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

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

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 »).

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

Integrand becomes

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

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:

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

Thus,

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

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