Integral of $$$\sqrt{\frac{1 - \sqrt{x}}{\sqrt{x}}}$$$
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Your Input
Find $$$\int \sqrt{\frac{1 - \sqrt{x}}{\sqrt{x}}}\, dx$$$.
Solution
The input is rewritten: $$$\int{\sqrt{\frac{1 - \sqrt{x}}{\sqrt{x}}} d x}=\int{\frac{\sqrt{1 - \sqrt{x}}}{\sqrt[4]{x}} d x}$$$.
Let $$$u=\sqrt[4]{x}$$$.
Then $$$du=\left(\sqrt[4]{x}\right)^{\prime }dx = \frac{1}{4 x^{\frac{3}{4}}} dx$$$ (steps can be seen »), and we have that $$$\frac{dx}{x^{\frac{3}{4}}} = 4 du$$$.
The integral can be rewritten as
$${\color{red}{\int{\frac{\sqrt{1 - \sqrt{x}}}{\sqrt[4]{x}} d x}}} = {\color{red}{\int{4 u^{2} \sqrt{1 - u^{2}} d u}}}$$
Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=4$$$ and $$$f{\left(u \right)} = u^{2} \sqrt{1 - u^{2}}$$$:
$${\color{red}{\int{4 u^{2} \sqrt{1 - u^{2}} d u}}} = {\color{red}{\left(4 \int{u^{2} \sqrt{1 - u^{2}} d u}\right)}}$$
Let $$$u=\sin{\left(v \right)}$$$.
Then $$$du=\left(\sin{\left(v \right)}\right)^{\prime }dv = \cos{\left(v \right)} dv$$$ (steps can be seen »).
Also, it follows that $$$v=\operatorname{asin}{\left(u \right)}$$$.
Thus,
$$$ u ^{2} \sqrt{1 - u ^{2}} = \sqrt{1 - \sin^{2}{\left( v \right)}} \sin^{2}{\left( v \right)}$$$
Use the identity $$$1 - \sin^{2}{\left( v \right)} = \cos^{2}{\left( v \right)}$$$:
$$$\sqrt{1 - \sin^{2}{\left( v \right)}} \sin^{2}{\left( v \right)}=\sqrt{\cos^{2}{\left( v \right)}} \sin^{2}{\left( v \right)}$$$
Assuming that $$$\cos{\left( v \right)} \ge 0$$$, we obtain the following:
$$$\sqrt{\cos^{2}{\left( v \right)}} \sin^{2}{\left( v \right)} = \sin^{2}{\left( v \right)} \cos{\left( v \right)}$$$
Therefore,
$$4 {\color{red}{\int{u^{2} \sqrt{1 - u^{2}} d u}}} = 4 {\color{red}{\int{\sin^{2}{\left(v \right)} \cos^{2}{\left(v \right)} d v}}}$$
Rewrite the integrand using the double angle formula $$$\sin\left( v \right)\cos\left( v \right)=\frac{1}{2}\sin\left( 2 v \right)$$$:
$$4 {\color{red}{\int{\sin^{2}{\left(v \right)} \cos^{2}{\left(v \right)} d v}}} = 4 {\color{red}{\int{\frac{\sin^{2}{\left(2 v \right)}}{4} d v}}}$$
Apply the constant multiple rule $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ with $$$c=\frac{1}{4}$$$ and $$$f{\left(v \right)} = \sin^{2}{\left(2 v \right)}$$$:
$$4 {\color{red}{\int{\frac{\sin^{2}{\left(2 v \right)}}{4} d v}}} = 4 {\color{red}{\left(\frac{\int{\sin^{2}{\left(2 v \right)} d v}}{4}\right)}}$$
Apply the power reducing formula $$$\sin^{2}{\left(\alpha \right)} = \frac{1}{2} - \frac{\cos{\left(2 \alpha \right)}}{2}$$$ with $$$\alpha=2 v $$$:
$${\color{red}{\int{\sin^{2}{\left(2 v \right)} d v}}} = {\color{red}{\int{\left(\frac{1}{2} - \frac{\cos{\left(4 v \right)}}{2}\right)d v}}}$$
Apply the constant multiple rule $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ with $$$c=\frac{1}{2}$$$ and $$$f{\left(v \right)} = 1 - \cos{\left(4 v \right)}$$$:
$${\color{red}{\int{\left(\frac{1}{2} - \frac{\cos{\left(4 v \right)}}{2}\right)d v}}} = {\color{red}{\left(\frac{\int{\left(1 - \cos{\left(4 v \right)}\right)d v}}{2}\right)}}$$
Integrate term by term:
$$\frac{{\color{red}{\int{\left(1 - \cos{\left(4 v \right)}\right)d v}}}}{2} = \frac{{\color{red}{\left(\int{1 d v} - \int{\cos{\left(4 v \right)} d v}\right)}}}{2}$$
Apply the constant rule $$$\int c\, dv = c v$$$ with $$$c=1$$$:
$$- \frac{\int{\cos{\left(4 v \right)} d v}}{2} + \frac{{\color{red}{\int{1 d v}}}}{2} = - \frac{\int{\cos{\left(4 v \right)} d v}}{2} + \frac{{\color{red}{v}}}{2}$$
Let $$$w=4 v$$$.
