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 \(\int \sin ^{6} x d x\)

Short answer

Question: Evaluate the integral of sin^6(x) dx. Answer: The integral of sin^6(x) dx is given by \(-\frac{\sin^{5} x \cos x}{6}-\frac{5\sin^{3} x \cos x}{24}-\frac{15\sin x \cos x}{48}+\frac{15x}{48}+C\), where C is the constant of integration.

Step-by-step solution

Apply integration by parts

Let's apply integration by parts to obtain a reduction formula for the integral of sin^n(x) dx. We'll use the integration by parts formula: \(\int u dv = uv - \int v du\) Let \(u=\sin ^{n-1} x\) and \(dv=\sin x d x\). Then, \(du = (n-1) \sin^{n-2} x \cos x dx\) and \(v = -\cos x\). Now, let's apply the integration by parts formula: \(\int \sin ^{n} x d x=-\sin ^{n-1} x \cos x+(n-1) \int \sin ^{n-2} x \cos ^{2} x d x\)

Use an identity to obtain the reduction formula

We can simplify the last integral using the identity \(\cos^2 x = 1-\sin^2 x\): \(\int \sin ^{n} x d x=-\sin ^{n-1} x \cos x+(n-1) \int \sin ^{n-2} x (1-\sin ^{2} x) d x\) Now, separate the last integral: \(\int \sin ^{n} x d x=-\frac{\sin ^{n-1} x \cos x}{n}+\frac{n-1}{n} \int \sin ^{n-2} x d x-\frac{n-1}{n} \int \sin^n x d x\)

Evaluate the given integral of sin^6(x) dx

Using the reduction formula, we can evaluate the given integral of sin^6(x) dx: First some simplifications, set n=6: \(\int \sin ^{6} x d x=-\frac{\sin ^{5} x \cos x}{6}+\frac{5}{6} \int \sin ^{4} x d x\) Now, set n=4: \(\int \sin^{4} x d x=-\frac{\sin^{3} x \cos x}{4}+\frac{3}{4} \int \sin^{2} x d x\) Finally, set n=2: \(\int \sin^{2} x d x=-\frac{\sin x \cos x}{2}+\frac{1}{2} \int d x\) Now plug in these three reduction formulas back into the original equation: \(\int \sin^{6} x d x=-\frac{\sin ^{5} x \cos x}{6}+\frac{5}{6}\left(-\frac{\sin^{3} x \cos x}{4}+\frac{3}{4} \left(-\frac{\sin x \cos x}{2}+\frac{1}{2} x\right)\right)\) Simplify the equation: \(\int \sin^{6} x d x=-\frac{\sin^{5} x \cos x}{6}-\frac{5\sin^{3} x \cos x}{24}-\frac{15\sin x \cos x}{48}+\frac{15x}{48}\) So, the final result is: \(\int \sin^{6} x d x=-\frac{\sin^{5} x \cos x}{6}-\frac{5\sin^{3} x \cos x}{24}-\frac{15\sin x \cos x}{48}+\frac{15x}{48}+C\) where C is the constant of integration.

Key concepts

Reduction Formula

In calculus, a reduction formula is a crucial tool when evaluating complex integrals, especially when dealing with powers of certain functions. The purpose of a reduction formula is to express an integral in terms of another similar integral with a reduced power. This simplification helps us solve the integral using previously known results or by evaluating basic integrals through repeated application.

In the exercise, we are tasked to derive a reduction formula for the integral of \( \int \sin^{n} x \, dx \) using integration by parts. This technique allows us to tackle these kinds of problems by breaking them into simpler integrals. Once we set proper variables for the integration by parts, we end up with a relationship that connects the original integral to one with a lower power. This step-by-step breakdown makes it easier to subsequently solve integrals of higher powers.

Reduction formulas are powerful because they offer a systematic method to tackle integrals of increasing complexity. By decomposing them into simpler, related integrations, we can solve for any power iteratively.

Trigonometric Integrals

Trigonometric Integrals involve the integration of trigonometric functions or their powers. These include functions like sine, cosine, tangent, and their powers, which often occur in various applications of calculus. These integrals may be straightforward in some cases, but they can become quite involved when raised to a power or combined in complex ways.

For example, \( \int \sin^{n} x \, dx \) requires more sophisticated techniques as seen in this exercise. To handle such integrals, methods like substitution, integration by parts, or trigonometric identities might be employed. In particular, for the integral of powers of sine, integration by parts proves handy. By setting variables accordingly, we equate the original integral to another with a reduced power, thereby simplifying the process.

Often, trigonometric identities are also beneficial. For instance, identities such as \( \cos^2 x = 1 - \sin^2 x \) allow us to modify and simplify the problem into a more manageable form. By leveraging such identities alongside calculus techniques, we transform intricate trigonometric integrals into those we already know how to solve.

Calculus Integration Techniques

Calculus offers several integration techniques, each serving to solve different kinds of integrals efficiently. When an integral cannot be solved instantly, knowledge of specific techniques allows us to find an appropriate approach.

**Integration by Parts**: This is derived from the product rule for differentiation and is handy for integrals of functions that are products of other functions. The formula is given by:

• \( \int u \, dv = uv - \int v \, du \)

This exercise demonstrates how integration by parts aids in developing reduction formulas.

**Substitution**: Commonly used for integrals involving a function and its derivative. It simplifies the integral by changing the variable of integration.

• \( \int f(g(x))g'(x)dx = \int f(u)du \)

**Trigonometric Identities**: Utilized for modifying and simplifying trigonometric integrals. These identities provide alternate expressions to make integrals more approachable.

Employing these diverse calculus integration techniques allows us to solve a wide array of integrals, turning complex integrals into simpler, more manageable forms.