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Chapter 7: Problem 16

Integrals of \(\sin x\) and \(\cos x\) Evaluate the following integrals. $$\int \sin ^{3} x \cos ^{5} x d x$$

Short Answer

Expert verified
Question: Calculate the integral \(\int \sin^{3}x \cos^{5}x dx\). Answer: \(\frac{1}{4}\sin^4 x - \frac{1}{3}\sin^6 x + \frac{1}{8}\sin^8 x + C\)

Step by step solution

01

Identify the substitution

Observe that the integral involves powers of sine and cosine functions. In order to simplify it, we can use the substitution \(u = \sin x\), so its differential, \(\frac{du}{dx} = \cos x\). Consequently, \(dx = \frac{1}{\cos x} du\).
02

Rewrite the integral in terms of u

Now, we substitute \(u\) and \(dx\) into the integral expression: $$\int \sin ^{3} x \cos ^{5} x d x = \int u^3 \cos^4 x (\cos x) \frac{1}{\cos x} du$$ Since \(u = \sin x\), we have \(\cos^2 x = 1 - u^2\). So, we can write $$\int u^3 \cos ^{5} x d x = \int u^3 (1-u^2)^2 du$$
03

Expand the expression

Expanding the expression inside the integral, we get $$\int u^3 (1-u^2)^2 du = \int u^3 (1 - 2u^2 + u^4) du$$
04

Distribute the u^3 term

Distribute the \(u^3\) term across each term of the expression: $$\int u^3 (1 - 2u^2 + u^4) du = \int (u^3 - 2u^5 + u^7) du$$
05

Integrate term by term

Now, integrate each term of the expression with respect to u: $$\int (u^3 - 2u^5 + u^7) du = \frac{1}{4}u^4 - \frac{1}{3}u^6 + \frac{1}{8}u^8 + C$$
06

Replace u with the original function

Finally, substitute back \(u = \sin x\) and simplify to obtain the solution: $$\frac{1}{4}\sin^4 x - \frac{1}{3}\sin^6 x + \frac{1}{8}\sin^8 x + C$$ Thus, the integral \(\int \sin^{3}x \cos^{5}x dx\) is equal to \(\frac{1}{4}\sin^4 x - \frac{1}{3}\sin^6 x + \frac{1}{8}\sin^8 x + C\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

U-Substitution
The technique of u-substitution is a fundamental method in calculus for finding the antiderivatives of more complex functions. It's essentially the reverse process of the chain rule for differentiation. With u-substitution, we select a portion of the integral's integrand, typically the most complicated part, and substitute it with a single variable, usually u, to simplify the integral.
  • Identify the 'inner function' within the integrand which, when differentiated, appears else where in the integrand.
  • Substitute that inner function with u and find the differential du based on the chosen u.
  • Rewrite the entire integral in terms of u, substituting in for the original variable and its differential.
  • Integrate with respect to u, which should be simpler than the original integral.
  • Finally, substitute the original function back in place of u to get the answer in terms of the original variable.
Improving understanding of u-substitution involves recognizing patterns and practicing with various functions. It is often used in combination with other integration techniques, particularly when integrating products of trigonometric functions with powers, as shown in the exercise.
Integrals of Trigonometric Functions
Trigonometric functions like sine and cosine often appear within integrals, presenting a unique challenge. The periodic nature of trigonometric functions means that their integrals also exhibit patterns.
Integrating trigonometric functions typically involves using identities to simplify the integrand to a form that is easier to integrate. When faced with powers of sine and cosine:
  • Use trigonometric identities, such as the Pythagorean identity, to express one function in terms of the other if necessary.
  • If the powers of sine and cosine are odd, save one sine or cosine term and use the Pythagorean identity to convert the remaining factors into the other trigonometric function.
  • If the powers are even, use power-reducing or half-angle formulas to simplify the expression before integrating.
In our exercise, we used the identity \( \[\cos^2(x) = 1 - \sin^2(x)\] \) to rewrite the integral in a simpler form using u-substitution.
Power of Sine and Cosine
When integrating powers of sine and cosine, the strategy often depends on whether the power is odd or even.
  • For odd powers, we can pull out one sine or cosine factor and convert the remaining even powers using the Pythagorean identity.
  • For even powers, trigonometric identities such as power-reduction formulas are handy to reduce the power of the trigonometric function before integrating.
In the given exercise, we see an integration involving both odd powers of sine and cosine. There we've put aside a single \( \[\cos(x)\] \) term, and then used the Pythagorean identity to express the remaining even power as a polynomial in terms of sine, which we substituted with u. This approach is particularly useful because it turns a difficult trigonometric integral into a simple polynomial integral.
Indefinite Integrals
Indefinite integrals represent the family of all antiderivatives of a function. They don't have specified limits of integration and include an arbitrary constant, C, since the derivative of a constant is zero. The indefinite integral of a function f(x) is denoted as \( \[\int f(x)\,dx\] \) and it represents all functions F(x) such that F'(x) = f(x).
  • An antiderivative is found through integration techniques such as u-substitution.
  • The result of an indefinite integral includes a constant of integration, symbolized by C.
  • Finding the indefinite integral involves reversing the process of differentiation.
For example, in our presented exercise, after integrating term by term with respect to u, we obtain an antiderivative in terms of u. By replacing u with the original trigonometric function, we get the antiderivatives in terms of x and add the constant of integration C, thus obtaining the indefinite integral of the original function.

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Most popular questions from this chapter

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