Problem 33 - Essential Calculus

Question

In Exercises 33-36, find or evaluate the integral. $$ \int_{0}^{\pi / 2} \frac{1}{1+\sin \theta+\cos \theta} d \theta $$

Step-by-Step Solution

Answer

The definite integral \( \int_{0}^{\pi / 2} \frac{1}{1+\sin \theta+\cos \theta} d \theta \) is equal to \( \ln|(2 + \sqrt{2})| \).

Step 1: Apply Trigonometric Identity

Take advantage of the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \). Substituting \( \sin \theta + \cos \theta = t \), then \( \sin \theta = t - \cos \theta \). Therefore, the integral becomes \( \int_{0}^{1 + \sqrt{2}} \frac{1}{1+t} dt \). Note that limits of the integral have also changed from [0, π/2] to [0, 1+ √2] due to the substitution.

Step 2: Compute the Integral

At this stage, the integral takes a simpler form that can be easily evaluated. So we compute the integral \( \int_{0}^{1 + \sqrt{2}} \frac{1}{1+t} dt \) which reduces to \( [\ln|1 + t|]_{0}^{1 + \sqrt{2}} \) after integration.

Step 3: Evaluate the Definite Integral

Now we evaluate the integral over the limits. This involves subtracting the obtained natural logarithmic terms evaluated at \( 1 + \sqrt{2} \) and \( 0 \). Hence, the final result is \( \ln|1 + 1 + \sqrt{2}| - \ln|1 + 0| = \ln|(2 + \sqrt{2})| - \ln|1| \). Remember that the natural logarithm of 1 equals 0.

Step 4: Simplify the Result

We simplify the result from the previous step to obtain the final answer. The expression reduces to \( \ln|(2 + \sqrt{2})| \).

Key Concepts

Trigonometric Substitution

Trigonometric substitution is a powerful integration technique used to simplify integrals involving trigonometric expressions.

By substituting a part of the expression with a trigonometric identity or function, the integral can become more manageable and easier to evaluate.

For instance, in the exercise involving the integral \( \int_{0}^{\pi / 2} \frac{1}{1+\sin \theta+\cos \theta} d \theta \), we utilize a form of substitution by setting \( \sin \theta + \cos \theta = t \). This changes the variables and the limits of integration.

It often uses identities like:

• The Pythagorean identity: \( \sin^2 \theta + \cos^2 \theta = 1 \)

By converting to a new variable \( t \), we focus on algebraic manipulation, and this can simplify integration steps.
This technique is especially useful for definite integrals as it can also adeptly adjust the limits of integration to new values corresponding to the substitution.

Definite Integrals

Definite integrals allow us to evaluate the area under a curve described by a function over a specified interval.

In the exercise, the definite integral is taken from \(0\) to \(\pi / 2\) for the expression \( \frac{1}{1+\sin \theta+\cos \theta} \).

When performing definite integrals:

• The limits of integration specify the boundary over which the area calculation is performed.
• Any substitution done changes these limits to suit the new variable.

In the solution, after substituting \( \sin \theta + \cos \theta = t \), these limits change from \([0, \pi/2]\) to \([0, 1+\sqrt{2}]\).

The definite integral is then computed with respect to this new variable \( t \) and evaluated between the new limits. This ability to adapt boundary conditions is what makes definite integrals a versatile tool in calculus.

Integration Techniques

Integration techniques refer to an array of methods used to evaluate integrals, particularly when standard methods like basic antiderivatives don't easily apply.

In the exercise provided, we employ trigonometric substitution as part of an integration technique. It simplifies our integrand into a form easier to integrate:

• The integral \( \int \frac{1}{1+t} dt \) results directly in the natural logarithm, \( \ln|1+t| \).

Once integrated, we apply the limits to find the solution \( \left[ \ln|1+t| \right]_{0}^{1+\sqrt{2}} \), a process that involves evaluating the definite form.

These techniques are an integral part of calculus, including:

• Substitution (trigonometric, algebraic, etc.)
• Integration by parts
• Partial fractions

Each technique tackles a specific kind of problem and can often be combined, as seen in this exercise, to arrive at the solution.