Problem 32 - Calculus for Biology and Medicine

Question

Use substitution to evaluate the indefinite integrals. $$ \int \sin ^{3} x \cos x d x $$

Step-by-Step Solution

Answer

The integral of \(\int \sin^3 x \cos x \, dx\) is \(-\frac{(\cos x)^4}{4} + C\).

Step 1: Identify Inner Function

To use substitution, first identify a part of the integrand to substitute with a single variable. Here, we notice how the \(\cos x\) part can be set as \(u\), since its derivative, \(\sin x dx\), appears in the integrand. We will use the substitution method by setting \(u = \cos x\).

Step 2: Differentiate the Inner Function

Differentiate the chosen inner function function, \(u = \cos x\), to find \(dx\) in terms of \(du\). The derivative is \(du = -\sin x \, dx\). From this, we can express \(\sin x \, dx\) as \(-du\).

Step 3: Express the Integral in Terms of u

Substitute \(\cos x\) with \(u\) and \(\sin x \, dx\) with \(-du\) in the integral.This transforms the integral into \(-\int u^{3} \, du\).

Step 4: Integrate with respect to u

Integrate \(-u^3\) with respect to \(u\). The result of the integration will be \(-\frac{u^4}{4} + C\), where \(C\) is the constant of integration.

Step 5: Substitute Back to Original Variable

Replace \(u\) with \(\cos x\) to write the integral in terms of \(x\).The integral becomes \(-\frac{(\cos x)^4}{4} + C\).

Key Concepts

Substitution Method

The substitution method is a powerful tool in calculus used to simplify the process of finding integrals. It allows you to change variables in an integral to make the integration process easier.
This technique is particularly useful when dealing with complex integrals where the integrand has a product of functions, especially when one function's derivative is present elsewhere in the integrand.

• First, identify a part of the integrand that can be substituted with a new variable, usually denoted as "u". This part is often chosen because its derivative is also present in the integral.
• Next, differentiate the chosen substitution to express the derivative in terms of du. This generally aligns with another part of the integral.
• Substitute these expressions into the integral, transforming it into a simpler form that is easier to evaluate.
• After integrating, substitute back the original expression for the variable to return to the original variable of the function.

Utilizing the substitution method helps tackle otherwise challenging integrals by reducing them into more manageable expressions for evaluation.

Trigonometric Functions

Trigonometric functions like sine and cosine are fundamental in calculus, especially within integration. When an integral involves trigonometric expressions, recognizing identities and derivatives of these functions can simplify integration work.
Sine and cosine have neat relationships with each other:

• The derivative of \({\cos x}\) is \({-sin x}\) – a fact used often in substitution.
• The derivative of \(sin x\) is \({cos x}\).

When faced with problems involving powers of these functions, such as \(\sin^3 x \, \cos x dx\), noting these derivatives helps in deciding substitution choices. For instance, setting \(\u = \cos x\) becomes advantageous as \(\d u = -\sin x \, dx\). This simplifies the function down to a polynomial form, easier for integration, leading to a neat closed-form solution.

Integration Techniques

Integration techniques, like substitution, are methods developed to evaluate integrals that are not immediately obvious. These techniques are essential because they provide various pathways to simplify and solve integration problems.

• Substitution is just one of many techniques and is often the go-to method when recognizing a part of the integrand that pairs well with its derivative.
• Another technique is integration by parts, which is helpful when dealing with products of functions.
• Partial fraction decomposition is effective for rational functions where the denominator can be factored.

For each of these techniques, identifying the structure of the integrand is crucial. By mastering these methods, you can tackle a wide range of integral problems, transforming complex expressions into simpler forms that are straightforward to integrate. Through practice, selecting the appropriate technique becomes easier, helping you to solve integrals efficiently.