Problem 64 - Calculus Early Transcendentals

Question

Use integration by parts to obtain a reduction formula for positive integers \(n\) : $$\int \sin ^{n} x d x=-\sin ^{n-1} x \cos x+(n-1) \int \sin ^{n-2} x \cos ^{2} x d x$$ Then use an identity to obtain the reduction formula $$\int \sin ^{n} x d x=-\frac{\sin ^{n-1} x \cos x}{n}+\frac{n-1}{n} \int \sin ^{n-2} x d x$$ Use this reduction formula to evaluate \(\int \sin ^{6} x d x\).

Step-by-Step Solution

Answer

To prove the given reduction formula and evaluate the given integral, we first used integration by parts to find the reduction formula for the integral of sine to the power of n. Then, we applied a trigonometric identity to simplify the result. Lastly, we applied the reduction formula repeatedly to evaluate the integral of sine to the power of 6. The final result is: $$\int \sin^6 x dx = -\frac{\sin^5 x \cos x}{6}-\frac{5\sin^3 x \cos x}{24}-\frac{15\sin x \cos x}{48}+\frac{5x}{8} + C$$

Step 1: Using Integration by Parts

In order to prove the given reduction formula, we can use Integration by Parts where \(u = \sin^{n-1}x\) and \(dv = \sin x dx\). Then, calculate the differential du and the antiderivative v. Using the formula: $$\int u dv = uv - \int v du$$ So, first we calculate \(du\) and \(v\): \begin{align*} du &= (n-1) \sin ^{n-2}(x) \cos(x) dx \\ v &= -\cos(x) \end{align*} Now, using integration by parts: \begin{align*} \int \sin^n x dx &= - \sin^{n-1}(x) \cos(x) + (n-1) \int \sin^{n-2}(x) \cos^2 (x) dx \end{align*}

Step 2: Using an Identity to Simplify

Now we need to apply the trigonometric identity \(\cos^2(x) = 1 - \sin^2(x)\): \begin{align*} \int \sin^n x dx &= -\sin^{n-1} x \cos x + (n-1) \int \sin^{n-2} x (1 - \sin^2(x)) dx \\ &= -\sin^{n-1} x \cos x + (n-1) \int \sin^{n-2} x dx - (n-1) \int \sin^{n} x dx \end{align*} Now, isolate the \(\int\sin^nx dx\) term: $$\int \sin^n x dx = -\frac{\sin^{n-1} x \cos x}{n}+\frac{n-1}{n} \int \sin^{n-2} x dx$$

Step 3: Evaluate the Integral \(\int \sin^6 x dx\)

Now we apply the reduction formula to evaluate the given integral: \begin{align*} \int \sin^6 x dx &= -\frac{\sin^5 x \cos x}{6}+\frac{5}{6} \int \sin^4 x dx \end{align*} Then, let's use the reduction formula again to find the antiderivative of \(\sin^4{x} dx\): \begin{align*} \int \sin^4 x dx &= -\frac{\sin^3 x \cos x}{4}+\frac{3}{4} \int \sin^2 x dx \end{align*} And once more for the antiderivative of \(\sin^2{x} dx\): \begin{align*} \int \sin^2 x dx &= -\frac{\sin x \cos x}{2}+\frac{1}{2} \int dx = -\frac{\sin x \cos x}{2}+\frac{1}{2} x \end{align*} Now substitute the values back: \begin{align*} \int \sin^6 x dx &= -\frac{\sin^5 x \cos x}{6}+\frac{5}{6} \left(-\frac{\sin^3 x \cos x}{4}+\frac{3}{4} \left(-\frac{\sin x \cos x}{2}+\frac{1}{2} x\right)\right) \\ &= -\frac{\sin^5 x \cos x}{6}-\frac{5\sin^3 x \cos x}{24}-\frac{15\sin x \cos x}{48}+\frac{5x}{8}+ C \end{align*} So, the final result is: $$\int \sin^6 x dx = -\frac{\sin^5 x \cos x}{6}-\frac{5\sin^3 x \cos x}{24}-\frac{15\sin x \cos x}{48}+\frac{5x}{8} + C$$

Key Concepts

Reduction Formula

A reduction formula is a powerful tool in calculus, particularly when dealing with repetitive integrals of similar forms. It simplifies the process of finding antiderivatives by reducing the power of the function with each iteration.

For example, if you're faced with integrating a trigonometric function raised to a power, like \( \sin^n x \), a direct approach might be complicated or even infeasible. A reduction formula, however, provides a recursive way to express the integral in terms of the integral of the same function to a lower power. You may notice the structure:\[ \int \sin^n x \, dx = \text{{terms without integrals}} + \text{{coefficient}} \times \int \sin^{n-2} x \, dx \]This pattern continues until you reach a base case that can be easily integrated, such as \( \sin x \) or \( \sin^2 x \).

Trigonometric Identity

A trigonometric identity is an equation that holds true for all values of the variable where both sides of the equation are defined. These identities are crucial for simplifying complex trigonometric expressions and for solving trigonometric equations.

One commonly used identity is \( \cos^2(x) = 1 - \sin^2(x) \), which originates from the Pythagorean identity \( \sin^2(x) + \cos^2(x) = 1 \). In the context of integration, such identities allow us to convert expressions into forms that are easier to integrate. The process usually involves recognizing patterns that match known identities and then substituting to simplify the integral before applying techniques like integration by parts or direct integration.

Antiderivative

An antiderivative, also known as an indefinite integral, of a function \( f(x) \) is a function \( F(x) \) whose derivative is \( f(x) \). In other words, \( F'(x) = f(x) \). The process of finding an antiderivative is called integration, and it plays a central role in calculus because of its relationship with the area under a curve.

Finding the antiderivative of a given function may involve several techniques, such as substitution, integration by parts, partial fraction decomposition, and sometimes, recognizing a pattern that leads to a reduction formula. Because antiderivatives can be non-unique (they differ by a constant), the most general form of an antiderivative includes a constant \( C \) known as the constant of integration.

Calculus

Calculus is a branch of mathematics focused on limits, functions, derivatives, integrals, and infinite series. This field has two major subdivisions — differential calculus and integral calculus. Differential calculus deals with the rate of change and slopes of curves, while integral calculus covers the accumulation of quantities and areas under and between curves.

Integration by parts is one of the many techniques used in calculus to find integrals. It originates from the product rule for differentiation and is usually applied when the integrand is a product of two functions where integration of each function separately is known. The main idea is to transform the integral of a product of functions into a simpler integral that can be evaluated more easily.