Tutorial to find integrals involving odd powers of \( \sin(x) \). Exercises with answers are at the bottom of the page. Integrating even powers of sine is generally more complex and requires power-reducing formulas.
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 \( \displaystyle \int \sin^3(x) \, dx \)
Solution to Example 1:
The main idea is to rewrite the power of \( \sin(x) \) as the product of a term with power 1 and a term with an even power:
\[ \sin^3(x) = \sin^2(x) \sin(x) \]Hence, the given integral may be written as:
\[ \int \sin^3(x) \, dx = \int \sin^2(x) \sin(x) \, dx \]Use the trigonometric identity \( \sin^2(x) = 1 - \cos^2(x) \) to write:
\[ \int \sin^3(x) \, dx = \int (1 - \cos^2(x)) \sin(x) \, dx \]Let \( u = \cos(x) \), which gives \( \dfrac{du}{dx} = -\sin(x) \) or \( -du = \sin(x) \, dx \). Substituting into the integral:
\[ \int \sin^3(x) \, dx = - \int (1 - u^2) \, du = \int (u^2 - 1) \, du = \dfrac{1}{3}u^3 - u + C \]Substitute back \( u = \cos(x) \):
\[ \int \sin^3(x) \, dx = \dfrac{1}{3}\cos^3(x) - \cos(x) + C \]Example 2: Evaluate \( \displaystyle \int \sin^5(x) \, dx \)
Solution to Example 2:
Rewrite \( \sin^5(x) \) as \( \sin^4(x) \sin(x) \). Hence, the integral becomes:
\[ \int \sin^5(x) \, dx = \int \sin^4(x) \sin(x) \, dx \]Use the identity \( \sin^2(x) = 1 - \cos^2(x) \) to rewrite \( \sin^4(x) \) in terms of powers of \( \cos(x) \):
\[ \int \sin^5(x) \, dx = \int (1 - \cos^2(x))^2 \sin(x) \, dx \]Let \( u = \cos(x) \), which gives \( du = -\sin(x) \, dx \) or \( -du = \sin(x) \, dx \). Substituting into the integral:
\[ \int \sin^5(x) \, dx = - \int (1 - u^2)^2 \, du \]Expand and calculate the integral on the right:
\[ = - \int (1 - 2u^2 + u^4) \, du = - \left( u - \dfrac{2}{3}u^3 + \dfrac{1}{5}u^5 \right) + C = -\dfrac{1}{5}u^5 + \dfrac{2}{3}u^3 - u + C \]Substitute back \( u = \cos(x) \):
\[ \int \sin^5(x) \, dx = -\left(\dfrac{1}{5}\cos^5(x) - \dfrac{2}{3}\cos^3(x) + \cos(x)\right) + C \]Exercises
Evaluate the following integrals. Click each exercise to reveal the step-by-step solution.
Exercise 1: Evaluate \( \displaystyle \int \sin^7(x) \, dx \)
Solution:
Rewrite as \( \int \sin^6(x)\sin(x) \, dx = \int (1 - \cos^2(x))^3 \sin(x) \, dx \).
Let \( u = \cos(x), du = -\sin(x) \, dx \):
\[ = -\int (1 - 3u^2 + 3u^4 - u^6) \, du = - \left( u - u^3 + \dfrac{3}{5}u^5 - \dfrac{1}{7}u^7 \right) + C \]Substitute back \( u = \cos(x) \):
\[ = \dfrac{1}{7}\cos^7(x) - \dfrac{3}{5}\cos^5(x) + \cos^3(x) - \cos(x) + C \]Exercise 2: Evaluate \( \displaystyle \int \sin^9(x) \, dx \)
Solution:
Rewrite as \( \int \sin^8(x)\sin(x) \, dx = \int (1 - \cos^2(x))^4 \sin(x) \, dx \).
Let \( u = \cos(x), du = -\sin(x) \, dx \):
\[ = -\int (1 - 4u^2 + 6u^4 - 4u^6 + u^8) \, du \] \[ = - \left( u - \dfrac{4}{3}u^3 + \dfrac{6}{5}u^5 - \dfrac{4}{7}u^7 + \dfrac{1}{9}u^9 \right) + C \]Substitute back \( u = \cos(x) \) and rearrange in descending powers:
\[ = -\dfrac{1}{9}\cos^9(x) + \dfrac{4}{7}\cos^7(x) - \dfrac{6}{5}\cos^5(x) + \dfrac{4}{3}\cos^3(x) - \cos(x) + C \]