Integral of $$$- 3 x \sqrt{5 - x^{2}}$$$

Find $$$\int \left(- 3 x \sqrt{5 - x^{2}}\right)\, dx$$$.

Let $$$u=5 - x^{2}$$$.

Then $$$du=\left(5 - x^{2}\right)^{\prime }dx = - 2 x dx$$$ (steps can be seen »), and we have that $$$x dx = - \frac{du}{2}$$$.

So,

$${\color{red}{\int{\left(- 3 x \sqrt{5 - x^{2}}\right)d x}}} = {\color{red}{\int{\frac{3 \sqrt{u}}{2} d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{3}{2}$$$ and $$$f{\left(u \right)} = \sqrt{u}$$$:

$${\color{red}{\int{\frac{3 \sqrt{u}}{2} d u}}} = {\color{red}{\left(\frac{3 \int{\sqrt{u} d u}}{2}\right)}}$$

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=\frac{1}{2}$$$:

$$\frac{3 {\color{red}{\int{\sqrt{u} d u}}}}{2}=\frac{3 {\color{red}{\int{u^{\frac{1}{2}} d u}}}}{2}=\frac{3 {\color{red}{\frac{u^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}}}}{2}=\frac{3 {\color{red}{\left(\frac{2 u^{\frac{3}{2}}}{3}\right)}}}{2}$$

Recall that $$$u=5 - x^{2}$$$:

$${\color{red}{u}}^{\frac{3}{2}} = {\color{red}{\left(5 - x^{2}\right)}}^{\frac{3}{2}}$$

Therefore,

$$\int{\left(- 3 x \sqrt{5 - x^{2}}\right)d x} = \left(5 - x^{2}\right)^{\frac{3}{2}}$$

Add the constant of integration:

$$\int{\left(- 3 x \sqrt{5 - x^{2}}\right)d x} = \left(5 - x^{2}\right)^{\frac{3}{2}}+C$$

$$$\int \left(- 3 x \sqrt{5 - x^{2}}\right)\, dx = \left(5 - x^{2}\right)^{\frac{3}{2}} + C$$$A