Integral of $$$\sqrt{\frac{x}{4 - x^{3}}}$$$
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Your Input
Find $$$\int \sqrt{\frac{x}{4 - x^{3}}}\, dx$$$.
Solution
The input is rewritten: $$$\int{\sqrt{\frac{x}{4 - x^{3}}} d x}=\int{\frac{\sqrt{x}}{\sqrt{4 - x^{3}}} d x}$$$.
Let $$$u=x^{\frac{3}{2}}$$$.
Then $$$du=\left(x^{\frac{3}{2}}\right)^{\prime }dx = \frac{3 \sqrt{x}}{2} dx$$$ (steps can be seen »), and we have that $$$\sqrt{x} dx = \frac{2 du}{3}$$$.
Thus,
$${\color{red}{\int{\frac{\sqrt{x}}{\sqrt{4 - x^{3}}} d x}}} = {\color{red}{\int{\frac{2}{3 \sqrt{4 - u^{2}}} d u}}}$$
Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{2}{3}$$$ and $$$f{\left(u \right)} = \frac{1}{\sqrt{4 - u^{2}}}$$$:
$${\color{red}{\int{\frac{2}{3 \sqrt{4 - u^{2}}} d u}}} = {\color{red}{\left(\frac{2 \int{\frac{1}{\sqrt{4 - u^{2}}} d u}}{3}\right)}}$$
Let $$$u=2 \sin{\left(v \right)}$$$.
Then $$$du=\left(2 \sin{\left(v \right)}\right)^{\prime }dv = 2 \cos{\left(v \right)} dv$$$ (steps can be seen »).
Also, it follows that $$$v=\operatorname{asin}{\left(\frac{u}{2} \right)}$$$.
So,
$$$\frac{1}{\sqrt{4 - u ^{2}}} = \frac{1}{\sqrt{4 - 4 \sin^{2}{\left( v \right)}}}$$$
Use the identity $$$1 - \sin^{2}{\left( v \right)} = \cos^{2}{\left( v \right)}$$$:
$$$\frac{1}{\sqrt{4 - 4 \sin^{2}{\left( v \right)}}}=\frac{1}{2 \sqrt{1 - \sin^{2}{\left( v \right)}}}=\frac{1}{2 \sqrt{\cos^{2}{\left( v \right)}}}$$$
Assuming that $$$\cos{\left( v \right)} \ge 0$$$, we obtain the following:
$$$\frac{1}{2 \sqrt{\cos^{2}{\left( v \right)}}} = \frac{1}{2 \cos{\left( v \right)}}$$$
Integral can be rewritten as
$$\frac{2 {\color{red}{\int{\frac{1}{\sqrt{4 - u^{2}}} d u}}}}{3} = \frac{2 {\color{red}{\int{1 d v}}}}{3}$$
Apply the constant rule $$$\int c\, dv = c v$$$ with $$$c=1$$$:
$$\frac{2 {\color{red}{\int{1 d v}}}}{3} = \frac{2 {\color{red}{v}}}{3}$$
Recall that $$$v=\operatorname{asin}{\left(\frac{u}{2} \right)}$$$:
$$\frac{2 {\color{red}{v}}}{3} = \frac{2 {\color{red}{\operatorname{asin}{\left(\frac{u}{2} \right)}}}}{3}$$
Recall that $$$u=x^{\frac{3}{2}}$$$:
$$\frac{2 \operatorname{asin}{\left(\frac{{\color{red}{u}}}{2} \right)}}{3} = \frac{2 \operatorname{asin}{\left(\frac{{\color{red}{x^{\frac{3}{2}}}}}{2} \right)}}{3}$$
Therefore,
$$\int{\frac{\sqrt{x}}{\sqrt{4 - x^{3}}} d x} = \frac{2 \operatorname{asin}{\left(\frac{x^{\frac{3}{2}}}{2} \right)}}{3}$$
Add the constant of integration:
$$\int{\frac{\sqrt{x}}{\sqrt{4 - x^{3}}} d x} = \frac{2 \operatorname{asin}{\left(\frac{x^{\frac{3}{2}}}{2} \right)}}{3}+C$$
Answer
$$$\int \sqrt{\frac{x}{4 - x^{3}}}\, dx = \frac{2 \operatorname{asin}{\left(\frac{x^{\frac{3}{2}}}{2} \right)}}{3} + C$$$A