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Chapter 8: Problem 58

Evaluating a Definite Integral In Exercises \(57-64\) , evaluate the definite integral. Use the integration capabilities of a graphing utility to verify your result. $$ \int_{0}^{\pi} \sin ^{2} t \cos t d t $$

Short Answer

Expert verified
The value of the integral \(\int_{0}^{\pi} \sin ^{2} t \cos t d t\) is \(0\).

Step by step solution

01

Apply the power-reducing identity

Trigonometric identities can be used to simplify integrals. In this case, apply the power-reducing identity for \(\sin ^{2}t\), which is \(\sin ^{2}t= 1/2 - 1/2 \cos (2t)\). So the integral becomes \(\int_{0}^{\pi} (1/2 - 1/2 \cos(2t)) \cos(t) dt\).
02

Split the integral

Split the integral into two parts: \(\int_{0}^{\pi} 1/2 \cos(t) dt - \int_{0}^{\pi} 1/2 \cos(2t) \cos(t) dt\).
03

Evaluate the first integral

The first part of the split integral \(\int_{0}^{\pi} 1/2 \cos(t) dt\) can be easily evaluated to be \([\sin(t)/2]_{0}^{\pi}\). Substituting \(\pi\) and \(0\) in \(t\) gives \(0 - 0 = 0\). So the result is \(0\).
04

Use U-substitution for the second integral

Use the substitution method for the second part. Set \(u = 2t - t\), and therefore \(du = t\). Then \(\int_{0}^{\pi} 1/2 \cos(u) du\) simplifies to \([\sin(u)/2]_{0}^{\pi}\). Substituting \(\pi\) and \(0\) into \(u\) gives \(0 - 0 = 0\). So the result is \(0\).
05

Add the solutions of the split integrals

By adding the solutions of the two integrals together, \(0 + 0\) we get the final answer \(0\).

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Key Concepts

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

Power-Reducing Identities
Power-reducing identities are essential tools in calculus for simplifying integrals involving powers of trigonometric functions. For instance, the power-reducing identity for \(\sin^2(t)\) is expressed as \(\sin^2(t) = \frac{1}{2} - \frac{1}{2} \cos(2t)\).

When you encounter powers of sine or cosine in an integral, using these identities can make the integral more manageable. To show their application, let's consider the integral \(\int_{0}^{\pi} \sin^2(t) \cos(t) dt\).
  • First, apply the identity to \(\sin^2(t)\), yielding \(\int_{0}^{\pi} (\frac{1}{2} - \frac{1}{2} \cos(2t)) \cos(t) dt\).
  • Next, split the integral into simpler parts.
This technique effectively reduces the integration problem into one that involves basic trigonometric functions, making the integration process more straightforward.
U-Substitution
U-substitution is akin to the chain rule in reverse and is a common technique used to simplify the process of evaluating integrals. It's particularly useful when integrating composite functions. Here's a step-by-step application:

  • Select a part of the integrand to replace with \(u\), generally the inner function of a composition, and differentiate it to find \(du\).
  • Rewrite the integral in terms of \(u\) and \(du\) to simplify the expression.
  • After simplification, integrate with respect to \(u\), then substitute back the original variable.
In our example, the substitution \(u = 2t\) is used to transform the integral \(\int_{0}^{\pi} \cos(2t)\cos(t) dt\) into a simpler form. Although the solution shows an incorrect application, the correct method involves replacing \(2t\) with \(u\), and noting that \(du = 2dt\), leading to the evaluation of a new integral in terms of \(u\) with adjusted limits of integration.
Integration Techniques
Several integration techniques exist to tackle a variety of integral forms encountered in calculus. After simplifying an integral with identities such as the power-reducing ones, you might need to employ techniques like:
  • Splitting the integral into parts that are more easily evaluated.

  • Using the method of partial fractions for rational functions.

  • Applying integration by parts for products of functions.
For the given integral, we split the expression after applying the power-reducing identity. The first part, \(\int_{0}^{\pi} \frac{1}{2}\cos(t) dt\), evaluates directly to \(0\) since \(\sin(t)\) is \(0\) at both \(t=0\) and \(t=\pi\). Such techniques, when combined, create a toolkit that allows us to navigate through complex integrals by reducing them to simpler forms that are more amenable to evaluation.
Trigonometric Integrals
Integration of trigonometric functions is a frequent task in calculus. It involves integrals of sine, cosine, tangent, and other trigonometric functions, sometimes in complex combinations.

Strategies for solving these include:
  • Using trigonometric identities to simplify the integral.
  • Employing methods like u-substitution when faced with trigonometric functions composed with linear functions.
  • For difficult integrals, techniques like trigonometric substitution or integration by parts may be required.
Our exercise dealt with integrating a sine squared function times a cosine function. By using a power-reducing identity and subsequently applying u-substitution (correctly identifying \(du\) and adjusting the limits of integration), we should be able to find the integral of this trigonometric expression effectively.

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Most popular questions from this chapter

True or False? In Exercises \(85-88\) , determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. If the graph of \(f\) is symmetric with respect to the origin or the \(y\) -axis, then \(\int_{0}^{\infty} f(x) d x\) converges if and only if \(\int_{-\infty}^{\infty} f(x) d x\) converges.

Finding Values In Exercises 49 and \(50,\) determine all values of \(p\) for which the improper integral converges. $$ \int_{1}^{\infty} \frac{1}{x^{p}} d x $$

Evaluating an Improper Integral In Exercises \(33-48\) determine whether the improper integral diverges or converges. Evaluate the integral if it converges, and check your results with the results obtained by using the integration capabilities of a graphing utility. $$ \int_{3}^{6} \frac{1}{\sqrt{36-x^{2}}} d x $$

Laplace Transforms Let \(f(t)\) be a function defined for all positive values of \(t .\) The Laplace Transform of \(f(t)\) is defined by $$F(s)=\int_{0}^{\infty} e^{-s t} f(t) d t$$ when the improper integral exists. Laplace Transforms are used to solve differential equations. In Exercises \(95-102,\) find the Laplace Transform of the function. $$ f(t)=\cos a t $$

Evaluating an Improper Integral In Exercises \(33-48\) determine whether the improper integral diverges or converges. Evaluate the integral if it converges, and check your results with the results obtained by using the integration capabilities of a graphing utility. $$ \int_{0}^{8} \frac{3}{\sqrt{8-x}} d x $$
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