Evaluating the Integral of Inverse Trigonometric Functions

We are asked to find the value of the integral: \(\int \dfrac{dx}{\sec^2({\tan}^{-1}x)}\).

This integral involves an inverse trigonometric function inside a trigonometric function. To simplify this, we can use properties of trigonometric functions and inverse trigonometric functions.

Step-by-Step Solution to the Integral

Let's evaluate the given integral step by step:

1. Identify the integrand: The integrand is \(\dfrac{1}{\sec^2({\tan}^{-1}x)}\).
2. Use a trigonometric identity: Recall the identity \(\dfrac{1}{\sec^2\theta} = \cos^2\theta\). Applying this identity, the integrand becomes \(\cos^2({\tan}^{-1}x)\).
3. Simplify the argument of the trigonometric function: Let \(y = \tan^{-1}x\). This means that \(\tan y = x\).
4. Express \(\cos^2 y\) in terms of \(x\): We have \(\tan y = x\). We can visualize this using a right-angled triangle where the angle is \(y\). The tangent is the ratio of the opposite side to the adjacent side. So, if the opposite side is \(x\) and the adjacent side is \(1\), the hypotenuse is given by the Pythagorean theorem: hypotenuse \(=\sqrt{x^2 + 1^2} = \sqrt{x^2+1}\).
5. Find \(\cos y\) from the triangle: \(\cos y = \dfrac{\text{Adjacent side}}{\text{Hypotenuse}} = \dfrac{1}{\sqrt{x^2+1}}\).
6. Find \(\cos^2 y\) in terms of \(x\): Squaring \(\cos y\), we get \(\cos^2 y = \left(\dfrac{1}{\sqrt{x^2+1}}\right)^2 = \dfrac{1}{x^2+1}\).
7. Substitute back into the integral: The integral \(\int \cos^2({\tan}^{-1}x) dx\) becomes \(\int \dfrac{1}{x^2+1} dx\).
8. Evaluate the standard integral: The integral \(\int \dfrac{1}{x^2+1} dx\) is a standard integral form. It is the derivative of \({\tan}^{-1}x\).
9. Write the final result: Therefore, \(\int \dfrac{1}{x^2+1} dx = {\tan}^{-1}x + C\), where \(C\) is the constant of integration.

Thus, the value of the integral \(\int \dfrac{dx}{\sec^2({\tan}^{-1}x)}\) is \({\tan}^{-1}x + C\).

Standard Integral Forms

It is helpful to remember some standard integral forms. The integral \(\int \dfrac{dx}{x^2+a^2}\) is a common one.

• \(\int \dfrac{dx}{x^2+a^2} = \dfrac{1}{a} {\tan}^{-1}\left(\dfrac{x}{a}\right) + C\)

In our case, the integral is \(\int \dfrac{dx}{x^2+1}\). This corresponds to the standard form with \(a=1\). So, \(\int \dfrac{dx}{x^2+1^2} = \dfrac{1}{1} {\tan}^{-1}\left(\dfrac{x}{1}\right) + C = {\tan}^{-1}x + C\).

Let's review the options provided:

• sin ^-1 x + c
• tan ^-1 x + c
• sec ^-1 x + c
• cos ^-1 x + c

Our calculated result is \({\tan}^{-1}x + C\), which matches one of the given options.

Revision Table: Key Concepts

Additional Information on Trigonometric Substitution

While we solved this by simplifying the integrand directly, sometimes integrals involving inverse trigonometric functions can be solved using trigonometric substitution. For example, if we had an expression like \(\sqrt{a^2+x^2}\), a substitution like \(x = a\tan\theta\) might be useful because \(a^2 + a^2\tan^2\theta = a^2(1+\tan^2\theta) = a^2\sec^2\theta\), and \(\sqrt{a^2\sec^2\theta} = a|\sec\theta|\).

In our problem, the expression involves \({\tan}^{-1}x\). The substitution \(y = {\tan}^{-1}x\) (or \(x = \tan y\)) allowed us to simplify the trigonometric function containing the inverse function. Then we converted the result back into a function of \(x\) before integrating.

Understanding the relationship between trigonometric functions and inverse trigonometric functions, as well as standard integral forms, is crucial for solving such problems effectively.