Question Statement

Evaluate the integral:

$\int x{\sec }^{-1}xdx$

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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 secant function and recognition of a substitution pattern for the remaining integral.

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Solution

We begin by denoting the integral as $I$ and applying integration by parts, which follows the formula $\int udv=uv-\int vdu$.

Let:

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

Applying integration by parts:

$I=\int x{\sec }^{-1}xdx={\sec }^{-1}x\cdot \frac{x^2}{2}-\int \frac{x^2}{2}\cdot \frac{1}{x\sqrt{x^2-1}}dx$

Simplifying the integrand in the second term:

$=\frac{x^2}{2}{\sec }^{-1}x-\frac{1}{2}\int \frac{x}{\sqrt{x^2-1}}dx$

Rewrite the remaining integral to prepare for substitution:

$=\frac{x^2}{2}{\sec }^{-1}x-\frac{1}{2}\int (x^2-1{)}^{-\frac{1}{2}}\cdot xdx$

Notice that the derivative of $(x^2-1)$ is $2x$, so we adjust the constant to match this pattern:

$=\frac{x^2}{2}{\sec }^{-1}x-\frac{1}{2}\cdot \frac{1}{2}\int (x^2-1{)}^{-\frac{1}{2}}\cdot 2xdx$

$=\frac{x^2}{2}{\sec }^{-1}x-\frac{1}{4}\int (x^2-1{)}^{-\frac{1}{2}}\cdot 2xdx$

Now apply the power rule for integration $\int [f(x){]}^n\cdot f^{\prime }(x)dx=\frac{[f(x){]}^{n+1}}{n+1}$ where $f(x)=x^2-1$ and $n=-\frac{1}{2}$:

$=\frac{x^2}{2}{\sec }^{-1}x-\frac{1}{4}\cdot \frac{(x^2-1{)}^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+c$

$=\frac{x^2}{2}{\sec }^{-1}x-\frac{1}{4}\cdot \frac{(x^2-1{)}^{\frac{1}{2}}}{\frac{1}{2}}+c$

Simplifying the fraction $\frac{1}{4}÷\frac{1}{2}=\frac{1}{4}\times 2=\frac{1}{2}$:

$=\frac{x^2}{2}{\sec }^{-1}x-\frac{1}{2}(x^2-1{)}^{\frac{1}{2}}+c$

Therefore, the final answer is:

$\boxed{\frac{x^2}{2}{\sec }^{-1}x-\frac{\sqrt{x^2-1}}{2}+c}$

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Key Formulas or Methods Used

• Integration by Parts: $\int udv=uv-\int vdu$
• Derivative of Inverse Secant: $\frac{d}{dx}({\sec }^{-1}x)=\frac{1}{x\sqrt{x^2-1}}$
• Power Rule for Integration: $\int [f(x){]}^n\cdot f^{\prime }(x)dx=\frac{[f(x){]}^{n+1}}{n+1}+c$ (where $n\ne -1$)
• Substitution Pattern: Recognizing that $xdx$ is proportional to $d(x^2-1)$

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Summary of Steps

1. Set up integration by parts with $u={\sec }^{-1}x$ and $dv=xdx$
2. Compute $v=\frac{x^2}{2}$ and $du=\frac{1}{x\sqrt{x^2-1}}dx$
3. Apply the formula to get $\frac{x^2}{2}{\sec }^{-1}x-\frac{1}{2}\int \frac{x}{\sqrt{x^2-1}}dx$
4. Rewrite the remaining integral as $\int (x^2-1{)}^{-1/2}\cdot xdx$
5. Multiply and divide by 2 to match the derivative pattern $2xdx=d(x^2-1)$
6. Apply the power rule to integrate: $\frac{(x^2-1{)}^{1/2}}{1/2}=2\sqrt{x^2-1}$
7. Simplify coefficients ($\frac{1}{4}\times 2=\frac{1}{2}$) and add constant of integration $c$