int{\frac{\cos{\left(w \right)}}{4} d w}}}}{2}$$

Aplica la regla del factor constante $$$\int c f{\left(w \right)}\, dw = c \int f{\left(w \right)}\, dw$$$ con $$$c=\frac{1}{4}$$$ y $$$f{\left(w \right)} = \cos{\left(w \right)}$$$:

$$\frac{v}{2} - \frac{{\color{red}{\int{\frac{\cos{\left(w \right)}}{4} d w}}}}{2} = \frac{v}{2} - \frac{{\color{red}{\left(\frac{\int{\cos{\left(w \right)} d w}}{4}\right)}}}{2}$$

La integral del coseno es $$$\int{\cos{\left(w \right)} d w} = \sin{\left(w \right)}$$$:

$$\frac{v}{2} - \frac{{\color{red}{\int{\cos{\left(w \right)} d w}}}}{8} = \frac{v}{2} - \frac{{\color{red}{\sin{\left(w \right)}}}}{8}$$

Recordemos que $$$w=4 v$$$:

$$\frac{v}{2} - \frac{\sin{\left({\color{red}{w}} \right)}}{8} = \frac{v}{2} - \frac{\sin{\left({\color{red}{\left(4 v\right)}} \right)}}{8}$$

Recordemos que $$$v=\operatorname{asin}{\left(u \right)}$$$:

$$- \frac{\sin{\left(4 {\color{red}{v}} \right)}}{8} + \frac{{\color{red}{v}}}{2} = - \frac{\sin{\left(4 {\color{red}{\operatorname{asin}{\left(u \right)}}} \right)}}{8} + \frac{{\color{red}{\operatorname{asin}{\left(u \right)}}}}{2}$$

Recordemos que $$$u=\sqrt[4]{x}$$$:

$$- \frac{\sin{\left(4 \operatorname{asin}{\left({\color{red}{u}} \right)} \right)}}{8} + \frac{\operatorname{asin}{\left({\color{red}{u}} \right)}}{2} = - \frac{\sin{\left(4 \operatorname{asin}{\left({\color{red}{\sqrt[4]{x}}} \right)} \right)}}{8} + \frac{\operatorname{asin}{\left({\color{red}{\sqrt[4]{x}}} \right)}}{2}$$

Por lo tanto,

$$\int{\frac{\sqrt{1 - \sqrt{x}}}{\sqrt[4]{x}} d x} = - \frac{\sin{\left(4 \operatorname{asin}{\left(\sqrt[4]{x} \right)} \right)}}{8} + \frac{\operatorname{asin}{\left(\sqrt[4]{x} \right)}}{2}$$

Simplificar:

$$\int{\frac{\sqrt{1 - \sqrt{x}}}{\sqrt[4]{x}} d x} = \sqrt[4]{x} \sqrt{1 - \sqrt{x}} \left(\sqrt{x} - \frac{1}{2}\right) + \frac{\operatorname{asin}{\left(\sqrt[4]{x} \right)}}{2}$$

Añade la constante de integración:

$$\int{\frac{\sqrt{1 - \sqrt{x}}}{\sqrt[4]{x}} d x} = \sqrt[4]{x} \sqrt{1 - \sqrt{x}} \left(\sqrt{x} - \frac{1}{2}\right) + \frac{\operatorname{asin}{\left(\sqrt[4]{x} \right)}}{2}+C$$

$$$\int \sqrt{\frac{1 - \sqrt{x}}{\sqrt{x}}}\, dx = \left(\sqrt[4]{x} \sqrt{1 - \sqrt{x}} \left(\sqrt{x} - \frac{1}{2}\right) + \frac{\operatorname{asin}{\left(\sqrt[4]{x} \right)}}{2}\right) + C$$$A