Integral de $$$\sqrt{1 - x^{2}}$$$

Encontre $$$\int \sqrt{1 - x^{2}}\, dx$$$.

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

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

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

So,

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

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

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

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

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

So,

$${\color{red}{\int{\sqrt{1 - x^{2}} d x}}} = {\color{red}{\int{\cos^{2}{\left(u \right)} d u}}}$$

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

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

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

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

Integrate term by term:

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

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

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

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

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

The integral can be rewritten as

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

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

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

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

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

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

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

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

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

Therefore,

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

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:

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

Add the constant of integration:

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

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