Question

Define \(J(m, n)\), for non-negative integers \(m\) and \(n\), by the integral $$ J(m, n)=\int_{0}^{\pi / 2} \cos ^{w} \theta \sin ^{n} \theta d \theta $$ (a) Evaluate \(J(0,0), J(0,1), J(1,0), J(1,1), J(m, 1), J(1, n)\) (b) Using integration by parts prove that, for \(m\) and \(n\) both \(>0\), $$ J(m, n)=\frac{m-1}{m+n} J(m-2, n) \quad \text { and } \quad J(m, n)=\frac{n-1}{m+n} J(m, n-2) $$ (c) Evaluate (i) \(J(5,3)\), (ii) \(J(6,5)\), (iii) \(J(4,8)\).

Short answer

J(5,3) = \frac{1}{24}, J(6,5)=\frac{5}{693}, J(4,8)=\frac{1}{360}

Step-by-step solution

Understanding the Integral

The integral provided is \[\begin{equation} J(m, n) = \int_{0}^{\pi / 2} \cos ^{m} \theta \sin ^{n} \theta \, d \theta \end{equation}\]. We need to evaluate this integral for different values of m and n.

Evaluate J(0,0)

Substitute m=0 and n=0 into the integral: \[\begin{equation} J(0,0) = \int_{0}^{\pi / 2} \cos^{0} \theta \sin^{0} \theta \, d \theta = \int_{0}^{\pi / 2} d \theta \end{equation}\]. This integrates to \[\begin{equation} \theta \bigg|_{0}^{\pi / 2} = \frac{\pi}{2} \end{equation}\]. So, \[\begin{equation} J(0,0) = \frac{\pi}{2} \end{equation}\].

Evaluate J(0,1)

Substitute m=0 and n=1 into the integral: \[\begin{equation} J(0,1) = \int_{0}^{\pi / 2} \sin \theta \, d \theta \end{equation}\]. This integrates to \[\begin{equation} -\cos \theta \bigg|_{0}^{\pi / 2} = 1 \end{equation}\]. So, \[\begin{equation} J(0,1) = 1 \end{equation}\].

Evaluate J(1,0)

Substitute m=1 and n=0 into the integral: \[\begin{equation} J(1,0) = \int_{0}^{\pi / 2} \cos \theta \, d \theta \end{equation}\]. This integrates to \[\begin{equation} \sin \theta \bigg|_{0}^{\pi / 2} = 1 \end{equation}\]. So, \[\begin{equation} J(1,0) = 1 \end{equation}\].

Evaluate J(1,1)

Substitute m=1 and n=1 into the integral: \[\begin{equation} J(1,1) = \int_{0}^{\pi / 2} \cos \theta \sin \theta \, d \theta \end{equation}\]. Use substitution \[\begin{equation} u = \sin \theta, \, du = \cos \theta \, d\theta \end{equation}\]. The integral becomes \[\begin{equation} \int_{0}^{1} u \, du = \frac{u^2}{2} \bigg|_{0}^{1} = \frac{1}{2} \end{equation}\]. So, \[\begin{equation} J(1,1) = \frac{1}{2} \end{equation}\].

General formula for J(m, 1)

Substitute n=1 into the integral: \[\begin{equation} J(m, 1) = \int_{0}^{\pi / 2} \cos ^{m} \theta \sin \theta \, d \theta \end{equation}\]. Use substitution \[\begin{equation} u = \cos \theta, \, du = -\sin \theta \, d\theta \end{equation}\]. The integral becomes \[\begin{equation} -\int_{1}^{0} u^{m} (-du) = \int_{0}^{1} u^{m} \, du \end{equation}\], which is \[\begin{equation} \frac{u^{m+1}}{m+1} \bigg|_{0}^{1} = \frac{1}{m+1} \end{equation}\]. So, \[\begin{equation} J(m, 1) = \frac{1}{m+1} \end{equation}\].

General formula for J(1, n)

Substitute m=1 into the integral: \[\begin{equation} J(1, n) = \int_{0}^{\pi / 2} \cos \theta \sin^{n} \theta \, d \theta \end{equation}\]. Substitute \[\begin{equation} u = \sin \theta, \, du = \cos \theta \, d\theta \end{equation}\]. The integral becomes \[\begin{equation} \int_{0}^{1} u^{n} \, du = \frac{u^{n+1}}{n+1} \bigg|_{0}^{1} = \frac{1}{n+1} \end{equation}\]. So, \[\begin{equation} J(1, n) = \frac{1}{n+1} \end{equation}\].

