Tutorial to find integrals involving the product of powers of \( \sin(x) \) and \( \cos(x) \) with one of the two having an odd power. Examples and exercises with solutions are included.
In what follows, \( C \) represents the constant of integration.
Examples with Detailed Solutions
Click on each example to view its detailed step-by-step solution.
Example 1: Evaluate \( \displaystyle \int \sin^3(x) \cos^2(x) \, dx \)
Solution to Example 1:
The main idea is to rewrite the integral by separating the term with the odd power into the product of an even power and a single power of 1:
\[ \sin^3(x) = \sin^2(x) \sin(x) \]Hence, the given integral may be written as:
\[ \int \sin^3(x) \cos^2(x) \, dx = \int \sin^2(x) \cos^2(x) \sin(x) \, dx \]We now use the Pythagorean identity \( \sin^2(x) = 1 - \cos^2(x) \) and substitute:
\[ \int \sin^3(x) \cos^2(x) \, dx = \int (1 - \cos^2(x)) \cos^2(x) \sin(x) \, dx \]Let \( u = \cos(x) \), which gives \( \dfrac{du}{dx} = -\sin(x) \) or \( -du = \sin(x) \, dx \). Substituting this into the integral:
\[ = -\int (1 - u^2) u^2 \, du \]Expand and calculate the integral:
\[ = \int (u^4 - u^2) \, du = \dfrac{1}{5}u^5 - \dfrac{1}{3}u^3 + C \]Substitute back \( u = \cos(x) \):
\[ \int \sin^3(x) \cos^2(x) \, dx = \dfrac{1}{5}\cos^5(x) - \dfrac{1}{3}\cos^3(x) + C \]Example 2: Evaluate \( \displaystyle \int \sin^{12}(x) \cos^5(x) \, dx \)
Solution to Example 2:
Rewrite \( \cos^5(x) \) as \( \cos^4(x) \cos(x) \):
\[ \int \sin^{12}(x) \cos^5(x) \, dx = \int \sin^{12}(x) \cos^4(x) \cos(x) \, dx \]Use the identity \( \cos^2(x) = 1 - \sin^2(x) \) to rewrite \( \cos^4(x) \) in terms of powers of \( \sin(x) \):
\[ = \int \sin^{12}(x) (1 - \sin^2(x))^2 \cos(x) \, dx \]Let \( u = \sin(x) \), which gives \( du = \cos(x) \, dx \). Substituting into the integral:
\[ = \int u^{12} (1 - u^2)^2 \, du \]Expand the integrand:
\[ = \int u^{12} (1 - 2u^2 + u^4) \, du = \int (u^{12} - 2u^{14} + u^{16}) \, du \]Integrate term by term:
\[ = \dfrac{1}{13}u^{13} - \dfrac{2}{15}u^{15} + \dfrac{1}{17}u^{17} + C \]Substitute back \( u = \sin(x) \):
\[ \int \sin^{12}(x) \cos^5(x) \, dx = \dfrac{1}{17}\sin^{17}(x) - \dfrac{2}{15}\sin^{15}(x) + \dfrac{1}{13}\sin^{13}(x) + C \]Exercises
Evaluate the following integrals. Click each exercise to reveal the step-by-step solution.
Exercise 1: Evaluate \( \displaystyle \int \cos^3(x) \sin^2(x) \, dx \)
Solution:
Rewrite \( \cos^3(x) = \cos^2(x)\cos(x) = (1 - \sin^2(x))\cos(x) \):
\[ \int \cos^3(x) \sin^2(x) \, dx = \int (1 - \sin^2(x))\sin^2(x) \cos(x) \, dx \]Let \( u = \sin(x) \), \( du = \cos(x) \, dx \):
\[ = \int (1 - u^2)u^2 \, du = \int (u^2 - u^4) \, du = \dfrac{1}{3}u^3 - \dfrac{1}{5}u^5 + C \]Substitute back \( u = \sin(x) \):
\[ = \dfrac{1}{3}\sin^3(x) - \dfrac{1}{5}\sin^5(x) + C \]Exercise 2: Evaluate \( \displaystyle \int \sin^3(x) \cos^{14}(x) \, dx \)
Solution:
Rewrite \( \sin^3(x) = \sin^2(x)\sin(x) = (1 - \cos^2(x))\sin(x) \):
\[ \int \sin^3(x) \cos^{14}(x) \, dx = \int (1 - \cos^2(x))\cos^{14}(x) \sin(x) \, dx \]Let \( u = \cos(x) \), \( du = -\sin(x) \, dx \implies -du = \sin(x) \, dx \):
\[ = -\int (1 - u^2)u^{14} \, du = \int (u^{16} - u^{14}) \, du = \dfrac{1}{17}u^{17} - \dfrac{1}{15}u^{15} + C \]Substitute back \( u = \cos(x) \):
\[ = \dfrac{1}{17}\cos^{17}(x) - \dfrac{1}{15}\cos^{15}(x) + C \]