Problem 64 - Calculus Early Transcendentals

Question

Use a change of variables to evaluate the following integrals. $$\int\left(\sin ^{5} x+3 \sin ^{3} x-\sin x\right) \cos x d x$$

Step-by-Step Solution

Answer

The integral of \(\left(\sin ^{5} x+3 \sin ^{3} x-\sin x\right) \cos x \, dx\) is \(\frac{1}{6}\sin^{6}x+\frac{3}{4}\sin^{4}x-\frac{1}{2}\sin^{2}x+C\), where \(C\) is the constant of integration.

Step 1: Substitution

Let \(u = \sin x\). Now, let's find the derivative of \(u\) with respect to \(x\). $$ \frac{d u}{d x} = \frac{d}{d x}(\sin x) = \cos x $$ Now, express the differential \(du\) in terms of \(dx\): $$ du = \cos x dx $$ Now we can rewrite the given integral in terms of \(u\): $$ \int\left(\sin ^{5} x+3 \sin ^{3} x-\sin x\right) \cos x d x =\int\left(u^{5}+3u^{3}-u\right) du $$

Step 2: Integrate with respect to \(u\)

Now, we have to integrate the expression in terms of \(u\): $$ \int\left(u^{5}+3u^{3}-u\right) du = \frac{1}{6}u^{6}+\frac{3}{4}u^{4}-\frac{1}{2}u^{2}+C $$ Here, \(C\) represents the constant of integration.

Step 3: Resubstitute \(u\) with \(\sin x\)

Finally, resubstitute \(u\) with \(\sin x\) to find the solution in terms of \(x\): $$ \frac{1}{6}u^{6}+\frac{3}{4}u^{4}-\frac{1}{2}u^{2}+C = \frac{1}{6}\sin^{6}x+\frac{3}{4}\sin^{4}x-\frac{1}{2}\sin^{2}x+C $$ Therefore, the evaluated integral is $$ \int\left(\sin ^{5} x+3 \sin ^{3} x-\sin x\right) \cos x d x = \frac{1}{6}\sin^{6}x+\frac{3}{4}\sin^{4}x-\frac{1}{2}\sin^{2}x+C $$

Key Concepts

Trigonometric Integrals

Trigonometric integrals involve integrating functions that are trigonometric in nature. This often includes expressions composed of sine, cosine, tangent, and other trigonometric functions.

• Key Idea: These integrals are treated using methods that exploit trigonometric identities or substitutions that simplify the expressions.

For the given integral \( \int(\sin^5 x + 3 \sin^3 x - \sin x) \cos x \, dx \), recognizing the pattern helps in effective substitution.
This specific integral is suitable for substitution because of the presence of \( \cos x \, dx \), which pairs well with functions of \( \sin x \). Thus, by recognizing this pairing with the derivative \( \cos x \), substitution becomes straightforward.
Using trigonometric identities can also simplify similar expressions before performing the integral, which is useful in other types of trigonometric integrals.

Change of Variables

Change of variables, often called substitution, is a method used in calculus to simplify the process of integration.

• Basic Concept: It involves substituting a part of the integrand with a new variable, thereby simplifying the integral into a basic form.

In our example, we used substitution by letting \( u = \sin x \). This means that the derivative \( \frac{du}{dx} = \cos x \), leading to the differential \( du = \cos x \, dx \).
This substitution was chosen because \( \cos x \, dx \) was part of the original integral. Changing the integrand based on this substitution simplified the equation significantly, transforming it into a polynomial form involving \( u \):\[ \int(u^{5} + 3u^{3} - u) \, du \].
Overall, substitution transforms complex integrals into easier ones by using the function and its derivative in the integration process.

Integral Calculus

Integral calculus is a branch of calculus focused on finding the accumulation of quantities, which is expressed as integrals. It involves definite and indefinite integrals, serving to measure areas, volumes, and other concepts that benefit from the summing of infinitesimally small items.

• Indefinite Integrals: These integrals do not have bounds and result in a general form solution with an integration constant \( C \). The solution to our exercise is an indefinite integral.
• Definite Integrals: In contrast, these have specific limits and calculate the area under a curve over a defined interval.

The method of evaluating integrals, such as we did with substitution, is a core tool in integral calculus. This exercise demonstrated how integral calculus is applied practically to solve problems containing trigonometric functions.
Once the substitution simplified the integral into a straightforward polynomial in \( u \), we used basic integral rules to evaluate it:\[ \int (u^{5} + 3u^{3} - u) \, du = \frac{1}{6} u^{6} + \frac{3}{4} u^{4} - \frac{1}{2} u^{2} + C \].
Finally, resubstituting \( u \) back to \( \sin x \) completed the solution in terms of the original variable.