Problem 297 - Calculus Volume 1

Question

In the following exercises, use a change of variables to evaluate the definite integral. $$ \int_{0}^{\pi / 4} \frac{\sin \theta}{\cos ^{4} \theta} d \theta $$

Step-by-Step Solution

Answer

\( \frac{\sqrt{2} - 1}{3} \)

Step 1: Identify a substitution

To evaluate the given integral, we will use a common integration technique called substitution. Here, we notice that the integrand involves both sine and cosine. A useful substitution is to let the denominator be a simple variable. Let \( u = \cos \theta \). This implies \( du = - \sin \theta \, d\theta \) or \( -du = \sin \theta \, d\theta \).

Step 2: Change variables in the integral

Substitute \( u \) and \( du \) into the integral. The integral becomes:\[ \int_{\cos(0)}^{\cos(\pi/4)} \frac{-du}{u^4} \]Substituting the limits of integration, we get:\[ \int_{1}^{\frac{\sqrt{2}}{2}} \frac{-du}{u^4} \]

Step 3: Simplify the integral

The integral simplifies by factoring out the negative sign:\[ -\int_{1}^{\frac{\sqrt{2}}{2}} u^{-4} \, du \]This is a standard power integral and can be integrated directly.

Step 4: Integrate the simplified function

Evaluate the integral using the power rule for integrals:\[ -\left[ \frac{u^{-3}}{-3} \right]_{1}^{\frac{\sqrt{2}}{2}} = \frac{1}{3} \left[ u^{-3} \right]_{1}^{\frac{\sqrt{2}}{2}} \]

Step 5: Evaluate the definite integral

Substitute the limits into the antiderivative:\[ \frac{1}{3} \left[ \left(\frac{\sqrt{2}}{2}\right)^{-3} - 1^{-3} \right] \]This gives:\[ \frac{1}{3} \left[ \left(\frac{2}{\sqrt{2}^3}\right) - 1 \right] = \frac{1}{3} \left[ 4\sqrt{2} - 1 \right] \]

Step 6: Simplify the result

Calculate \( \left(\frac{2}{\sqrt{2}^3}\right) \). Since \( \sqrt{2}^3 = 2\sqrt{2} \), it simplifies to \( \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}} \), and hence we calculate:\[ \frac{1}{3} \left[ \frac{1}{\sqrt{2}} - 1 \right] = \frac{1}{3} \left[ \sqrt{2} - 1 \right] \]Finally, the integral evaluates to \( \frac{\sqrt{2} - 1}{3} \).

Key Concepts

Definite Integrals

Definite integrals allow us to calculate the area under a curve for a specific range of values. These types of integrals are used to determine the net "accumulated" value, whether this is distance, area, or another quantity, over a particular interval. When you see the notation \( \int_{a}^{b} f(x) \, dx \), it represents a definite integral from \( a \) to \( b \). Here, \( a \) and \( b \) are the limits of integration.

To evaluate a definite integral:

• Find the antiderivative of the function, known as the indefinite integral.
• Apply the Fundamental Theorem of Calculus, which states: \( \int_{a}^{b} f(x) \, dx = F(b) - F(a) \).

This means you substitute the upper limit into your antiderivative and subtract the result of substituting the lower limit.

For our exercise, we worked through a definite integral involving a trigonometric function. We calculated it over the interval \( [0, \pi/4] \).

Trigonometric Integration

Trigonometric integration is a technique used to integrate functions involving trigonometric functions like \( \sin(\theta) \) and \( \cos(\theta) \). These integrals often require special methods, such as using identities or substitutions, due to the periodic and oscillating nature of trigonometric functions.

To handle integrals like the one in our problem, here are some tips:

• Utilize trigonometric identities to simplify the integrand. Identities can sometimes reduce complex expressions into simpler, more integrable forms.
• Consider appropriate substitutions. For instance, expressing sine in terms of cosine or vice-versa might simplify the expression.

In the original exercise, we substituted \( u = \cos(\theta) \), which transformed the original problem and made integrating the expression more straightforward. The substitution also facilitated changing the integration limits, via \( u \), simplifying our task further.

Change of Variables in Calculus

The change of variables method, also known as substitution, is a clever and powerful integration technique. It involves replacing the integration variable with another variable, which can make a complex integral easier to solve. This is especially useful in cases like trigonometric or exponential integrals.

There are a few essential steps in using substitution correctly:

• Choose a substitution \( u = g(x) \) that simplifies the integral's composition.
• Determine \( du \) by differentiating your substitution with respect to \( x \), resulting in \( du = g'(x) \, dx \).
• Rewrite the integral in terms of \( u \) and \( du \).
• Don't forget to adjust the limits of integration if you're solving a definite integral.

In our given exercise, we replaced \( \cos(\theta) \) with \( u \), and converted our differential, \( d\theta \), into \( du \). This altered the definite integral, creating a simpler expression to evaluate and perfectly demonstrated the power of substitution in making complex integrals solvable.