Problem 22 - Calculus: One and Several Variables

Question

Calculate. (If you run out of ideas, use the examples as models.) $$\int \cos ^{5} x \sin ^{5} x d x$$.

Step-by-Step Solution

Answer

The short answer is that the integral of \(\cos^{5}(x) \sin^{5}(x) dx\) is \(\frac{\sin^6(x)}{6} - \frac{2\sin^8(x)}{8} + \frac{\sin^{10}(x)}{10} + C\).

Step 1: Let u = sin(x)

To simplify the problem, we will make the substitution \(u = \sin(x)\). Taking the derivative of this substitution, we have \(du = \cos(x)dx\). Now, let's rewrite the integral with this substitution: \(\int \cos^{5}(x) \sin^{5}(x) dx = \int (1 - u^2)^2 u^5 du\) Step 2: Expand the integrand

Step 2: Expand the polynomial

Next, we will expand the polynomial \((1 - u^2)^2\) to simplify the integrand. \((1 - u^2)^2 = 1 - 2u^2 + u^4\) Now, the integral becomes: \(\int (1 - 2u^2 + u^4)u^5 du\) Step 3: Distribute u^5

Step 3: Distribute the u^5

We can now distribute the \(u^5\) to each term inside the parentheses. \(\int (1 - 2u^2 + u^4)u^5 du = \int(u^5 - 2u^7 + u^9) du\) Step 4: Integrate term by term

Step 4: Integrate each term

We can now integrate each term in the polynomial separately. \(\int(u^5 - 2u^7 + u^9) du = \frac{u^6}{6} - \frac{2u^8}{8} + \frac{u^{10}}{10} + C\) Step 5: Back substitute sin(x) for u

Step 5: Substitute back sin(x) for u

Now, let's substitute back sin(x) for u in the integration result to get our final answer. \[\frac{u^6}{6} - \frac{2u^8}{8} + \frac{u^{10}}{10} + C = \frac{\sin^6(x)}{6} - \frac{2\sin^8(x)}{8} + \frac{\sin^{10}(x)}{10} + C\] Thus, the integral of \(\cos^{5}(x) \sin^{5}(x) dx\) is \(\frac{\sin^6(x)}{6} - \frac{2\sin^8(x)}{8} + \frac{\sin^{10}(x)}{10} + C\).

Key Concepts

Trigonometric Substitution

Trigonometric substitution is a technique used in integration to simplify expressions involving trigonometric functions. In the original problem, we used the substitution \(u = \sin(x)\) to help transform the integrand \(\cos^5(x) \sin^5(x)\) into a polynomial form involving \(u\). This technique leverages the basic trigonometric identity \(\sin^2(x) + \cos^2(x) = 1\). By making this substitution,

• the complexity of the integral is reduced,
• allowing easier polynomial manipulation,
• which in turn simplifies the integration process.

Trigonometric substitution is particularly useful when dealing with integrals that involve powers of sine and cosine functions. It's a strategic approach that transforms challenging problems into more manageable algebraic forms. Once the integration is completed in terms of \(u\), the final step involves substituting back the original trigonometric function to express the result in the same terms as the original integral.

Polynomial Expansion

Polynomial expansion is a vital step when integrating complex polynomials. In the exercise, we expanded the expression \((1 - u^2)^2\) as part of transforming the original trigonometric integral. The expanded polynomial is

• \((1 - 2u^2 + u^4)\).

This step breaks down a compounded term into simpler components which can be easily integrated term by term. By writing the polynomial as the sum of individual monomials,

• it becomes straightforward to integrate,
• as each term can be handled separately.

The polynomial expansion simplifies not just the computation but also improves clarity and understanding of the integral's transformation from trigonometric terms to a manageable algebraic expression. This strategy is widely used across calculus to handle a variety of integration challenges.

Definite and Indefinite Integrals

Understanding the difference between definite and indefinite integrals is crucial in calculus. In the given problem, we worked with an indefinite integral, which involves finding the antiderivative without specific bounds.

• The result includes a constant \(C\), representing the family of all antiderivatives.

Indefinite integrals are central to understanding the general behaviour of functions within calculus, especially in situations where the boundary conditions aren't predefined. They provide a general formula which can be later adapted into definite integrals by specifying limits.

• Definite integrals calculate the area under a curve between two points,
• resulting in a fixed value rather than a function.

While this exercise involves indefinite integration, understanding both concepts is key as they complement each other in mathematical problems and real-world applications.

Integration by Parts

Integration by parts is another powerful technique for solving integrals, particularly when the standard method does not lead to a simple solution. The formula used for this technique is based on the product rule for differentiation: \[ \int u \, dv = uv - \int v \, du \]However, in this specific problem, the application wasn't directly necessary as trigonometric substitution and polynomial expansion sufficiently simplified the integral. This method is extremely useful when you have a product of two functions, where one becomes simpler upon differentiation and the other becomes more complex, or stays manageable, when integrated.

• Identify parts \(u\) and \(dv\) from the original integral's product,
• then compute the derivatives and integrals of these parts,
• finally substitute into the integration by parts formula to solve.

Using integration by parts wisely can turn a challenging integral into one composed of simpler pieces, much like using substitution and polynomial expansion together can simplify trigonometric integrals, as done in the problem at hand.