Question

For the following exercises, find a general formula for the integrals. \(\int \sin ^{2} a x \cos a x d x\)

Short answer

The integral is \( \frac{\sin(ax)}{4a} - \frac{\sin(3ax)}{12} + C \).

Step-by-step solution

Apply a Trigonometric Identity

To simplify the integral \( \int \sin^2(ax) \cos(ax) \, dx \), we start by applying the trigonometric identity \( \sin^2(x) = \frac{1 - \cos(2x)}{2} \). This gives us \( \sin^2(ax) = \frac{1 - \cos(2ax)}{2} \). Substitute this into the integral, yielding: \[ \int \frac{1 - \cos(2ax)}{2} \cos(ax) \, dx. \]

Simplify the Integral

Distribute \( \cos(ax) \) across the numerator: \[ \int \left( \frac{1}{2} \cos(ax) - \frac{1}{2} \cos(2ax) \cos(ax) \right) dx. \] Split the integral as follows: \[ \int \frac{1}{2} \cos(ax) \, dx - \frac{1}{2} \int \cos(2ax) \cos(ax) \, dx. \]

Solve the First Integral

The first integral is \( \int \frac{1}{2} \cos(ax) \, dx \). Since the integral of \( \cos(ax) \) is \( \frac{\sin(ax)}{a} \), the first integral becomes \[ \frac{1}{2} \times \frac{\sin(ax)}{a} = \frac{\sin(ax)}{2a}. \]

Apply Product-to-Sum Formula

For the second integral \( \int \cos(2ax) \cos(ax) \, dx \), apply the product-to-sum formula: \( \cos(A) \cos(B) = \frac{1}{2} \left( \cos(A - B) + \cos(A + B) \right) \). Here, \( A = 2ax \) and \( B = ax \), so \[ \cos(2ax) \cos(ax) = \frac{1}{2}(\cos(ax) + \cos(3ax)). \] This results in the integral: \[ \frac{1}{2} \int \cos(ax) \, dx + \frac{1}{2} \int \cos(3ax) \, dx. \]

Solve Each Part of the Second Integral

Solve \( \frac{1}{2} \int \cos(ax) dx \) as \( \frac{\sin(ax)}{2a} \), and \( \frac{1}{2} \int \cos(3ax) dx \) as \( \frac{1}{6} \sin(3ax) \). Therefore, \[ \frac{1}{2} \int \cos(ax) \, dx + \frac{1}{2} \int \cos(3ax) \, dx = \frac{\sin(ax)}{2a} + \frac{\sin(3ax)}{6}. \]

Combine All Parts

Combine the solved integrals together. From Step 3, we have \( \frac{\sin(ax)}{2a} \), and from Step 5, we have \[ -\frac{1}{2} \left( \frac{\sin(ax)}{2a} + \frac{\sin(3ax)}{6} \right). \] So, the final integral expression is \[ \frac{\sin(ax)}{2a} - \frac{1}{4a} \sin(ax) - \frac{1}{12} \sin(3ax) + C. \] Collect like terms for a simplified solution.

Key concepts

Trigonometric Identities

Trigonometric identities are essential tools in simplifying complex integrals and expressions. In calculus, these identities help transform difficult trigonometric expressions into simpler forms. To solve the integral \( \int \sin^2(ax) \cos(ax) \, dx \), we use the identity \( \sin^2(x) = \frac{1 - \cos(2x)}{2} \). This identity allows us to express \( \sin^2(ax) \) in terms of \( \cos(2ax) \), making the integral more manageable. By substituting \( \sin^2(ax) \) with its equivalent expression, we transform the integral into \( \int \frac{1 - \cos(2ax)}{2} \cos(ax) \, dx \), simplifying the problem significantly. Trigonometric identities like this one are vital for reducing the complexity of integrals, enabling us to solve them with the use of other calculus techniques.

Product-to-Sum Formula

The product-to-sum formula is another valuable strategy when integrating functions involving products of trigonometric terms. It converts products of sines and cosines into sums, which are often easier to integrate. In this exercise, we encounter the integral \( \int \cos(2ax) \cos(ax) \, dx \). To handle this, we apply the product-to-sum formula: \( \cos(A)\cos(B) = \frac{1}{2} \left( \cos(A-B) + \cos(A+B) \right) \).For our specific integral, \( A = 2ax \) and \( B = ax \), resulting in \( \cos(2ax)\cos(ax) = \frac{1}{2}(\cos(ax) + \cos(3ax)) \). By substituting this back into the integral, it becomes easier to split into two simpler integrals: \( \frac{1}{2} \int \cos(ax) \, dx \) and \( \frac{1}{2} \int \cos(3ax) \, dx \).This step is crucial as it breaks down the complex product into manageable parts, each of which can be integrated using basic trigonometric integro-differential formulae.

Integration Techniques

Integration techniques are the various strategies used to tackle integrals that are not immediately solvable by basic formulas. In this exercise, we deployed several methods such as substitution and the use of trigonometric identities to simplify the problem. Starting with \( \int \frac{1}{2} \cos(ax) \, dx \), we recognize a standard integral. The antiderivative of \( \cos(ax) \) is \( \frac{\sin(ax)}{a} \), so multiplying by the constant \( \frac{1}{2} \) results in \( \frac{\sin(ax)}{2a} \).For \( \int \cos(3ax) \, dx \), the process is similar, with an adjustment for the different frequency inside the cosine function, yielding \( \frac{1}{6} \sin(3ax) \).Combining these results allows us to simplify and consolidate the terms, resulting in a final expression for the integral: \( \frac{\sin(ax)}{2a} - \frac{1}{4a} \sin(ax) - \frac{1}{12} \sin(3ax) + C \). These methodologies emphasize the power of diverse integration strategies in solving complex trigonometric integrals effectively.