Question

Use a calculator or CAS to evaluate the following integrals. $$ \int_{0}^{2 / \sqrt{3}} \frac{1}{4+9 x^{2}} d x $$

Short answer

The integral evaluates to \( \frac{\pi}{6} \).

Step-by-step solution

Recognize the Type of Integral

We see that the integral is of the form \( \int \frac{1}{a^2 + b^2x^2} \, dx \), which resembles the formula for the inverse tangent function: \( \frac{1}{a} \tan^{-1}\left(\frac{b}{a}x\right) + C \).

Identify Constants and Apply Substitution Formula

In the integral \( \int \frac{1}{4 + 9x^2} \, dx \), we identify \( a^2 = 4 \) and \( b^2 = 9 \). Thus, \( a = 2 \) and \( b = 3 \). Using the inverse tangent formula, \( \int \frac{1}{a^2+b^2x^2}dx = \frac{1}{a} \tan^{-1} \left( \frac{b}{a} x \right) + C \), the antiderivative is \( \frac{1}{2} \tan^{-1}(\frac{3}{2}x) + C \).

Evaluate Definite Integral using the Antiderivative

Now that we have the antiderivative, we evaluate the definite integral from \( 0 \) to \( \frac{2}{\sqrt{3}} \):\[\left[ \frac{1}{2} \tan^{-1} \left( \frac{3}{2}x \right) \right]_0^{\frac{2}{\sqrt{3}}} = \frac{1}{2} \left[ \tan^{-1} \left( \frac{3}{2} \times \frac{2}{\sqrt{3}} \right) - \tan^{-1}(0) \right]\]

Simplify the Expression

Calculate the terms inside the \( \tan^{-1} \):\( \frac{3}{2} \times \frac{2}{\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3} \).The inverse tangent \( \tan^{-1}(0) \) is \( 0 \).Therefore, the expression becomes:\[ \frac{1}{2} \left( \tan^{-1}(\sqrt{3}) - 0 \right) \]Using known values, \( \tan^{-1}(\sqrt{3}) = \frac{\pi}{3} \). Thus the result is:\[ \frac{1}{2} \times \frac{\pi}{3} = \frac{\pi}{6} \].

Key concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are vital in calculus, especially in integral calculus. These functions help when dealing with expressions where regular trigonometric functions are reversed. For instance, when you see an integral like \( \int \frac{1}{a^2 + b^2x^2} \, dx \), this is a perfect spot to apply an inverse trigonometric function. Specifically, the expression begins to mirror the arctan function, \( \tan^{-1}(x) \). Understanding these functions assists in recognizing integral forms and solving them accurately.

• \( \sin^{-1}(x) \) or arcsin: Inverse of sine
• \( \cos^{-1}(x) \) or arccos: Inverse of cosine
• \( \tan^{-1}(x) \) or arctan: Inverse of tangent

Whenever you're tackling problems with forms like \( a^2 + b^2x^2 \), think inverse tangent. The connection between the integral and \( \tan^{-1} \) helps us transition seamlessly from antiderivative to evaluating definite integrals.

Integration by Substitution

Integration by substitution is like finding a key to unlock a more complex problem. It helps in simplifying integrals to make them solvable with known formulas. Particularly, when dealing with inverse trigonometric functions, it often involves identifying a suitable substitution that transforms the integral into a comprehensible form.The method involves making a substitution where you choose \( u = g(x) \), which simplifies \( du = g'(x) dx \). This substitution should transform the integral into an easier one where standard antiderivative formulas can be applied. For example, in solving \( \int \frac{1}{4 + 9x^2} \, dx \), you might identify parts of the expression as substitution candidates, given the pattern matches that of an arctan derivation. Remember:

• Identify a substitutable part within the integral.
• Find \( du \) based on your substitution \( u \).
• Replace and simplify the original integral.

Integration becomes easier with substitution by breaking down complex expressions using known patterns and antiderivatives.

Evaluating Antiderivatives

Evaluating antiderivatives means finding the original function given its derivative. Mastering this bridges the gap to solving definite integrals. Once you compute an antiderivative, you're essentially at the halfway mark to evaluating a definite integral.To properly evaluate the antiderivative of expressions containing inverse trigonometric functions, recognize standard formula forms. In the expression \( \int \frac{1}{4 + 9x^2} \, dx \), we used the arctan function formula: \[ \int \frac{1}{a^2 + b^2x^2}dx = \frac{1}{a} \tan^{-1} \left( \frac{b}{a} x \right) + C \]With this, after substitution, the antiderivative simplifies into a trigonometric inverse form. You then substitute back any replaced variables if needed and insert limits of integration for definite evaluation. The antiderivative helps in calculating the value of the function over specific intervals by applying the fundamental theorem of calculus.

• Establish the correct antiderivative form with respect to the integral type.
• Reintegrate any substitutions made earlier then apply limits.
• Simplify your outcome to arrive at the definite integral's value.

Evaluating the antiderivative provides the core function needed to determine the definite integral's value, essentially completing the solution.