Question

Compute the following integrals using the guidelines for integrating powers of trigonometric functions. Use a CAS to check the solutions. (Note: Some of the problems may be done using techniques of integration learned previously.) \(\int \sin ^{5} x \cos ^{2} x d x\)

Short answer

The integral is \( -\frac{\cos^7 x}{7} + \frac{2\cos^5 x}{5} - \frac{\cos^3 x}{3} + C \).

Step-by-step solution

Identify Integration Technique

The integral involves powers of sine and cosine functions. Since there is an odd power, apply the strategy of saving one sine factor to use substitution later. This falls under the technique of using trigonometric identities and substitution for integrals of powers of trigonometric functions.

Save a Sine Factor

Rewrite the integral separating one sine factor as \( \int \sin^5 x \cos^2 x \, dx = \int (\sin^4 x)(\sin x \cos^2 x) \, dx \).

Apply Trigonometric Identity

Use the identity \( \sin^2 x = 1 - \cos^2 x \) to rewrite \( \sin^4 x \) as \( (1 - \cos^2 x)^2 \). Now the integral becomes \( \int (1 - \cos^2 x)^2 \sin x \cos^2 x \, dx \).

Use Substitution

Let \( u = \cos x \), thus \( du = -\sin x \, dx \) or \( -du = \sin x \, dx \). The integral now transforms to \( \int (1 - u^2)^2 u^2 (-du) \).

Simplify and Integrate

Distribute the terms: \( -(1 - 2u^2 + u^4)u^2 \) becomes \( -u^2 + 2u^4 - u^6 \). So, integrate \( \int (u^6 - 2u^4 + u^2) \, du \).

Calculate the Integral

Integrate each term: \( \int u^6 \, du = \frac{u^7}{7} \), \( \int (-2u^4) \, du = \frac{-2u^5}{5} \), and \( \int u^2 \, du = \frac{u^3}{3} \). Combine these to get \( -\frac{u^7}{7} + \frac{2u^5}{5} - \frac{u^3}{3} + C \).

Back-Substitute

Replace \( u \) with \( \cos x \) to back-substitute: \( -\frac{\cos^7 x}{7} + \frac{2\cos^5 x}{5} - \frac{\cos^3 x}{3} + C \).

Key concepts

Trigonometric Identities

Trigonometric identities are fundamental tools used in integration, especially when dealing with integrals involving powers of sine and cosine. They help simplify expressions and make integrals more manageable. One of the most commonly used identities is the Pythagorean identity: \[\sin^2 x + \cos^2 x = 1.\]This identity can be rearranged to express either \(\sin^2 x\) or \(\cos^2 x\) in terms of the other function:

• \(\sin^2 x = 1 - \cos^2 x\)
• \(\cos^2 x = 1 - \sin^2 x\)

These rearrangements are particularly useful when we need to convert higher powers of sine or cosine into forms that can be easily integrated.In the problem with \(\int \sin^5 x \cos^2 x \, dx\), utilizing \(\sin^2 x = 1 - \cos^2 x\) allowed us to express \(\sin^4 x\) as \((1 - \cos^2 x)^2\). This conversion set the stage for using substitution later, simplifying the integration process.

Substitution Method in Integration

The substitution method is a critical technique in calculus for solving integrals. It involves changing variables to transform an integral into a more recognizable or solvable form. This method is akin to the reverse of the chain rule for differentiation. Here's how substitution works in the context of our given integral:We aim to simplify the integral \(\int \sin^5 x \cos^2 x \, dx\). After expressing \(\sin^4 x\) using the identity, the next step is to choose a substitution that simplifies the integral significantly. Choosing \(u = \cos x\) simplifies the integral because its differential, \(du = -\sin x \, dx\), matches the remaining sine factor. This transforms the integral into terms of \(u\):\[\int (1 - u^2)^2 u^2 (-du).\]The sign change due to \(-du\) ensures that the limits of integration properly reflect the orientation, maintaining the integral's value. Substitution often simplifies complex trigonometric power integrals into polynomial integrals, which are more straightforward to evaluate.

Integration Techniques

Integration techniques encompass various methods used to solve integrals that may not be straightforward. These methods include, but are not limited to:

• Basic integration involving standard functions
• Integration by parts
• Partial fraction decomposition
• Trigonometric substitution and identities
• Trigonometric powers and reference to identities

For the given exercise, we use the integration of power functions by choosing substitutions and applying trigonometric identities. After substitution, the integral transforms to a polynomial form. The transformed form of \(\int (u^6 - 2u^4 + u^2) \, du\) in the problem requires the application of basic integration techniques for polynomials:

• Integrate the polynomial term by term: \(\int u^n \, du = \frac{u^{n+1}}{n+1}\) for any power \(n\).

Applying this to each of the terms, we achieve a solution that is polynomial in \(u\). Once solved, back-substitution returns the solution to the original variable \(x\).This structured approach showcases how trigonometric identities and substitution together make solving complex integrals more tractable.