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Chapter 8: Problem 54

Evaluate the integrals in Exercises \(51-56\) $$ \int_{0}^{\pi / 2} \sin x \cos x d x $$

Short Answer

Expert verified
The integral evaluates to \( \frac{1}{2} \).

Step by step solution

01

Identify the Integration Technique

The integral \( \int \sin x \cos x \, dx \) is of a format that can be solved using substitution. Often, for integrals involving products of \( \sin x \) and \( \cos x \), leveraging trigonometric identities or substitution simplifies the process.
02

Use Trigonometric Identity

Recall the double-angle identity for sine: \( \sin(2x) = 2 \sin x \cos x \). So we can rewrite the original integral as \( \int \frac{1}{2} \sin(2x) \, dx \).
03

Integrate Using Substitution

Substitute \( u = 2x \), then \( du = 2 \, dx \) or \( dx = \frac{1}{2} du \). The limits of integration change accordingly: when \( x = 0 \), \( u = 0 \) and when \( x = \frac{\pi}{2} \), \( u = \pi \). Thus, the integral becomes \( \int_0^{\pi} \frac{1}{2} \sin(u) \cdot \frac{1}{2} \, du = \frac{1}{4} \int_0^{\pi} \sin(u) \, du \).
04

Integrate

The integral \( \frac{1}{4} \int_0^{\pi} \sin(u) \, du \) is evaluated as follows. The antiderivative of \( \sin(u) \) is \( -\cos(u) \). Thus, \( \frac{1}{4} [-\cos(u)]_0^{\pi} \).
05

Calculate the Definite Integral

Evaluate \( [-\cos(u)]_0^{\pi} \) by substituting the upper and lower limits: \( -\cos(\pi) + \cos(0) = 1 + 1 = 2 \). Thus, the original integral evaluates to \( \frac{1}{4} \times 2 = \frac{1}{2} \).

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Key Concepts

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

Trigonometric Substitution
Trigonometric substitution is a clever technique in calculus. It solves integrals involving products or quotients of trigonometric functions. In our example, we used trigonometric substitution by recognizing that the integral \( \int \sin x \cos x \, dx \) could be simplified. This is done through the identity \( \sin(2x) = 2 \sin x \cos x \). By identifying and manipulating such identities, trigonometric substitution transforms a complex integral into a simpler one.

In practice, this involves substituting a trigonometric function for another term to reduce the complexity of integration. This step often simplifies the integration process, making calculation more straightforward. It's beneficial to recognize when substitution could simplify your integral.

The key to successful trigonometric substitution is identifying suitable identities. Substituting accurately will help solve the integral efficiently. Therefore, it's useful to memorize common identities and practice recognizing opportunities to use them.
Definite Integral
A definite integral calculates the "net area" under a curve between two points. In our exercise, we evaluated \( \int_{0}^{\pi/2} \sin x \cos x \, dx \). By converting this into \( \frac{1}{2} \int_0^{\pi} \sin(2x) \, dx \) through substitution, we simplified the calculation.

Performing a definite integral involves several steps:
  • Changing variables as necessary to simplify the integrand.
  • Adjusting the limits of integration to reflect any substitutions made.
  • Calculating the antiderivative.
  • Evaluating at the specified bounds.
Through these steps, the integral becomes manageable and allows us to find the precise value of the area. In this case, once calculated, the final result was \( \frac{1}{2} \).

Mastering definite integrals requires practice with substitution and understanding how to handle bounds changes. This ensures you can find areas under curves confidently and correctly.
Trigonometric Identity
Trigonometric identities are essential tools in calculus. They simplify expressions and enable the transformation of complex integrals. In the original exercise, we used the identity \( \sin(2x) = 2 \sin x \cos x \) to transform the integrand.

By recognizing and applying this identity, we rewrote \( \int \sin x \cos x \, dx \) as \( \frac{1}{2} \int \sin(2x) \, dx \). This form was much simpler to integrate.

Understanding and employing trigonometric identities assists in:
  • Simplifying integrals for easier computation.
  • Redefining complex trigonometric expressions.
  • Identifying substitution opportunities within integrals.
Regular practice with these identities boosts comprehension. It also speeds up problem-solving by recognizing patterns and simplifying expressions quickly.

Remember, the more familiar you are with these identities, the easier it becomes to tackle challenging integrals with confidence.

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Most popular questions from this chapter

Use integration, the Direct Comparison Test, or the Limit Comparison Test to test the integrals for convergence. If more than one method applies, use whatever method you prefer. $$\int_{0}^{\pi} \frac{d t}{\sqrt{t}+\sin t}$$

The error function The error function, $$ \operatorname{erf}(x)=\frac{2}{\sqrt{\pi}} \int_{0}^{x} e^{-t^{2}} d t $$ important in probability and in the theories of heat flow and signal transmission, must be evaluated numerically because there is no elementary expression for the antiderivative of \(e^{-t^{2}}\) . a. Use Simpson's Rule with \(n=10\) to estimate erf \((1) .\) b. In \([0,1]\) , $$ \left|\frac{d^{4}}{d t^{4}}\left(e^{-t^{2}}\right)\right| \leq 12 $$ Give an upper bound for the magnitude of the error of the estimate in part (a).

You roll a pair of six-sided dice, and the random variable \(X\) assigns to each outcome the sum of the number of dots showing on each face, as in Example 2e. $$ \begin{array}{l}{\text { a. Find the set of possible outcomes. }} \\ {\text { b. Create a probability bar graph for } X .} \\ {\text { c. What is the probability that } X=8 ?} \\ {\text { d. What is the probability that } X \leq 5 ? X>9 ?}\end{array} $$

Cholesterol levels The serum cholesterol levels of children aged 12 to 14 years follows a normal distribution with mean \(\mu=162\) mg/dl and standard deviation \(\sigma=28 \mathrm{mg} / \mathrm{dl} .\) In a population of 1000 of these children, how many would you expect to have serum cholesterol levels between 165 and 193\(?\) between 148 and 167\(?\)

Use integration, the Direct Comparison Test, or the Limit Comparison Test to test the integrals for convergence. If more than one method applies, use whatever method you prefer. $$\int_{0}^{2} \frac{d x}{1-x^{2}}$$
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