Question

The following integrals may require more than one table look-up. Evaluate the integrals using a table of integrals, and then check your answer with a computer algebra system. $$\int \frac{\sin ^{-1} a x}{x^{2}} d x, a>0$$

Short answer

Question: Evaluate the integral $$\int \frac{\sin^{-1}(ax)}{x^2} dx.$$ Answer: The integral $$\int \frac{\sin^{-1}(ax)}{x^2} dx = P - \frac{\sin^{-1}(ax)}{x} + C$$ where P is the principal part of the integral and C is the integration constant.

Step-by-step solution

Set up the integration by parts

Integration by parts formula is given as: $$\int u dv=uv - \int vdu$$ For our given integral, set: $$u = \sin ^{-1} a x \quad \text{and} \quad dv = \frac{dx}{x^{2}}$$ Now, we differentiate u and integrate dv: $$du = \frac{adx}{\sqrt{1-(ax)^2}}dx \quad \text{and} \quad v = \frac{-1}{x}$$ The given integral becomes: $$\int \frac{\sin ^{-1} a x}{x^{2}} d x=-\frac{\sin ^{-1} a x}{x}+\int\frac{a}{\sqrt{1-(ax)^2}}dx$$

Substitution to simplify integrand

Now we need to handle the remaining integral. We'll use the following substitution: $$w = a x$$ $$dw = a dx$$ $$dx = \frac{dw}{a}$$ The integral now becomes: $$\int \frac{a}{\sqrt{1-(ax)^ 2}}dx = \int \frac{1}{\sqrt{1-w^2}} dw$$

Integrate and rewrite in terms of x

Now, we consult the table of integrals or use basic trigonometric knowledge to evaluate the integral: $$\int \frac{1}{\sqrt{1-w^2}} dw = \sin^{-1} w + C$$ Now, substitute back \(w = ax\): $$\int \frac{1}{\sqrt{1-(ax)^ 2}}dx = \sin^{-1} (ax) + C$$

Combine the results

Now we need to combine the results of the integration by parts and the remaining integral. The final answer is: $$\int \frac{\sin ^{-1} a x}{x^{2}} d x = -\frac{\sin ^{-1} a x}{x}+\sin^{-1} (ax) + C = P-\frac{\sin ^{-1} a x}{x} + C$$ where P is the principal part of the integral. To check with a computer algebra system, you can use software like Mathematica or online options such as Wolfram Alpha or Symbolab.

Key concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are key to solving certain types of integrals, especially those involving arc functions like arcsin, arccos, and arctan. They are the reverse operations of the regular trigonometric functions.
These functions help to determine the angle whose trigonometric ratio is a given value. For example:

• \(\sin^{-1}(x)\) means what angle gives \(\sin(\theta) = x\).
• Similarly, \(\cos^{-1}(x)\) and \(\tan^{-1}(x)\) solve for the respective angles.

These functions are particularly handy in integration techniques, as they allow you to transform and simplify complex expressions through integration by parts and substitutions.
Understanding the differentiation and integration of inverse trigonometric functions is crucial because they appear frequently across calculus problems.
For instance, derivatives of these functions often use expressions like \( \frac{1}{\sqrt{1-x^2}} \) for arcsine, where their derivatives match the antiderivatives required in integration problems, as we see in the integration from the exercise.

Substitution Method

The substitution method is one of the most important and useful techniques for solving integrals. It involves changing variables to make the integral easier to solve. The idea is to substitute a part of the integrand with a new variable, transforming the problem into a basic integral that is easier to handle.
In the original exercise, substitution was used to simplify an integral involving an inverse trigonometric function:

• Let's say you have an integral \(\int f(ax) \, dx\). Here, you set \(w = ax\), which transforms \(dx\) to \(\frac{dw}{a}\).
• This helps reduce complex expressions and is especially useful when dealing with functions like \(\frac{1}{\sqrt{1-(ax)^2}}\), turning it into a simpler form like \(\frac{1}{\sqrt{1-w^2}}\), which is more straightforward to integrate.

The substitution not only simplifies integration but also provides a path to convert back to the original variables, which is necessary for solving problems accurately and entirely.
By mastering substitution, students can tackle a wider range of problems and improve their integration skills greatly.

Integration Techniques

Integration techniques are essential tools in calculus for finding antiderivatives of functions. One primary technique is integration by parts, useful when the product of functions is involved.
The integration by parts formula is derived from the product rule of differentiation:

• The formula is \(\int u \, dv = uv - \int v \, du\), where you select \(u\) and \(dv\) from the integrand.
• The task is then to differentiate \(u\) to find \(du\) and integrate \(dv\) to find \(v\), substituting them into the equation.

In our case from the exercise, \(u = \sin^{-1}(ax)\) and \(dv = \frac{dx}{x^2}\) were cleverly chosen to ease solving of the integral.
Integration by parts often requires creativity and experience since choosing \(u\) and \(dv\) is not always straightforward.
It's paired with other techniques like substitution to manage more involved problems effectively. Understanding these techniques prepares you to deal with any complexity in integral calculus problems.