Integral of $$$\sqrt{- x^{2} + 3 x}$$$

Find $$$\int \sqrt{- x^{2} + 3 x}\, dx$$$.

Complete the square (steps can be seen »): $$$- x^{2} + 3 x = \frac{9}{4} - \left(x - \frac{3}{2}\right)^{2}$$$:

$${\color{red}{\int{\sqrt{- x^{2} + 3 x} d x}}} = {\color{red}{\int{\sqrt{\frac{9}{4} - \left(x - \frac{3}{2}\right)^{2}} d x}}}$$

Let $$$u=x - \frac{3}{2}$$$.

Then $$$du=\left(x - \frac{3}{2}\right)^{\prime }dx = 1 dx$$$ (steps can be seen »), and we have that $$$dx = du$$$.

Therefore,

$${\color{red}{\int{\sqrt{\frac{9}{4} - \left(x - \frac{3}{2}\right)^{2}} d x}}} = {\color{red}{\int{\sqrt{\frac{9}{4} - u^{2}} d u}}}$$

Let $$$u=\frac{3 \sin{\left(v \right)}}{2}$$$.

Then $$$du=\left(\frac{3 \sin{\left(v \right)}}{2}\right)^{\prime }dv = \frac{3 \cos{\left(v \right)}}{2} dv$$$ (steps can be seen »).

Also, it follows that $$$v=\operatorname{asin}{\left(\frac{2 u}{3} \right)}$$$.

Therefore,

$$$\sqrt{\frac{9}{4} - u ^{2}} = \sqrt{\frac{9}{4} - \frac{9 \sin^{2}{\left( v \right)}}{4}}$$$

Use the identity $$$1 - \sin^{2}{\left( v \right)} = \cos^{2}{\left( v \right)}$$$:

$$$\sqrt{\frac{9}{4} - \frac{9 \sin^{2}{\left( v \right)}}{4}}=\frac{3 \sqrt{1 - \sin^{2}{\left( v \right)}}}{2}=\frac{3 \sqrt{\cos^{2}{\left( v \right)}}}{2}$$$

Assuming that $$$\cos{\left( v \right)} \ge 0$$$, we obtain the following:

$$$\frac{3 \sqrt{\cos^{2}{\left( v \right)}}}{2} = \frac{3 \cos{\left( v \right)}}{2}$$$

Thus,

$${\color{red}{\int{\sqrt{\frac{9}{4} - u^{2}} d u}}} = {\color{red}{\int{\frac{9 \cos^{2}{\left(v \right)}}{4} d v}}}$$

Apply the constant multiple rule $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ with $$$c=\frac{9}{4}$$$ and $$$f{\left(v \right)} = \cos^{2}{\left(v \right)}$$$:

$${\color{red}{\int{\frac{9 \cos^{2}{\left(v \right)}}{4} d v}}} = {\color{red}{\left(\frac{9 \int{\cos^{2}{\left(v \right)} d v}}{4}\right)}}$$

Apply the power reducing formula $$$\cos^{2}{\left(\alpha \right)} = \frac{\cos{\left(2 \alpha \right)}}{2} + \frac{1}{2}$$$ with $$$\alpha= v $$$:

$$\frac{9 {\color{red}{\int{\cos^{2}{\left(v \right)} d v}}}}{4} = \frac{9 {\color{red}{\int{\left(\frac{\cos{\left(2 v \right)}}{2} + \frac{1}{2}\right)d v}}}}{4}$$

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 $$$f{\left(v \right)} = \cos{\left(2 v \right)} + 1$$$:

$$\frac{9 {\color{red}{\int{\left(\frac{\cos{\left(2 v \right)}}{2} + \frac{1}{2}\right)d v}}}}{4} = \frac{9 {\color{red}{\left(\frac{\int{\left(\cos{\left(2 v \right)} + 1\right)d v}}{2}\right)}}}{4}$$

Integrate term by term:

$$\frac{9 {\color{red}{\int{\left(\cos{\left(2 v \right)} + 1\right)d v}}}}{8} = \frac{9 {\color{red}{\left(\int{1 d v} + \int{\cos{\left(2 v \right)} d v}\right)}}}{8}$$

Apply the constant rule $$$\int c\, dv = c v$$$ with $$$c=1$$$:

$$\frac{9 \int{\cos{\left(2 v \right)} d v}}{8} + \frac{9 {\color{red}{\int{1 d v}}}}{8} = \frac{9 \int{\cos{\left(2 v \right)} d v}}{8} + \frac{9 {\color{red}{v}}}{8}$$

Let $$$w=2 v$$$.

Then $$$dw=\left(2 v\right)^{\prime }dv = 2 dv$$$ (steps can be seen »), and we have that $$$dv = \frac{dw}{2}$$$.

