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Chapter 5: Problem 48

In Problems 35-62, use the Substitution Rule for Definite Integrals to evaluate each definite integral. \(\int_{0}^{\pi / 6} \frac{\sin \theta}{\cos ^{3} \theta} d \theta\)

Short Answer

Expert verified
The value of the integral is \(\frac{7}{6}\).

Step by step solution

01

Identify Substitution Variable

Identify a substitution that simplifies the integral. In this case, let us set the substitution variable as \(u = \cos \theta\). This choice is made because it will simplify the denominator \(\cos^3 \theta\).
02

Determine du and adjust limits

Differentiate \(u = \cos \theta\) to find \(du\): \(-\sin \theta d\theta = du\), or equivalently, \(d\theta = \frac{-du}{\sin \theta}\). Substitute \(\theta = 0\) and \(\theta = \pi/6\) into \(u = \cos \theta\) to find the new limits: \(u(0) = 1\) and \(u(\pi/6) = \sqrt{3}/2\).
03

Substitute and Simplify Integral

Substitute \(u = \cos \theta\) and \(d\theta = \frac{-du}{\sin \theta}\) into the integral. This gives: \[\int_{1}^{\sqrt{3}/2} \frac{\sin \theta}{u^3} \left(\frac{-du}{\sin \theta}\right) = -\int_{1}^{\sqrt{3}/2} \frac{du}{u^3}\]The \(\sin \theta\) terms cancel each other out.
04

Integrate with respect to u

Integrate \(-\int_{1}^{\sqrt{3}/2} \frac{du}{u^3}\). This simplifies to: \[-\int_{1}^{\sqrt{3}/2} u^{-3} \, du\]The antiderivative of \(u^{-3}\) is \(-\frac{1}{2u^2}\). Evaluate this from 1 to \(\sqrt{3}/2\).
05

Calculate Definite Integral

Calculate the evaluated antiderivative: \[-\left( \left[ -\frac{1}{2u^2} \right]_{1}^{\sqrt{3}/2} \right)\]This simplifies to:\[-\left( \frac{-1}{2(\sqrt{3}/2)^2} + \frac{1}{2(1)^2} \right) = -\frac{-1}{2 \cdot 3/4} + \frac{1}{2}\]\[= \frac{2}{3} + \frac{1}{2} = \frac{4}{6} + \frac{3}{6} = \frac{7}{6}\]
06

Final Result

Thus, the value of the definite integral is \(\frac{7}{6}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Definite Integral
Definite integrals represent the area under a curve within given limits on the x-axis. In our exercise, we've evaluated one such integral from \(\int_{0}^{\pi/6} \frac{\sin \theta}{\cos^3 \theta} d\theta\). This involves understanding the concept of definite integrals in calculus:
  • Definite integrals calculate the net area between a function and the x-axis over a specified interval.
  • Unlike indefinite integrals, which include a constant of integration, definite integrals yield a single numerical value.
  • The limits of integration are essential; they are the boundaries for which the area under the curve is measured.
For our exercise, we are interested in the range from \(0\) to \(\frac{\pi}{6}\). By solving the definite integral, we determine the precise area under the curve \(\frac{\sin \theta}{\cos^3 \theta}\) in this interval.
Substitution Method in Calculus
The substitution method, also called u-substitution, is a powerful technique for evaluating integrals. It essentially reverses the chain rule for derivatives. In our exercise, we used substitution to simplify the integral:
  • Identify a substitution that will simplify the integral. Here, choosing \(u = \cos \theta\) simplifies \(\cos^3 \theta\) in the denominator.
  • Calculate \(du\) by differentiating \(u = \cos \theta\), resulting in \(du = -\sin \theta d\theta\).
  • Change the limits of integration to match the new substitution, transforming \(\theta\) limits to \(u\) limits.
After substitution, the integral looks much simpler and involves a straightforward algebraic form. This makes the calculation of the integral much more manageable.
Antiderivatives in Calculus
Antiderivatives play a crucial role in solving integrals, especially when converting definite integrals into simpler forms. When we talk about antiderivatives:
  • An antiderivative is a function whose derivative is the original function.
  • Finding an antiderivative allows us to apply the Fundamental Theorem of Calculus, linking antiderivatives and definite integrals.
  • In our case, we found the antiderivative of \(u^{-3}\), which is \(-\frac{1}{2u^2}\).
Once the antiderivative is determined, evaluating it over the new limits provides the solution to our integral. The stepwise application of antiderivatives enables us to transition from the modified u-integral back to our original \(\theta\) terms, giving us the final numeric result: \(\frac{7}{6}\).

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