Integral of $$$\left(4 - x^{2}\right)^{\frac{3}{2}}$$$

Find $$$\int \left(4 - x^{2}\right)^{\frac{3}{2}}\, dx$$$.

Let $$$x=2 \sin{\left(u \right)}$$$.

Then $$$dx=\left(2 \sin{\left(u \right)}\right)^{\prime }du = 2 \cos{\left(u \right)} du$$$ (steps can be seen »).

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

So,

$$$\left(4 - x^{2}\right)^{\frac{3}{2}} = \left(4 - 4 \sin^{2}{\left( u \right)}\right)^{\frac{3}{2}}$$$

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

$$$\left(4 - 4 \sin^{2}{\left( u \right)}\right)^{\frac{3}{2}}=8 \left(1 - \sin^{2}{\left( u \right)}\right)^{\frac{3}{2}}=8 \left(\cos^{2}{\left( u \right)}\right)^{\frac{3}{2}}$$$

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

$$$8 \left(\cos^{2}{\left( u \right)}\right)^{\frac{3}{2}} = 8 \cos^{3}{\left( u \right)}$$$

So,

$${\color{red}{\int{\left(4 - x^{2}\right)^{\frac{3}{2}} d x}}} = {\color{red}{\int{16 \cos^{4}{\left(u \right)} d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=16$$$ and $$$f{\left(u \right)} = \cos^{4}{\left(u \right)}$$$:

$${\color{red}{\int{16 \cos^{4}{\left(u \right)} d u}}} = {\color{red}{\left(16 \int{\cos^{4}{\left(u \right)} d u}\right)}}$$

Apply the power reducing formula $$$\cos^{4}{\left(\alpha \right)} = \frac{\cos{\left(2 \alpha \right)}}{2} + \frac{\cos{\left(4 \alpha \right)}}{8} + \frac{3}{8}$$$ with $$$\alpha= u $$$:

$$16 {\color{red}{\int{\cos^{4}{\left(u \right)} d u}}} = 16 {\color{red}{\int{\left(\frac{\cos{\left(2 u \right)}}{2} + \frac{\cos{\left(4 u \right)}}{8} + \frac{3}{8}\right)d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{8}$$$ and $$$f{\left(u \right)} = 4 \cos{\left(2 u \right)} + \cos{\left(4 u \right)} + 3$$$:

$$16 {\color{red}{\int{\left(\frac{\cos{\left(2 u \right)}}{2} + \frac{\cos{\left(4 u \right)}}{8} + \frac{3}{8}\right)d u}}} = 16 {\color{red}{\left(\frac{\int{\left(4 \cos{\left(2 u \right)} + \cos{\left(4 u \right)} + 3\right)d u}}{8}\right)}}$$

Integrate term by term:

$$2 {\color{red}{\int{\left(4 \cos{\left(2 u \right)} + \cos{\left(4 u \right)} + 3\right)d u}}} = 2 {\color{red}{\left(\int{3 d u} + \int{4 \cos{\left(2 u \right)} d u} + \int{\cos{\left(4 u \right)} d u}\right)}}$$

Apply the constant rule $$$\int c\, du = c u$$$ with $$$c=3$$$:

$$2 \int{4 \cos{\left(2 u \right)} d u} + 2 \int{\cos{\left(4 u \right)} d u} + 2 {\color{red}{\int{3 d u}}} = 2 \int{4 \cos{\left(2 u \right)} d u} + 2 \int{\cos{\left(4 u \right)} d u} + 2 {\color{red}{\left(3 u\right)}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=4$$$ and $$$f{\left(u \right)} = \cos{\left(2 u \right)}$$$:

$$6 u + 2 \int{\cos{\left(4 u \right)} d u} + 2 {\color{red}{\int{4 \cos{\left(2 u \right)} d u}}} = 6 u + 2 \int{\cos{\left(4 u \right)} d u} + 2 {\color{red}{\left(4 \int{\cos{\left(2 u \right)} d u}\right)}}$$

Let $$$v=2 u$$$.

