Problem 48 - Thomas Calculus

Question

Evaluate the integrals. \begin{equation}\int_{\pi / 4}^{\pi / 2}\left(1+e^{\cot \theta}\right) \csc ^{2} \theta d \theta\end{equation}

Step-by-Step Solution

Answer

The value of the integral is \( e \).

Step 1: Substitution

First, we will perform a substitution to simplify the integral. Let \( u = \cot \theta \). Then, the derivative \( du = -\csc^2 \theta \, d\theta \). Thus, \( d\theta = -\frac{1}{\csc^2 \theta} \, du \). However, to keep the integral in the same format, substitute \( \csc^2 \theta \, d\theta \) with \( -du \). We also need to change the limits of integration: when \( \theta = \pi/4 \), \( u = 1 \), and when \( \theta = \pi/2 \), \( u = 0 \).

Step 2: Transform the Integral

Replace \( \cot \theta \) by \( u \) and \( \csc^2 \theta \ d\theta \) by \( -du \). Therefore, the integral becomes: \[\int_{\pi/4}^{\pi/2} (1+e^{\cot \theta}) \csc^2 \theta \, d\theta = \int_{1}^{0} (1 + e^{u}) (-du) = \int_{0}^{1} (1 + e^{u}) \, du\] where we have reversed the limits of integration by introducing a negative sign.

Step 3: Integrate the Function

Now, evaluate the integral \( \int_{0}^{1} (1 + e^{u}) \, du \). This integral can be split into two separate integrals: \[\int_{0}^{1} 1 \, du + \int_{0}^{1} e^{u} \, du\]The first integral is \( \int_{0}^{1} 1 \, du = [u]_{0}^{1} = 1 \). The second integral is \( \int_{0}^{1} e^{u} \, du = [e^{u}]_{0}^{1} = e - 1 \).

Step 4: Combine and Simplify the Results

Add the results from the separate integrals: \(1 + (e - 1) = e\). Thus, the value of the integral is \( e \).

Key Concepts

Substitution Method

The substitution method is a technique to simplify complex integrals by changing the variable of integration.
This approach involves substituting a part of the integral with a new variable, typically called "u". This new variable replaces part of the integrand, making the integral easier to solve.
In this task, we used the substitution \( u = \cot \theta \), which simplified the original expression by transforming the trigonometric part into a more manageable algebraic form.
When using substitution:

• Identify a substitution that will simplify the integral.
• Find the differential \( du \) in terms of \( d\theta \) or another appropriate variable.
• Replace each instance of the original variable and differential with the new variable and its differential.
• Update the limits of integration if dealing with a definite integral.

Trigonometric Substitution

Trigonometric substitution is particularly useful for integrals involving trigonometric functions like sine and cosine.
It's a bit different than the general substitution method because it leverages trigonometric identities to simplify the integral. Typical choices for trigonometric substitution include expressions like \( \sin \theta \), \( \cos \theta \), \( \tan \theta \), etc.
In the given exercise, we initially dealt with \( \csc^2 \theta \), leading us to use \( u = \cot \theta \) due to their intrinsic trigonometric relations.
Key points about trigonometric substitution:

• Use relevant trigonometric identities to simplify integrals, such as \( \sin^2 \theta + \cos^2 \theta = 1 \).
• Changes the problem into an algebraic form or another simple integrable form.
• It's usually used in cases where direct integration is too complex or not possible.

Integration Limits

Integration limits are crucial when solving definite integrals because they define the range over which you are integrating.
These limits can change if you use a substitution method, as the substitution alters the variable being integrated.
In our example, the initial integration limits were from \( \theta = \pi/4 \) to \( \theta = \pi/2 \).
After the substitution \( u = \cot \theta \), and calculating these limits in terms of \( u \), the new limits became from \( u = 1 \) to \( u = 0 \).
Important considerations:

• Calculate new limits using the substitution relation before performing the integration.
• If the limits are reversed after substitution, change the sign of the integral to correct direction.
• Accurately applying limits is essential for getting the correct result of the definite integral.

Definite Integrals

Definite integrals compute the net area under the curve of the function over a specific interval.
This is in contrast with indefinite integrals, which represent a family of functions. The bounds on definite integrals are specified by the integration limits.
The solution for definite integrals includes:

• Solving the integral function within given limits.
• Calculating the difference between the antiderivative evaluated at the upper and lower bounds.

In the exercise, after applying trigonometric substitution and simplifying the integral, we computed the integral from 0 to 1.
We broke it into two separate integrals: \( \int_0^1 1 \ du \) and \( \int_0^1 e^u \ du \), finding the combined result to be \( e \).
Utilizing definite integrals:

• Provides exact area calculations over a bounded region.
• Ensures simplification and evaluation all occur within given limits.
• Makes calculation finite, providing concrete numerical results.