Integral of $$$\sqrt{16 - x^{2}}$$$

Find $$$\int \sqrt{16 - x^{2}}\, dx$$$.

Let $$$x=4 \sin{\left(u \right)}$$$.

Then $$$dx=\left(4 \sin{\left(u \right)}\right)^{\prime }du = 4 \cos{\left(u \right)} du$$$ (steps can be seen »).

Also, it follows that $$$u=\operatorname{asin}{\left(\frac{x}{4} \right)}$$$.

Integrand becomes

$$$\sqrt{16 - x^{2}} = \sqrt{16 - 16 \sin^{2}{\left( u \right)}}$$$

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

$$$\sqrt{16 - 16 \sin^{2}{\left( u \right)}}=4 \sqrt{1 - \sin^{2}{\left( u \right)}}=4 \sqrt{\cos^{2}{\left( u \right)}}$$$

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

$$$4 \sqrt{\cos^{2}{\left( u \right)}} = 4 \cos{\left( u \right)}$$$

So,

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

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=16$$$ and $$$f{\left(u \right)} = \cos^{2}{\left(u \right)}$$$:

$${\color{red}{\int{16 \cos^{2}{\left(u \right)} d u}}} = {\color{red}{\left(16 \int{\cos^{2}{\left(u \right)} d u}\right)}}$$

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

$$16 {\color{red}{\int{\cos^{2}{\left(u \right)} d u}}} = 16 {\color{red}{\int{\left(\frac{\cos{\left(2 u \right)}}{2} + \frac{1}{2}\right)d u}}}$$

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

$$16 {\color{red}{\int{\left(\frac{\cos{\left(2 u \right)}}{2} + \frac{1}{2}\right)d u}}} = 16 {\color{red}{\left(\frac{\int{\left(\cos{\left(2 u \right)} + 1\right)d u}}{2}\right)}}$$

Integrate term by term:

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

Apply the constant rule $$$\int c\, du = c u$$$ with $$$c=1$$$:

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

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

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

The integral can be rewritten as

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

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

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

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

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

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

$$8 u + 4 \sin{\left({\color{red}{v}} \right)} = 8 u + 4 \sin{\left({\color{red}{\left(2 u\right)}} \right)}$$

Recall that $$$u=\operatorname{asin}{\left(\frac{x}{4} \right)}$$$:

$$4 \sin{\left(2 {\color{red}{u}} \right)} + 8 {\color{red}{u}} = 4 \sin{\left(2 {\color{red}{\operatorname{asin}{\left(\frac{x}{4} \right)}}} \right)} + 8 {\color{red}{\operatorname{asin}{\left(\frac{x}{4} \right)}}}$$

Therefore,

$$\int{\sqrt{16 - x^{2}} d x} = 4 \sin{\left(2 \operatorname{asin}{\left(\frac{x}{4} \right)} \right)} + 8 \operatorname{asin}{\left(\frac{x}{4} \right)}$$

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:

$$\int{\sqrt{16 - x^{2}} d x} = 2 x \sqrt{1 - \frac{x^{2}}{16}} + 8 \operatorname{asin}{\left(\frac{x}{4} \right)}$$

Simplify further:

$$\int{\sqrt{16 - x^{2}} d x} = \frac{x \sqrt{16 - x^{2}}}{2} + 8 \operatorname{asin}{\left(\frac{x}{4} \right)}$$

Add the constant of integration:

$$\int{\sqrt{16 - x^{2}} d x} = \frac{x \sqrt{16 - x^{2}}}{2} + 8 \operatorname{asin}{\left(\frac{x}{4} \right)}+C$$

$$$\int \sqrt{16 - x^{2}}\, dx = \left(\frac{x \sqrt{16 - x^{2}}}{2} + 8 \operatorname{asin}{\left(\frac{x}{4} \right)}\right) + C$$$A