Problem 53 - Calculus: One and Several Variables

Question

(a) Use integration by parts to show that for \(n>2\) $$\int \sin ^{n} x d x=-\frac{1}{n} \sin ^{n-1} x \cos x+\frac{n-1}{n} \int \sin ^{n-2} x d x.$$ (b) Then show that $$\int_{0}^{\pi / 2} \sin ^{\prime \prime} x d x=\frac{n-1}{n} \int_{0}^{\pi / 2} \sin ^{n-2} x d x.$$ (c) Verify the Wallis sine formulas: for even \(n \geq 2\) $$\int_{0}^{\pi / 2} \sin ^{n} x d x=\frac{(n-1) \cdots 5 \cdot 3 \cdot 1}{n \cdots 6 \cdot 4 \cdot 2} \cdot \frac{\pi}{2}.$$ for odd \(n \geq 3,\) $$\int_{0}^{\pi / 2} \sin ^{5} x d x=\frac{(n-1) \cdots 4 \cdot 2}{n \cdots 5 \cdot 3}.$$

Step-by-Step Solution

Answer

In this exercise, we first found an expression for the integral of \(\sin^n x\) using integration by parts. We then evaluated the given integral within the specified limits and derived the Wallis sine formulas for both even and odd powers of \(\sin x\). For even powers, the formula is: \[\int_{0}^{\pi/2} \sin^{2k} x dx = \frac{(2k-1)(2k-3) \cdots 1}{(2k)(2k-2) \cdots 2} \cdot \frac{\pi}{2}\] For odd powers, the formula is: \[\int_{0}^{\pi/2} \sin^{2k+1} x dx = \frac{(2k)(2k-2) \cdots 2}{(2k + 1)(2k - 1) \cdots 3}\]

Step 1: Integration by Parts of \(\sin^n x\)

First, let's represent the integral of \(\sin^n x\) using integration by parts: Integration by parts formula: \[\int u dv = uv - \int v du.\] In our case, we take \(u = \sin^{n-1} x\) and \(dv = \sin x dx\). Now we find \(du\) and \(v\): \(du = (n-1) \sin^{n-2} x \cos x dx\) \(v = -\cos x\) Now we apply the formula: \[\int \sin^n x dx = -\sin^{n-1} x \cos x - \int -(n-1) \sin^{n-2} x \cos^2 x dx\] Next, we use the trigonometric identity \(\cos^{2} x = 1 - \sin^{2} x\): \[\int \sin^n x dx = -\sin^{n-1}x \cos x + (n-1) \int \sin^{n-2}x(1-\sin^{2}x)dx\] Separating the integrals, we have: \[\int \sin^n x dx = -\sin^{n-1}x \cos x + (n-1)\int \sin^{n-2}x dx - (n-1) \int \sin^n x dx\] Now, isolate \(\int \sin^n x dx\) as the exercise asks: \[\int \sin^n x dx = -\frac{1}{n} \sin^{n-1} x \cos x + \frac{n-1}{n} \int \sin^{n-2}x dx\]

Step 2: Evaluate the Given Integral

Next, we find the given integral: \[\int_{0}^{\pi/2} \sin^{n'} x dx = \frac{n'-1}{n'} \int_{0}^{\pi/2} \sin^{n'-2} x dx\] Since we are taking limits, we can evaluate: \[-\frac{1}{n'} \left[ \sin^{n'-1} x \cos x \right]_0^{\pi/2} = \frac{n'-1}{n'} \int_{0}^{\pi/2} \sin^{n'-2} x dx\] When \(x = 0\) and \(x = \frac{\pi}{2}\), \(\sin^{n'-1} x \cos x = 0\). Therefore, we have: \[\int_{0}^{\pi/2} \sin^{n'} x dx = \frac{n'-1}{n'} \int_{0}^{\pi/2} \sin^{n'-2} x dx\]

