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

The calculator will find the integral/antiderivative of $$$\operatorname{acos}{\left(x \right)}$$$, with steps shown.

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

Solution

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 here) and $$$\operatorname{v}=\int{1 d x}=x$$$ (steps can be seen here).

Thus,

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

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

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

Therefore,

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

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:

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

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