So,

$$\frac{9 v}{8} + \frac{9 {\color{red}{\int{\cos{\left(2 v \right)} d v}}}}{8} = \frac{9 v}{8} + \frac{9 {\color{red}{\int{\frac{\cos{\left(w \right)}}{2} d w}}}}{8}$$

Apply the constant multiple rule $$$\int c f{\left(w \right)}\, dw = c \int f{\left(w \right)}\, dw$$$ with $$$c=\frac{1}{2}$$$ and $$$f{\left(w \right)} = \cos{\left(w \right)}$$$:

$$\frac{9 v}{8} + \frac{9 {\color{red}{\int{\frac{\cos{\left(w \right)}}{2} d w}}}}{8} = \frac{9 v}{8} + \frac{9 {\color{red}{\left(\frac{\int{\cos{\left(w \right)} d w}}{2}\right)}}}{8}$$

The integral of the cosine is $$$\int{\cos{\left(w \right)} d w} = \sin{\left(w \right)}$$$:

$$\frac{9 v}{8} + \frac{9 {\color{red}{\int{\cos{\left(w \right)} d w}}}}{16} = \frac{9 v}{8} + \frac{9 {\color{red}{\sin{\left(w \right)}}}}{16}$$

Recall that $$$w=2 v$$$:

$$\frac{9 v}{8} + \frac{9 \sin{\left({\color{red}{w}} \right)}}{16} = \frac{9 v}{8} + \frac{9 \sin{\left({\color{red}{\left(2 v\right)}} \right)}}{16}$$

Recall that $$$v=\operatorname{asin}{\left(\frac{2 u}{3} \right)}$$$:

$$\frac{9 \sin{\left(2 {\color{red}{v}} \right)}}{16} + \frac{9 {\color{red}{v}}}{8} = \frac{9 \sin{\left(2 {\color{red}{\operatorname{asin}{\left(\frac{2 u}{3} \right)}}} \right)}}{16} + \frac{9 {\color{red}{\operatorname{asin}{\left(\frac{2 u}{3} \right)}}}}{8}$$

Recall that $$$u=x - \frac{3}{2}$$$:

$$\frac{9 \sin{\left(2 \operatorname{asin}{\left(\frac{2 {\color{red}{u}}}{3} \right)} \right)}}{16} + \frac{9 \operatorname{asin}{\left(\frac{2 {\color{red}{u}}}{3} \right)}}{8} = \frac{9 \sin{\left(2 \operatorname{asin}{\left(\frac{2 {\color{red}{\left(x - \frac{3}{2}\right)}}}{3} \right)} \right)}}{16} + \frac{9 \operatorname{asin}{\left(\frac{2 {\color{red}{\left(x - \frac{3}{2}\right)}}}{3} \right)}}{8}$$

Therefore,

$$\int{\sqrt{- x^{2} + 3 x} d x} = \frac{9 \sin{\left(2 \operatorname{asin}{\left(\frac{2 x}{3} - 1 \right)} \right)}}{16} + \frac{9 \operatorname{asin}{\left(\frac{2 x}{3} - 1 \right)}}{8}$$

Using the formulas $$$\sin{\left(2 \operatorname{asin}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{1 - \alpha^{2}}$$$, $$$\sin{\left(2 \operatorname{acos}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{1 - \alpha^{2}}$$$, $$$\cos{\left(2 \operatorname{asin}{\left(\alpha \right)} \right)} = 1 - 2 \alpha^{2}$$$, $$$\cos{\left(2 \operatorname{acos}{\left(\alpha \right)} \right)} = 2 \alpha^{2} - 1$$$, $$$\sinh{\left(2 \operatorname{asinh}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{\alpha^{2} + 1}$$$, $$$\sinh{\left(2 \operatorname{acosh}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{\alpha - 1} \sqrt{\alpha + 1}$$$, $$$\cosh{\left(2 \operatorname{asinh}{\left(\alpha \right)} \right)} = 2 \alpha^{2} + 1$$$, $$$\cosh{\left(2 \operatorname{acosh}{\left(\alpha \right)} \right)} = 2 \alpha^{2} - 1$$$, simplify the expression:

$$\int{\sqrt{- x^{2} + 3 x} d x} = \frac{9 \sqrt{1 - \left(\frac{2 x}{3} - 1\right)^{2}} \left(\frac{2 x}{3} - 1\right)}{8} + \frac{9 \operatorname{asin}{\left(\frac{2 x}{3} - 1 \right)}}{8}$$

Simplify further:

$$\int{\sqrt{- x^{2} + 3 x} d x} = \frac{\sqrt{9 - \left(2 x - 3\right)^{2}} \left(2 x - 3\right)}{8} + \frac{9 \operatorname{asin}{\left(\frac{2 x}{3} - 1 \right)}}{8}$$

Add the constant of integration:

$$\int{\sqrt{- x^{2} + 3 x} d x} = \frac{\sqrt{9 - \left(2 x - 3\right)^{2}} \left(2 x - 3\right)}{8} + \frac{9 \operatorname{asin}{\left(\frac{2 x}{3} - 1 \right)}}{8}+C$$

$$$\int \sqrt{- x^{2} + 3 x}\, dx = \left(\frac{\sqrt{9 - \left(2 x - 3\right)^{2}} \left(2 x - 3\right)}{8} + \frac{9 \operatorname{asin}{\left(\frac{2 x}{3} - 1 \right)}}{8}\right) + C$$$A