Step 3: Wallis Sine Formulas

Now, we can find the Wallis sine formulas for even \((n = 2k)\) and odd \((n = 2k+1)\) powers of \(\sin x\): Even Powers: By plugging \(n = 2k\) into Step 2, we can integrate multiple times until we get to the base case \(\int_{0}^{\pi/2} \sin x dx\): \[\int_{0}^{\pi/2} \sin^{2k} x dx = \frac{2k-1}{2k} \int_{0}^{\pi/2} \sin^{2k-2} x dx = \frac{(2k-1)(2k-3) \cdots 1}{(2k)(2k-2) \cdots 2} \int_{0}^{\pi/2} \sin^0 x dx\] Since \(\sin^0 x = 1\), we have: \[\int_{0}^{\pi/2} \sin^{2k} x dx = \frac{(2k-1)(2k-3) \cdots 1}{(2k)(2k-2) \cdots 2} \cdot \frac{\pi}{2}\] Odd Powers: Similarly, we can plug \(n = 2k+1\) into Step 2 and integrate multiple times until we get to \(\int_{0}^{\pi/2} \sin x dx\): \[\int_{0}^{\pi/2} \sin^{2k+1} x dx = \frac{2k}{2k + 1} \int_{0}^{\pi/2} \sin^{2k-1} x dx = \frac{(2k)(2k-2) \cdots 2}{(2k + 1)(2k - 1) \cdots 3} \int_{0}^{\pi/2} \sin x dx\] Now, recall that \(\int_{0}^{\pi/2} \sin x dx = 1\), so we find: \[\int_{0}^{\pi/2} \sin^{2k+1} x dx = \frac{(2k)(2k-2) \cdots 2}{(2k + 1)(2k - 1) \cdots 3}\] These are the desired Wallis sine formulas for even and odd powers of \(\sin x\).

Key Concepts

Wallis Sine Formulas

The Wallis Sine Formulas provide a fascinating way to evaluate definite integrals of powers of the sine function. These formulas are particularly useful for both educational and practical purposes in mathematical analysis.
For even values of \[n\], the formula is \[\int_{0}^{\pi / 2} \sin^{n} x \, dx = \frac{(n-1)(n-3)\cdots 5 \cdot 3 \cdot 1}{n(n-2)\cdots 6 \cdot 4 \cdot 2} \cdot \frac{\pi}{2}\]. This formula simplifies the evaluation of definite integration by reducing it to a product of integers and continues until the factors run out.
For odd values of \[n\], the formula becomes \[\int_{0}^{\pi / 2} \sin^{n} x \, dx = \frac{(n-1)(n-3)\cdots 4 \cdot 2}{n(n-2)\cdots 5 \cdot 3}\]. This expression doesn't include the traditional \(\frac{\pi}{2}\) multiplier and is resolved by direct multiplication.
These formulas are derived through repeated application of integration by parts, highlighting how the integral of a power is related to integrals of smaller powers.

Trigonometric Functions Integration

Integrating trigonometric functions, like \(\sin x\) and \(\cos x\), often requires specific techniques due to the cyclic nature and unique properties of these functions.
One approach is the use of trigonometric identities to simplify the integral before applying standard techniques. For example, utilizing \(\cos^{2}x=1-\sin^{2}x\) can help break down more complex expressions into simpler ones.

• Integration by parts is a common technique applied to trigonometric functions, especially when dealing with powers of sine and cosine.
• Sometimes substituting trigonometric functions with other variables, like \(u = \sin x\), helps facilitate easier integration.

Through practice, students can become familiar with evaluating integral expressions involving trigonometric functions by breaking them down, simplifying, and using appropriate methodologies.

Definite Integration Techniques

Definite integration involves calculating the integral of a function within specified limits or boundaries. This process provides valuable information like the total area under a curve between two points.
Techniques for definite integration include:

• Utilization of known results for basic integrals. Understanding integral derivatives of simple functions is always a good starting point.
• Employing symmetry properties of functions, especially for even and odd functions to simplify computations.
• Partial fraction decomposition is useful when dealing with rational functions, after which standard integrative techniques can be applied.

Applying these techniques ensures that even challenging integrative tasks are handled with precision and clarity on the defined intervals.