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Chapter 6: Problem 81

Evaluate the definite integral. \(\int_{-\pi / 2}^{\pi / 2} \sin ^{2} x \cos x d x\)

Short Answer

Expert verified
The integral evaluates to \( \frac{2}{3} \).

Step by step solution

01

Choose a Suitable Substitution

Notice that the integral is of the form \( \int \sin^2 x \cos x \, dx \). This suggests a substitution that can simplify the expression. Consider setting \( u = \sin x \), which makes \( du = \cos x \, dx \). The limits of integration will change according to the substitution as follows: when \( x = -\pi/2 \), \( u = \sin(-\pi/2) = -1 \), and when \( x = \pi/2 \), \( u = \sin(\pi/2) = 1 \). Thus, our integral becomes \( \int_{-1}^{1} u^2 \, du \).
02

Integrate with Respect to \( u \)

Now, we need to integrate \( u^2 \) with respect to \( u \). The integral \( \int u^2 \, du \) is straightforward and is given by the antiderivative \( \frac{u^3}{3} + C \).
03

Evaluate the Definite Integral

Substitute the limits of integration into the antiderivative: \[ \left[ \frac{u^3}{3} \right]_{-1}^{1} = \frac{1^3}{3} - \frac{(-1)^3}{3}. \] Simplifying, we find: \[ \frac{1}{3} - \frac{-1}{3} = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}. \]
04

Final Result

Having evaluated the definite integral, we conclude that the integral \( \int_{-\pi / 2}^{\pi / 2} \sin^{2} x \cos x \, dx \) equals \( \frac{2}{3} \).

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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 technique used to simplify integrals involving trigonometric functions. It's quite powerful when dealing with expressions like \( \int \sin^2 x \cos x \, dx \). Here, we substitute a part of the trigonometric function with a new variable. This helps to transform the problem into a more straightforward integral. In our exercise, we chose \( u = \sin x \), which results in \( du = \cos x \, dx \).
  • This substitution simplifies the setup since \( \sin^2 x \) becomes \( u^2 \) and \( \cos x \, dx \) becomes \( du \).
  • It changes the approach from trigonometric to polynomial, making the integration much simpler.
These substitutions are vital tools in calculus, helping us translate complex problems into solvable ones.
Limits of Integration
When performing a definite integral, the limits of integration define the range over which you're integrating. In definite integrals involving substitution, it's also crucial to adjust these limits based on your substitution.
  • In this exercise, the original limits were given in terms of \( x \) and were \(-\pi/2\) to \(\pi/2\).
  • Mindfully, we adapt these to our new variable \( u = \sin x \). Therefore, when \( x = -\pi/2 \), \( u \) becomes \(-1\); and when \( x = \pi/2 \), \( u \) becomes \(1\).
This transformation is crucial because it ensures that our definite integral is computed correctly through the change of variables. We are still considering the same segment of the function but expressed through our substitution variable.
Antiderivative
Finding the antiderivative is a key step in solving integrals. It involves determining a function whose derivative yields the original function you started with. In our example, integrating \( u^2 \) leads us to its antiderivative.
  • The integral, \( \int u^2 \, du \), is relatively straightforward and results in \( \frac{u^3}{3} + C \).
  • The expression \( \frac{u^3}{3} \) must then be evaluated using our limits of integration, from \(-1\) to \(1\), turning it into \( \left[ \frac{u^3}{3} \right]_{-1}^{1} \).
  • Substituting these limits into the antiderivative results in \( \frac{1}{3} - \frac{-1}{3} = \frac{2}{3} \).
By applying the antiderivative, we successfully evaluate the definite integral, achieving a clear and definite result. Understanding how to calculate antiderivatives makes solving integrals significantly more manageable.

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