Question

Evaluate the integrals $$ \int_{1}^{\sqrt{3}} x y d y, \text { where } x=\sec t, y=\tan t $$

Short answer

\( \frac{8 - 2\sqrt{2}}{3} \)

Step-by-step solution

Parameterize the bounds

First, let's determine the bounds of integration in terms of the variable \( t \). Given \( y = \tan t \), the lower bound \( y = 1 \) occurs when \( t = \frac{\pi}{4} \) and the upper bound \( y = \sqrt{3} \) occurs when \( t = \frac{\pi}{3} \). Thus, the bounds in terms of \( t \) are from \( t = \frac{\pi}{4} \) to \( t = \frac{\pi}{3} \).

Convert the integral to a single variable

Since \( x = \sec t \) and \( y = \tan t \), the integral \( \int x y \ dy \) can be rewritten as \( \int \sec t \tan t \ dy \). However, we need \( dy \) in terms of \( dt \), so differentiate \( y = \tan t \) to obtain \( dy = \sec^2 t \ dt \). The integral now becomes \( \int \sec t \tan t \sec^2 t \ dt \).

Simplify the integral

Simplify the integral \( \int \sec t \tan t \sec^2 t \ dt \) to \( \int \sec^3 t \tan t \ dt \).

Evaluate the integral

The integral \( \int \sec^3 t \tan t \ dt \) can be tricky, but it can be approached with a substitution. Let \( u = \sec t \), then \( du = \sec t \tan t \ dt \). This changes the integral to \( \int u^2 \ du \). Integrating gives \( \frac{u^3}{3} + C \). Substitute back \( u = \sec t \) to get \( \frac{\sec^3 t}{3} + C \).

Apply the bounds

Evaluate \( \left[ \frac{\sec^3 t}{3} \right]_{t=\frac{\pi}{4}}^{t=\frac{\pi}{3}} \). Calculate the definite integral: first at \( \frac{\pi}{3} \) gives \( \left( \frac{2^3}{3} \right) = \frac{8}{3} \), and at \( \frac{\pi}{4} \) gives \( \left( \frac{\sqrt{2}^3}{3} \right) = \frac{2\sqrt{2}}{3} \). Subtract these results to obtain \( \frac{8}{3} - \frac{2\sqrt{2}}{3} \).

Simplify the final expression

Simplify the expression \( \frac{8}{3} - \frac{2\sqrt{2}}{3} \) to reach the final simplified result, which is \( \frac{8 - 2\sqrt{2}}{3} \).

Key concepts

Trigonometric Integrals

Trigonometric integrals, as the name suggests, involve trigonometric functions like sine, cosine, tangent, and their inverses. These integrals are very common in calculus, especially when dealing with problems involving angles and periodic phenomena. In this exercise, we encounter trigonometric functions because of the transformation from Cartesian to polar-like coordinates where \( x = \sec t \) and \( y = \tan t \).

• Here, \( \sec t \) (secant) is the reciprocal of \( \cos t \), and \( \tan t \) (tangent) is the ratio of \( \sin t \) to \( \cos t \).
• Trigonometric integrals often require specific strategies, such as substitutions, to simplify and solve them effectively.
• For instance, in this problem, the integration of \( \sec^3 t \tan t \ dt \) is reduced by a substitution method utilizing the derivative properties of trigonometric functions.

Understanding these trigonometric functions and their integrals is crucial in solving calculus problems efficiently, especially those involving rotations, vibrations, and waves.

Definite Integrals

Definite integrals allow us to calculate the net area under a curve between two specific points. In our exercise, we use definite integration to find the area with specific trigonometric bounds.

• The main difference between definite and indefinite integrals is that definite integrals have specified integration limits. This gives a numeric result rather than a function with constant \( C \).
• In the problem, the limits are transformed from \( y \)-coordinates to \( t \)-coordinates, changing from \( t = \frac{\pi}{4} \) to \( t = \frac{\pi}{3} \).
• The final step involves evaluating the antiderivative at these limits and finding their difference. This ultimately provides the area or total accumulation between these two bounds.

When working with definite integrals, accuracy during transformation and application of the limits is key to ensure the right outcome.

Substitution Method Integration

The substitution method is a powerful tool in integration used to simplify integrals by transforming variables. The primary goal of substitution is to make the integral easier to evaluate by turning it into a form we recognize and can solve.

• In this method, you choose a substitution that simplifies the integral, often turning a complex function into a simpler polynomial form.
• For example, in our exercise, the substitution \( u = \sec t \) transforms the complex integral \( \int \sec^3 t \tan t \ dt \) into a simpler polynomial form \( \int u^2 \ du \).
• By differentiating \( u \), you express \( dt \) in terms of \( du \), facilitating the substitution process.

Using substitution effectively requires practice and intuition to choose transformations that simplify the expression significantly, often involving derivatives and inverse relationships in trigonometry.