Then $$$dw=\left(4 v\right)^{\prime }dv = 4 dv$$$ (steps can be seen »), and we have that $$$dv = \frac{dw}{4}$$$.
The integral becomes
$$\frac{v}{2} - \frac{{\color{red}{\int{\cos{\left(4 v \right)} d v}}}}{2} = \frac{v}{2} - \frac{{\color{red}{\int{\frac{\cos{\left(w \right)}}{4} d w}}}}{2}$$
Apply the constant multiple rule $$$\int c f{\left(w \right)}\, dw = c \int f{\left(w \right)}\, dw$$$ with $$$c=\frac{1}{4}$$$ and $$$f{\left(w \right)} = \cos{\left(w \right)}$$$:
$$\frac{v}{2} - \frac{{\color{red}{\int{\frac{\cos{\left(w \right)}}{4} d w}}}}{2} = \frac{v}{2} - \frac{{\color{red}{\left(\frac{\int{\cos{\left(w \right)} d w}}{4}\right)}}}{2}$$
The integral of the cosine is $$$\int{\cos{\left(w \right)} d w} = \sin{\left(w \right)}$$$:
$$\frac{v}{2} - \frac{{\color{red}{\int{\cos{\left(w \right)} d w}}}}{8} = \frac{v}{2} - \frac{{\color{red}{\sin{\left(w \right)}}}}{8}$$
Recall that $$$w=4 v$$$:
$$\frac{v}{2} - \frac{\sin{\left({\color{red}{w}} \right)}}{8} = \frac{v}{2} - \frac{\sin{\left({\color{red}{\left(4 v\right)}} \right)}}{8}$$
Recall that $$$v=\operatorname{asin}{\left(u \right)}$$$:
$$- \frac{\sin{\left(4 {\color{red}{v}} \right)}}{8} + \frac{{\color{red}{v}}}{2} = - \frac{\sin{\left(4 {\color{red}{\operatorname{asin}{\left(u \right)}}} \right)}}{8} + \frac{{\color{red}{\operatorname{asin}{\left(u \right)}}}}{2}$$
Recall that $$$u=\sqrt[4]{x}$$$:
$$- \frac{\sin{\left(4 \operatorname{asin}{\left({\color{red}{u}} \right)} \right)}}{8} + \frac{\operatorname{asin}{\left({\color{red}{u}} \right)}}{2} = - \frac{\sin{\left(4 \operatorname{asin}{\left({\color{red}{\sqrt[4]{x}}} \right)} \right)}}{8} + \frac{\operatorname{asin}{\left({\color{red}{\sqrt[4]{x}}} \right)}}{2}$$
Therefore,
$$\int{\frac{\sqrt{1 - \sqrt{x}}}{\sqrt[4]{x}} d x} = - \frac{\sin{\left(4 \operatorname{asin}{\left(\sqrt[4]{x} \right)} \right)}}{8} + \frac{\operatorname{asin}{\left(\sqrt[4]{x} \right)}}{2}$$
Simplify:
$$\int{\frac{\sqrt{1 - \sqrt{x}}}{\sqrt[4]{x}} d x} = \sqrt[4]{x} \sqrt{1 - \sqrt{x}} \left(\sqrt{x} - \frac{1}{2}\right) + \frac{\operatorname{asin}{\left(\sqrt[4]{x} \right)}}{2}$$
Add the constant of integration:
$$\int{\frac{\sqrt{1 - \sqrt{x}}}{\sqrt[4]{x}} d x} = \sqrt[4]{x} \sqrt{1 - \sqrt{x}} \left(\sqrt{x} - \frac{1}{2}\right) + \frac{\operatorname{asin}{\left(\sqrt[4]{x} \right)}}{2}+C$$
Answer
$$$\int \sqrt{\frac{1 - \sqrt{x}}{\sqrt{x}}}\, dx = \left(\sqrt[4]{x} \sqrt{1 - \sqrt{x}} \left(\sqrt{x} - \frac{1}{2}\right) + \frac{\operatorname{asin}{\left(\sqrt[4]{x} \right)}}{2}\right) + C$$$A