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

So,

$$6 u + 2 \int{\cos{\left(4 u \right)} d u} + 8 {\color{red}{\int{\cos{\left(2 u \right)} d u}}} = 6 u + 2 \int{\cos{\left(4 u \right)} d u} + 8 {\color{red}{\int{\frac{\cos{\left(v \right)}}{2} d v}}}$$

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

$$6 u + 2 \int{\cos{\left(4 u \right)} d u} + 8 {\color{red}{\int{\frac{\cos{\left(v \right)}}{2} d v}}} = 6 u + 2 \int{\cos{\left(4 u \right)} d u} + 8 {\color{red}{\left(\frac{\int{\cos{\left(v \right)} d v}}{2}\right)}}$$

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

$$6 u + 2 \int{\cos{\left(4 u \right)} d u} + 4 {\color{red}{\int{\cos{\left(v \right)} d v}}} = 6 u + 2 \int{\cos{\left(4 u \right)} d u} + 4 {\color{red}{\sin{\left(v \right)}}}$$

Recall that $$$v=2 u$$$:

$$6 u + 2 \int{\cos{\left(4 u \right)} d u} + 4 \sin{\left({\color{red}{v}} \right)} = 6 u + 2 \int{\cos{\left(4 u \right)} d u} + 4 \sin{\left({\color{red}{\left(2 u\right)}} \right)}$$

Let $$$v=4 u$$$.

Then $$$dv=\left(4 u\right)^{\prime }du = 4 du$$$ (steps can be seen »), and we have that $$$du = \frac{dv}{4}$$$.

So,

$$6 u + 4 \sin{\left(2 u \right)} + 2 {\color{red}{\int{\cos{\left(4 u \right)} d u}}} = 6 u + 4 \sin{\left(2 u \right)} + 2 {\color{red}{\int{\frac{\cos{\left(v \right)}}{4} d v}}}$$

Apply the constant multiple rule $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ with $$$c=\frac{1}{4}$$$ and $$$f{\left(v \right)} = \cos{\left(v \right)}$$$:

$$6 u + 4 \sin{\left(2 u \right)} + 2 {\color{red}{\int{\frac{\cos{\left(v \right)}}{4} d v}}} = 6 u + 4 \sin{\left(2 u \right)} + 2 {\color{red}{\left(\frac{\int{\cos{\left(v \right)} d v}}{4}\right)}}$$

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

$$6 u + 4 \sin{\left(2 u \right)} + \frac{{\color{red}{\int{\cos{\left(v \right)} d v}}}}{2} = 6 u + 4 \sin{\left(2 u \right)} + \frac{{\color{red}{\sin{\left(v \right)}}}}{2}$$

Recall that $$$v=4 u$$$:

$$6 u + 4 \sin{\left(2 u \right)} + \frac{\sin{\left({\color{red}{v}} \right)}}{2} = 6 u + 4 \sin{\left(2 u \right)} + \frac{\sin{\left({\color{red}{\left(4 u\right)}} \right)}}{2}$$

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

$$4 \sin{\left(2 {\color{red}{u}} \right)} + \frac{\sin{\left(4 {\color{red}{u}} \right)}}{2} + 6 {\color{red}{u}} = 4 \sin{\left(2 {\color{red}{\operatorname{asin}{\left(\frac{x}{2} \right)}}} \right)} + \frac{\sin{\left(4 {\color{red}{\operatorname{asin}{\left(\frac{x}{2} \right)}}} \right)}}{2} + 6 {\color{red}{\operatorname{asin}{\left(\frac{x}{2} \right)}}}$$

Therefore,

$$\int{\left(4 - x^{2}\right)^{\frac{3}{2}} d x} = 4 \sin{\left(2 \operatorname{asin}{\left(\frac{x}{2} \right)} \right)} + \frac{\sin{\left(4 \operatorname{asin}{\left(\frac{x}{2} \right)} \right)}}{2} + 6 \operatorname{asin}{\left(\frac{x}{2} \right)}$$

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:

$$\int{\left(4 - x^{2}\right)^{\frac{3}{2}} d x} = 4 x \sqrt{1 - \frac{x^{2}}{4}} + \frac{\sin{\left(4 \operatorname{asin}{\left(\frac{x}{2} \right)} \right)}}{2} + 6 \operatorname{asin}{\left(\frac{x}{2} \right)}$$

Simplify further:

$$\int{\left(4 - x^{2}\right)^{\frac{3}{2}} d x} = - \frac{x^{3} \sqrt{4 - x^{2}}}{4} + \frac{5 x \sqrt{4 - x^{2}}}{2} + 6 \operatorname{asin}{\left(\frac{x}{2} \right)}$$

Add the constant of integration:

$$\int{\left(4 - x^{2}\right)^{\frac{3}{2}} d x} = - \frac{x^{3} \sqrt{4 - x^{2}}}{4} + \frac{5 x \sqrt{4 - x^{2}}}{2} + 6 \operatorname{asin}{\left(\frac{x}{2} \right)}+C$$

$$$\int \left(4 - x^{2}\right)^{\frac{3}{2}}\, dx = \left(- \frac{x^{3} \sqrt{4 - x^{2}}}{4} + \frac{5 x \sqrt{4 - x^{2}}}{2} + 6 \operatorname{asin}{\left(\frac{x}{2} \right)}\right) + C$$$A