How do you integrate #int arccosx# by integration by parts method?

Question

Christopher Agresta · Accepted Answer

see here

https://tutor.hix.ai

Luke Alvarez · Answer

undefined To integrate $\int \arccos(x) \,dx$ using the integration by parts formula, we use the formula $\int u \,dv = uv - \int v \,du$, where one part is differentiated ($du$) and the other is integrated ($dv$).

Let:
- $u = \arccos(x)$ → $du = -\frac{1}{\sqrt{1-x^2}} dx$
- $dv = dx$ → $v = x$

Now, substitute $u$, $du$, $dv$, and $v$ into the integration by parts formula:

$$
\int \arccos(x) dx = x \arccos(x) - \int x \left(-\frac{1}{\sqrt{1-x^2}}\right) dx
$$

This simplifies to:
$$
x \arccos(x) + \int \frac{x}{\sqrt{1-x^2}} dx
$$

Next, solve the integral $\int \frac{x}{\sqrt{1-x^2}} dx$. Let's use a substitution method for this part:

Let $s = 1-x^2$, thus $ds = -2x dx$. We adjust the integral accordingly:

$$
\frac{1}{2} \int \frac{-ds}{\sqrt{s}} = \frac{1}{2} \int s^{-\frac{1}{2}} (-ds)
$$

Solving this integral gives:

$$
\frac{1}{2} \times (-2) s^{\frac{1}{2}} = -\sqrt{s} = -\sqrt{1-x^2}
$$

Bringing it all together:

$$
\int \arccos(x) dx = x \arccos(x) - \sqrt{1-x^2} + C
$$

where $C$ is the constant of integration.