Question

Indefinite integrals Use a change of variables or Table 5.6 to evaluate the following indefinite integrals. Check your work by differentiating. $$\int\left(\sin ^{5} x+3 \sin ^{3} x-\sin x\right) \cos x \, d x$$

Short answer

Question: Find the indefinite integral of the function \(\int(\sin^5x + 3\sin^3x - \sin x)\cos x \, dx\), and check your work by differentiating the result. Answer: The indefinite integral of the function is \(\frac{1}{6}\sin^6x + \frac{3}{4}\sin^4x - \frac{1}{2}\sin^2x + C\).

Step-by-step solution

Choose the substitution

Let's choose the substitution: \(u = \sin x\).

Calculate du in terms of dx

Now, let's differentiate both sides of the substitution equation with respect to \(x\) to obtain \(du\): $$\frac{du}{dx} = \cos x$$ Then, solve for \(dx\): $$dx = \frac{1}{\cos x}du$$

Substitute u and du into the integral

Replace \(\sin x\) with \(u\) and \(dx\) with \(\frac{1}{\cos x}du\): $$\int (u^5 + 3u^3 - u) \cos x \frac{1}{\cos x}du$$ Now, the \(\cos x\) terms cancel out: $$\int (u^5 + 3u^3 - u) \, du$$

Integrate with respect to u

Now, we can easily integrate the function with respect to \(u\): $$\int u^5 \, du + 3\int u^3 \, du - \int u \, du$$ $$=\frac{1}{6}u^6+\frac{3}{4}u^4-\frac{1}{2}u^2+C$$ Where \(C\) is a constant of integration.

Substitute back in terms of x

Now, let's replace \(u\) back to \(\sin x\) to obtain the antiderivative in terms of \(x\): $$\frac{1}{6}\sin^6x + \frac{3}{4}\sin^4x - \frac{1}{2}\sin^2x + C$$

Check the result by differentiating

Now, to check our work, let's differentiate the antiderivative we found with respect to \(x\): $$\frac{d}{dx}\left(\frac{1}{6}\sin^6x + \frac{3}{4}\sin^4x - \frac{1}{2}\sin^2x + C\right)$$ $$=\left(\sin^5x\cos x + 3\sin^3x\cos x - \sin x \cos x\right)$$ As we can see, this is the same as the given function inside the integral. Therefore, our solution is correct. The indefinite integral is: $$\int\left(\sin ^{5} x+3 \sin ^{3} x-\sin x\right) \cos x \, d x = \frac{1}{6}\sin^6x + \frac{3}{4}\sin^4x - \frac{1}{2}\sin^2x + C$$

Key concepts

Integration by Substitution

Integration by substitution is a powerful technique used to simplify complex integrals. When faced with a function that seems difficult to integrate, we look for a substitution that can change the form of the integral into something more manageable.

In this exercise, the substitution chosen is \( u = \sin x \). This choice is strategic because it transforms a part of the integral involving \( \sin x \) and its derivative \( \cos x \) into a simpler expression. The key is to select a substitution where the differential, \( du \), aligns with existing parts of the integrand, making it easier to integrate.

• First, differentiate the substitution: the derivative of \( \sin x \) with respect to \( x \) is \( \cos x \), giving us \( du = \cos x \, dx \).
• Then, solve for \( dx \) obtaining \( dx = \frac{du}{\cos x} \).
• Substitute both \( u \) and \( du \) into the integral.

This transforms the integral into one purely in terms of \( u \), which is generally easier to solve.

Antiderivative

The antiderivative is essentially the reverse of differentiation.

When you find an antiderivative, you're looking for a function that differentiates to give the integrand. In the given problem, after substitution, the integral becomes:
\[ \int (u^5 + 3u^3 - u) \, du \]

• This can be easily separated into three integrals: \( \int u^5 \, du \), \( 3\int u^3 \, du \), and \( -\int u \, du \).
• Calculate each antiderivative: \( \int u^5 \, du = \frac{1}{6}u^6 \), \( 3 \int u^3 \, du = \frac{3}{4}u^4 \), and \( -\int u \, du = -\frac{1}{2}u^2 \).

Finally, combine these results and add a constant \( C \) to get the general antiderivative in terms of \( u \). Remember, the constant of integration is crucial because it represents an infinite number of possible solutions.

Trigonometric Functions

Trigonometric functions like sine and cosine are integral in many calculus problems.

These functions and their derivatives often appear in problems involving periodic behavior. When integrating expressions involving trigonometric functions, it's important to recognize these patterns and relationships.

• Here, \( \sin x \) was chosen as \( u \), simplifying the integral into powers of \( u \).
• The function \( \cos x \) supports the substitution, as \( du \) involves \( \cos x \, dx \).
• Transforming to \( u \) streamlines the function by eliminating the trigonometric element temporarily.

This method reduces the complexity, allowing easier integration of polynomial expressions – leading back eventually to trigonometric terms upon substituting back in \( x \). These steps illustrate how trigonometric identities and integration techniques interplay to solve integrals.

Check by Differentiation

Checking your work by differentiating, also known as verification, is the last important step.

Once you've found an antiderivative, differentiate it to ensure it matches the original integrand. This helps confirm that the integration process was executed correctly. Here's how you can approach this validation:

• Take the antiderivative found, \( \frac{1}{6}\sin^6x + \frac{3}{4}\sin^4x - \frac{1}{2}\sin^2x + C \), and differentiate it with respect to \( x \).
• Apply the chain rule where necessary. Notice that when differentiating \( \sin nx \), the result employs the \( \cos x \) factor.
• The differentiated expression, \( \sin^5x\cos x + 3\sin^3x\cos x - \sin x \cos x \), should match the original integrand exactly.

By verifying your result through differentiation, you reinforce your solution's accuracy. This step ties together the concepts of integration and differentiation, showcasing their complementary nature.