Question

A sine reduction formula Use integration by parts to obtain the 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 $$ \int \sin ^{n} x d x=-\frac{\sin ^{n-1} x \cos x}{n}+\frac{n-1}{n} \int \sin ^{n-2} x d $$ Use this reduction formula to evaluate \(\int \sin ^{6} x d x\)

Short answer

The reduction formula derived for integrating \(\sin^n(x)\) is: $$\int \sin^n(x) dx =-\frac{\sin^{n-1}(x) \cos(x)}{n} + \frac{(n-1)(1 - \sin^2(x))}{n} \int \sin^{n-2}(x) dx$$

Step-by-step solution

Apply Integration by Parts

To start, we need to apply integration by parts on the given formula: \(\int \sin^{n}(x) dx\). Consider \(u = \sin^{n-1}(x)\) and \(dv = \sin(x) dx\), so \( du = (n-1) \sin^{n-2}(x) \cos(x) dx\) and \(v = -\cos(x)\). Using integration by parts formula, \(\int u dv = uv - \int v du\), we get: $$ \int \sin^n(x) dx = -\sin^{n-1}(x) \cos(x) + \int(n-1) \sin^{n-2}(x) \cos^2(x) dx $$

Use a Trigonometric Identity

Now, we need to simplify the integral using a trigonometric identity. Notice that \(\cos^2(x) = 1 - \sin^2(x)\). Therefore, we can write our integral as: $$ \int \sin^{n-2}(x) \cos^2(x) dx = \int \sin^{n-2}(x) (1 - \sin^2(x)) dx $$ Now, let's substitute this back into the formula from Step 1: $$ \int \sin^n(x) dx =-\sin^{n-1}(x) \cos(x) + (n-1) \int \sin^{n-2}(x) (1 - \sin^2(x)) dx $$ Simplify by separating the integrals: $$=-\frac{\sin^{n-1}(x) \cos(x)}{n} + \frac{(n-1)(1 - \sin^2(x))}{n} \int \sin^{n-2}(x) dx$$

Apply the Reduction Formula to Evaluate the Integral

Now we need to use the derived reduction formula to evaluate the integral \(\int \sin^{6}(x) dx\). Let \(n = 6\), so the formula becomes: $$-\frac{\sin^5(x) \cos(x)}{6} + \frac{5}{6} \int \sin^4(x) dx$$ Apply the reduction formula again with \(n = 4\): $$= -\frac{\sin^5(x) \cos(x)}{6} + \frac{5}{6}\left(-\frac{\sin^3(x) \cos(x)}{4} + \frac{3}{4} \int \sin^2(x) dx\right)$$ Apply the reduction formula one last time with \(n = 2\): $$= -\frac{\sin^5(x) \cos(x)}{6} + \frac{5}{6}\left(-\frac{\sin^3(x) \cos(x)}{4} + \frac{3}{4} \left(-\frac{\sin(x) \cos(x)}{2} + \frac{1}{2} \int dx\right)\right)$$ Simplify and integrate the only remaining integral: $$= -\frac{\sin^5(x) \cos(x)}{6} - \frac{5\sin^3(x) \cos(x)}{24} - \frac{15\sin(x) \cos(x)}{48} + \frac{15}{48} \int dx$$ Finally, find the antiderivative: $$= -\frac{\sin^5(x) \cos(x)}{6} - \frac{5\sin^3(x) \cos(x)}{24} - \frac{15\sin(x) \cos(x)}{48} + \frac{15}{48}x + C$$ Thus, the integral \(\int \sin^6(x) dx\) is equal to: $$ \int \sin^6(x) dx = -\frac{\sin^5(x) \cos(x)}{6} - \frac{5\sin^3(x) \cos(x)}{24} - \frac{15\sin(x) \cos(x)}{48} + \frac{15}{48}x + C $$

Key concepts

Integration by Parts

When we come across an integral that is a product of two functions, sometimes we can simplify it using a technique called integration by parts. This method is based on the product rule for differentiation and provides a way to break down complicated integrals into more manageable ones. The formula for integration by parts is \[ \int u dv = uv - \int v du \] where \( u \) is a function of \( x \) and \( dv \) is the derivative of another function \( v \) with respect to \( x \). To apply this method, we choose \( u \) and \( dv \) from the integral such that the derivative of \( u \) and the antiderivative of \( dv \) are simpler to work with.

To effectively use integration by parts, it's essential to recognize which part of the integral to choose as \( u \) and as \( dv \). Typically, we use the LIATE rule (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, and Exponential functions) to decide. This rule suggests a priority order for choosing \( u \), decreasing the likelihood of increasing the integral's complexity. In the exercise provided, \( u = \sin^{n-1}(x) \) and \( dv = \sin(x) dx \), which results in a more straightforward integral after applying the formula.

Trigonometric Identities

Trigonometric identities are equations that are true for all values of the variables involved and involve trigonometric functions. They are essential tools for simplifying trigonometric expressions and solving trigonometric equations. Identities like \( \cos^2(x) + \sin^2(x) = 1 \) can be rearranged to express one function in terms of another, such as \( \cos^2(x) = 1 - \sin^2(x) \), which is the identity used in the exercise solution.

Knowing these identities allows us to transform the integral into a form that might be more amenable to integration, especially when dealing with powers of sine and cosine functions. By substituting \( \cos^2(x) \) with \( 1 - \sin^2(x) \) in the exercise, we were able to reduce the integral into a form that could be integrated using the reduction formula. Without using these identities, the integration process could become significantly more challenging or even unfeasible.

Antiderivatives

An antiderivative of a function \( f(x) \) is another function \( F(x) \) such that \( F'(x) = f(x) \). In other words, the antiderivative is a function that reverses the process of differentiation. When we integrate a function, we're essentially finding its antiderivative. The most general antiderivative of a function is represented as \( F(x) + C \), where \( C \) is the constant of integration, reflecting the fact that there are infinitely many functions that can serve as antiderivatives differing only by a constant.

To solve an integral, we look for an antiderivative that, when differentiated, gives us the original function under the integral sign. In the context of our example exercise, the antiderivatives of \( \sin^{n}x \) are obtained step-by-step using the reduction formula. The final step involves finding the antiderivative of a constant, which is simply \( x \) times the constant, plus the integration constant \( C \). This process highlights the importance of antiderivatives in integration, serving as the foundation for finding the area under the curve of a function or the accumulated quantity represented by the integral.