Integration by Parts for J(m,n)

To prove \[\begin{equation} J(m, n) = \frac{m-1}{m+n} J(m-2, n) \end{equation}\], integration by parts is used. Let \[\begin{equation} u = \cos^{m-1} \theta, \, dv = \cos \theta \sin^n \theta \, d\theta \end{equation}\]. Then, \[\begin{equation} du = -(m-1) \cos^{m-2} \theta \sin \theta \, d\theta, \, v = \frac{\sin^{n+1} \theta}{n+1} \end{equation}\]. By parts: \[\begin{equation} \int_{0}^{\pi / 2} \cos^m \theta \sin^n \theta d\theta = \cos^{m-1} \theta \frac{\sin^{n+1} \theta}{n+1} \bigg|_{0}^{\pi/2} + (m-1) \int_{0}^{\pi / 2} \cos^{m-2} \theta \sin^{n+2} \theta \frac{d\theta}{n+1} = \frac{(m-1)}{m+n} J(m-2, n) \end{equation}\]. Similarly, \[\begin{equation} J(m, n) = \frac{n-1}{m+n} J(m, n-2) \end{equation}\].

Calculate J(5,3)

Using the formula \[\begin{equation} J(m, n) = \frac{m-1}{m+n} J(m-2, n) \end{equation}\]: \[\begin{equation} J(5,3) = \frac{4}{8} J(3, 3) = \frac{1}{2} \cdot \frac{2}{6} J(1, 3) = \frac{1}{6} \cdot \frac{1}{4} = \frac{1}{24} \end{equation}\].

Calculate J(6,5)

Using the formula \[\begin{equation} J(m, n) = \frac{m-1}{m+n} J(m-2, n) \end{equation}\]: \[\begin{equation} J(6,5) = \frac{5}{11} J(4, 5) = \frac{5}{11} \cdot \frac{3}{9} J(2, 5) = \frac{5}{11} \cdot \frac{1}{3} \cdot \frac{2}{7} J(0, 5) = \frac{5}{11} \cdot \frac{1}{3} \cdot \frac{2}{7} \cdot \frac{1}{6} = \frac{10}{1386} = \frac{5}{693} \end{equation}\].

Calculate J(4,8)

Using the formula \[\begin{equation} J(m, n) = \frac{m-1}{m+n} J(m-2, n) \end{equation}\]: \[\begin{equation} J(4,8) = \frac{3}{12} J(2, 8) = \frac{1}{4} \cdot \frac{1}{10} J(0, 8) = \frac{1}{4} \cdot \frac{1}{10} \cdot \frac{1}{9} = \frac{1}{360} \end{equation}\].

Key concepts

Integral Calculus

Integral calculus is a subfield of calculus concerned with the concept of the integral. It plays a significant role in finding quantities such as areas, volumes, and many other values related to the accumulation of quantities. In this exercise, we are dealing with a specific type of integral involving trigonometric functions.

Trigonometric Integrals

Trigonometric integrals involve the integration of products of trigonometric functions. These integrals can often be simplified using various techniques such as substitutions and trigonometric identities. For example, the integral given in the exercise is \[J(m, n) = \int_{0}^{\pi / 2} \cos^{m} \theta \sin^{n} \theta \, d \theta\]. This type of integral can be solved by using substitution methods that make the integration process easier.

Integration by Parts

Integration by parts is a technique based on the product rule for differentiation. It's particularly useful when dealing with the product of two functions. The formula is given by \[\int u \, dv = uv - \int v \, du\]. In the given exercise, we can use this technique to derive a recursive formula for \ J(m, n)\. For instance, if we let \ u = \cos^{m-1}\theta\ and \ dv = \cos\theta \sin^{n}\theta d\theta, we can find a relation that connects integrals with different powers, helping us solve complex trigonometric integrals step by step.

Special Functions

Special functions are particular mathematical functions which have more specific definitions than general functions. In this exercise, the Beta function can be directly related to the given integrals. The Beta function, B(x, y), is defined as \[B(x, y) = \int_{0}^{1} t^{x-1} (1-t)^{y-1} dt\]. The integral in our problem has similar properties and can often be converted or directly related to these special functions, thus making it easier to compute.