Integral of $\operatorname{acos}{\left(x \right)}$

Find $\int \operatorname{acos}{\left(x \right)}\, dx$.

For the integral $\int{\operatorname{acos}{\left(x \right)} d x}$, use integration by parts $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$.

Let $\operatorname{u}=\operatorname{acos}{\left(x \right)}$ and $\operatorname{dv}=dx$.

Then $\operatorname{du}=\left(\operatorname{acos}{\left(x \right)}\right)^{\prime }dx=- \frac{1}{\sqrt{1 - x^{2}}} dx$ (steps can be seen ») and $\operatorname{v}=\int{1 d x}=x$ (steps can be seen »).

So,

$${\color{red}{\int{\operatorname{acos}{\left(x \right)} d x}}}={\color{red}{\left(\operatorname{acos}{\left(x \right)} \cdot x-\int{x \cdot \left(- \frac{1}{\sqrt{1 - x^{2}}}\right) d x}\right)}}={\color{red}{\left(x \operatorname{acos}{\left(x \right)} - \int{\left(- \frac{x}{\sqrt{1 - x^{2}}}\right)d x}\right)}}$$

Let $u=1 - x^{2}$.

Then $du=\left(1 - x^{2}\right)^{\prime }dx = - 2 x dx$ (steps can be seen »), and we have that $x dx = - \frac{du}{2}$.

The integral becomes

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

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 $f{\left(u \right)} = \frac{1}{\sqrt{u}}$:

$$x \operatorname{acos}{\left(x \right)} - {\color{red}{\int{\frac{1}{2 \sqrt{u}} d u}}} = x \operatorname{acos}{\left(x \right)} - {\color{red}{\left(\frac{\int{\frac{1}{\sqrt{u}} d u}}{2}\right)}}$$

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

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

Recall that $u=1 - x^{2}$:

$$x \operatorname{acos}{\left(x \right)} - \sqrt{{\color{red}{u}}} = x \operatorname{acos}{\left(x \right)} - \sqrt{{\color{red}{\left(1 - x^{2}\right)}}}$$

Therefore,

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

Add the constant of integration:

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

$\int \operatorname{acos}{\left(x \right)}\, dx = \left(x \operatorname{acos}{\left(x \right)} - \sqrt{1 - x^{2}}\right) + C$A