/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 65 Evaluate. $$\int_{-\pi}^{\pi} ... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 65

Evaluate. $$\int_{-\pi}^{\pi} \sin ^{4} x \cos x d x$$

Short Answer

Expert verified
The result of \(\int_{-\pi}^{\pi} \sin^{4} x \cos x dx\) is 0.

Step by step solution

01

Apply substitution

Assign \(u = \sin x\) and calculate \(du\), the differential of \(u\). Here \(du = \cos x dx\). So the integral transforms to \(\int u^{4} du\)
02

Carry out the Integration

Perform integration to get \(\frac{1}{5}u^{5}+C\). Substituting back \(u = \sin x\) and our constant for the indefinite integral \(C = 0\) (since we are doing a definite integral), we have \(\frac{1}{5}\sin^{5}x + 0\) from \(-\pi\) to \(\pi\) as our result.
03

Apply Odd Function Property

Recognising that \(\sin^{5}x\) is an odd function on the interval \([-\pi, \pi]\) and we know that the integral of an odd function over an interval symmetric about the origin is zero. Therefore, the final evaluation returns 0.

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Key Concepts

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

Integration Techniques
Understanding integration techniques is key to solving calculus problems involving the calculation of the area under a curve or the total accumulated value. In calculus, integration is the reverse process of differentiation, and it comes with various methods such as the Substitution Method, Integration by Parts, Partial Fractions, and Trigonometric Integration, among others.

The purpose of these techniques is to transform complex integrals into a simpler form that can be readily evaluated. For instance, when faced with a difficult integral, a student may opt for the Substitution Method to simplify and make the integral more tractable. With each technique tailored to specific types of functions and equations, knowing when and how to apply each is crucial for any calculus student.
Substitution Method
The substitution method works wonders for integrating functions that are the product of a function and its derivative, or for functions that can be made to fit this form through algebraic manipulation.

Here's how it unfolds:
  • Choose a part of the integral to be the new variable u, which often simplifies the integral.
  • Calculate the differential du which replaces another part of the integral.
  • Rewrite the integral in terms of u and du, simplifying it greatly.
  • Finally, solve the simpler integral and substitute back the original variable.
In the given exercise, the Substitution Method is used by setting u to \(\sin x\). After finding \(du\), the integral \(\int \sin ^{4} x \cos x d x\) becomes a much simpler integral in terms of u, \(\int u^{4} du\).
Odd Function Properties
The odd function properties are particularly intriguing in calculus due to their symmetrical nature about the origin. An odd function, mathematically, is one which satisfies the condition \(f(-x) = -f(x)\) for all x in the function's domain.

When integrating an odd function over a symmetric interval, such as from -a to a, the result is always zero. This is because the area under the curve above the x-axis over [0, a] is cancelled out by the area under the curve below the x-axis over [-a, 0].

How does it apply to the exercise?

In our case, \sin^{5}x represents an odd function. By recognizing this property, we quickly conclude that the integral over the symmetric interval [-蟺, 蟺] is zero, thus simplifying the problem and avoiding unnecessary calculations.

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Most popular questions from this chapter

A particle moves along the \(x\) -axis with velocity \(v(t)=\) \(At\) \(B\). Determine \(A\) and \(B\) given that the initial velocity of the particle is 2 units per second and the position of the particle after 2 seconds of motion is 1 unit to the left of the initial position.

Reverse the roles of \(x\) and \(u\) in \((5.7 .2)\) and write $$\int_{x(a)}^{x(b)} f(x) d x=\int_{a}^{b} f(x(u)) x^{\prime}(u) d u$$ (The area of a circular region) The circle \(x^{2}+y^{2}=r^{2}\) encloses a circular disc of radius \(r\). Justify the familiar formula \(\therefore=\pi r^{2}\) by integration. HINT: The quarter-disk in the first quadrant is the region below the curve \(y=\sqrt{r^{2} - x^{2}}, x \in\) [0,\(r\) ]. Therefore $$A=4 \int_{0}^{r} \sqrt{r^{2}-x^{2}} d x$$ Set \(x=r \sin u, d x=r \cos u d u\).

Calculate. $$\frac{d}{d x}\left(\int_{0}^{x^{3}} \cdot \frac{d t}{\sqrt{1+t^{2}}}\right)$$

Calculate. $$\int \frac{\sin x}{\sqrt{1+\cos x}} d x$$

Evaluate the integral. $$\int_{0}^{2} f(x) d r ; \quad f(x)=\left\\{\begin{aligned} 2 x+1, & 0 \leq x \leq 1 \\ 4-x, & 1< x \leq 4 \end{aligned}\right.$$
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