Question Statement

Evaluate the integral:

$\int {\cos }^{-1}xdx$

---

Background and Explanation

This problem requires integration by parts, a technique used when integrating products of functions. You'll also need the derivative of the inverse cosine function and recognition of a substitution pattern for the remaining integral.

---

Solution

We begin by setting up the integral:

$I=\int {\cos }^{-1}xdx$

Applying Integration by Parts

We use the integration by parts formula: $\int udv=uv-\int vdu$

Let us choose:

• $u={\cos }^{-1}x$ (so that $du=\frac{-1}{\sqrt{1-x^2}}dx$)
• $dv=dx$ (so that $v=x$)

Applying the formula:

$\begin{array}{rl}I& ={\cos }^{-1}x\cdot \int 1dx-\int \left(\int 1dx\right)\cdot \frac{d}{dx}\left({\cos }^{-1}x\right)dx\\ & ={\cos }^{-1}x\cdot x-\int x\cdot \frac{-1}{\sqrt{1-x^2}}dx\\ & =x{\cos }^{-1}x+\int \frac{x}{\sqrt{1-x^2}}dx\end{array}$

Evaluating the Remaining Integral

Now we focus on $\int \frac{x}{\sqrt{1-x^2}}dx$. We can rewrite this as:

$=x{\cos }^{-1}x+\int {\left(1-x^2\right)}^{-\frac{1}{2}}\cdot xdx$

To integrate $\int (1-x^2{)}^{-1/2}\cdot xdx$, we use the substitution method. Notice that the derivative of $(1-x^2)$ is $-2x$, so we manipulate the expression:

$\begin{array}{rl}& =x{\cos }^{-1}x+\left(\frac{-1}{2}\right)\int {\left(1-x^2\right)}^{-\frac{1}{2}}\cdot (-2x)dx\\ & =x{\cos }^{-1}x-\frac{1}{2}\cdot \frac{{\left(1-x^2\right)}^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+c\\ & =x{\cos }^{-1}x-\frac{1}{2}\cdot \frac{{\left(1-x^2\right)}^{\frac{1}{2}}}{\frac{1}{2}}+c\end{array}$

Final Simplification

Simplifying the fraction $\frac{1/2}{1/2}=1$:

$\begin{array}{rl}& =x{\cos }^{-1}x-\frac{1}{2}\cdot 2\sqrt{1-x^2}+c\\ & =x{\cos }^{-1}x-\sqrt{1-x^2}+c\end{array}$

Therefore, the final answer is:

$\int {\cos }^{-1}xdx=x{\cos }^{-1}x-\sqrt{1-x^2}+c$

---

Key Formulas or Methods Used

• Integration by Parts: $\int udv=uv-\int vdu$
• Derivative of Inverse Cosine: $\frac{d}{dx}({\cos }^{-1}x)=\frac{-1}{\sqrt{1-x^2}}$
• Power Rule for Integration: $\int x^{n}dx=\frac{x^{n+1}}{n+1}+c$ (for $n\ne -1$)
• Substitution Method: Recognizing that $xdx$ is proportional to the derivative of the inner function $(1-x^2)$

---

Summary of Steps

1. Set up integration by parts with $u={\cos }^{-1}x$ and $dv=dx$
2. Apply the formula to get $x{\cos }^{-1}x-\int x\cdot \frac{-1}{\sqrt{1-x^2}}dx$
3. Simplify signs to obtain $x{\cos }^{-1}x+\int \frac{x}{\sqrt{1-x^2}}dx$
4. Rewrite the integrand as $(1-x^2{)}^{-1/2}\cdot x$ to prepare for substitution
5. Adjust constants by factoring out $-\frac{1}{2}$ to match the derivative of the inner function $(1-x^2)$
6. Integrate using power rule to get $-\frac{1}{2}\cdot \frac{(1-x^2{)}^{1/2}}{1/2}$
7. Simplify the final expression to arrive at $x{\cos }^{-1}x-\sqrt{1-x^2}+c$