Problem 2 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Evaluate the integral. \(\int_{0}^{\pi} \sin ^{3} x d x\)

Step-by-Step Solution

Answer

The integral evaluates to 0.

Step 1: Expressing \( \sin^{3} (x) \) as the product of other trigonometric functions

Rewrite the integrand \( \sin^{3} (x) \) as \( \sin (x) \cdot \sin^{2} (x) \). Then using the identity \( \sin^{2} (x) = 1 - \cos^{2} (x) \), we get \(\sin (x) \cdot (1 - \cos^2(x))\) as the expression to integrate.

Step 2: Substitution for simplification

To simplify the integration, let \( u = \cos(x) \). Then \( du = -\sin(x) dx \). Hence our integral transforms to \( -\int u^{2} - 1 du \).

Step 3: Compute the integral

Now break down \( -\int u^{2} - 1 du \) into two separate integrals, \( -\int u^{2} du - \int du \). The integral of \( u^{2} \) is \( \frac{1}{3} u^{3} \) and the integral of 1 is u. Replace the variable u back with \( \cos(x) \) after computing the integrals.

Step 4: Evaluate the definite integral

Finally, we evaluate the new expressions at the limits of the integral. This is done by replacing \( x \) with the upper limit \( \pi \) and then the lower limit 0, and subtract the two results due to the Fundamental Theorem of Calculus. The difference gives us the value of the integral.

Key Concepts

Trigonometric Functions

Understanding trigonometric functions is crucial when dealing with integrals, especially those involving powers of sine and cosine functions. A fundamental set of functions, trigonometric functions like sine, cosine, and tangent, relate the angles of a triangle to the lengths of its sides in a right triangle. But beyond geometry, these functions are also periodic, making them essential in the study of waves, oscillations, and circular motion.

When faced with an integral involving a trigonometric function raised to a power, such as the given exercise involving \(\sin^3(x)\), it’s often helpful to use trigonometric identities to rewrite the expression in a form that’s easier to integrate. For instance, the identity \(\sin^2(x) = 1 - \cos^2(x)\) is used to transform \(\sin^3(x)\) into a format that allows for a straightforward application of integration techniques. This act of manipulation within the trigonometric realm sets the stage for efficient integration.

Integration by Substitution

Integration by substitution, often referred to as u-substitution, is the counterpart of the chain rule for differentiation. It allows us to transform a tough integral into a simpler one by changing variables. This technique is particularly effective when dealing with compositions of functions or integrands involving products of functions.

In the exercise at hand, the substitution of \(u = \cos(x)\) conveniently converts the function involving a sine term into one involving u, thereby simplifying the integral. After substitution, we need to consider the derivatives involved so that all instances of \(x\) are replaced with \(u\), and \(dx\) is accounted for by \(du\). The beauty of u-substitution lies in its ability to reduce complex integrals into basic polynomials or simple functions that can be integrated readily, as demonstrated in the provided exercise steps.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges the concepts of differentiation and integration, showing that they are essentially inverse processes. It consists of two parts: the first part guarantees the existence of antiderivatives for functions, while the second part, often used in evaluating definite integrals, states that if \(F\) is an antiderivative of \(f\) over an interval \[a, b\], then the definite integral of \(f\) from \(a\) to \(b\) is given by \(F(b) - F(a)\).

This theorem is regularly used after finding the indefinite integral of a function – the antiderivative is evaluated at the upper and lower limits of integration, and the difference of these values gives us the area under the curve, or the total accumulation, represented by the original definite integral. As illustrated in the exercise solution, after integrating and substituting back the original variable, we apply the Fundamental Theorem of Calculus to find the exact value of the integral from \(0\) to \(\pi\).