Tutorial to find integrals involving even powers of \( \sin(x) \), using reducing power formulas, are presented. Exercises with answers are at the bottom of the page. Integrating odd powers of sine is much simpler.
In what follows, \( C \) represents the constant of integration.
Review Power Reducing Formulas
For \( n \) a positive integer, we have the following formulas in trigonometry that may be used to reduce the power of \( \sin(x) \):
- (a) \( \sin^2(x) = \dfrac{1}{2}(1 - \cos(2x)) \)
- (b) \( \sin^3(x) = \dfrac{1}{4}(3\sin(x) - \sin(3x)) \)
- (c) \( \sin^4(x) = \dfrac{1}{8}(3 - 4\cos(2x) + \cos(4x)) \)
- (d) \( \sin^5(x) = \dfrac{1}{16}(\sin(5x) - 5\sin(3x) + 10\sin(x)) \)
- (e) \( \sin^6(x) = \dfrac{1}{32}(10 - 15\cos(2x) + 6\cos(4x) - \cos(6x)) \)
Examples with Detailed Solutions
Click on each example to view its detailed step-by-step solution.
Example 1: Evaluate \( \displaystyle \int \sin^2(x) \, dx \)
Solution to Example 1:
The main idea is to use the identity \( \sin^2(x) = \dfrac{1}{2}(1 - \cos(2x)) \) to reduce the power and make the integral easily evaluated:
\[ \int \sin^2(x) \, dx = \int \dfrac{1}{2}(1 - \cos(2x)) \, dx \] \[ = \dfrac{1}{2}\int 1 \, dx - \dfrac{1}{2}\int \cos(2x) \, dx \] \[ = \dfrac{1}{2}x - \dfrac{1}{4}\sin(2x) + C \]Example 2: Evaluate \( \displaystyle \int [\sin^2(x) - 16\sin^6(x)] \, dx \)
Solution to Example 2:
Use the power reducing formulas to rewrite the integral:
\[ \int [\sin^2(x) - 16\sin^6(x)] \, dx \] \[ = \int \left[\dfrac{1}{2}(1 - \cos(2x)) - 16\left(\dfrac{1}{32}\right)(10 - 15\cos(2x) + 6\cos(4x) - \cos(6x))\right] \, dx \] \[ = \dfrac{1}{2} \int [1 - \cos(2x) - 10 + 15\cos(2x) - 6\cos(4x) + \cos(6x)] \, dx \] \[ = \dfrac{1}{2} \int [-9 + 14\cos(2x) - 6\cos(4x) + \cos(6x)] \, dx \] \[ = \dfrac{1}{2} \left( -9x + 7\sin(2x) - \dfrac{3}{2}\sin(4x) + \dfrac{1}{6}\sin(6x) \right) + C \]Exercises
Evaluate the following integrals. Click each exercise to reveal the step-by-step solution.
Exercise (a): Evaluate \( \displaystyle \int [8\sin^6(x) - 2\sin^2(x)] \, dx \)
Solution:
Substitute power-reducing formulas for \( \sin^6(x) \) and \( \sin^2(x) \):
\[ 8 \cdot \dfrac{1}{32}(10 - 15\cos(2x) + 6\cos(4x) - \cos(6x)) - 2 \cdot \dfrac{1}{2}(1 - \cos(2x)) \] \[ = \dfrac{1}{4}(10 - 15\cos(2x) + 6\cos(4x) - \cos(6x)) - (1 - \cos(2x)) \] \[ = \dfrac{5}{2} - \dfrac{15}{4}\cos(2x) + \dfrac{3}{2}\cos(4x) - \dfrac{1}{4}\cos(6x) - 1 + \cos(2x) \] \[ = \dfrac{3}{2} - \dfrac{11}{4}\cos(2x) + \dfrac{3}{2}\cos(4x) - \dfrac{1}{4}\cos(6x) \]Integrating term by term yields:
\[ \dfrac{3}{2}x - \dfrac{11}{8}\sin(2x) + \dfrac{3}{8}\sin(4x) - \dfrac{1}{24}\sin(6x) + C \]Exercise (b): Evaluate \( \displaystyle \int [4\sin^4(x) + \sin^2(x)] \, dx \)
Solution:
Substitute power-reducing formulas for \( \sin^4(x) \) and \( \sin^2(x) \):
\[ 4 \cdot \dfrac{1}{8}(3 - 4\cos(2x) + \cos(4x)) + \dfrac{1}{2}(1 - \cos(2x)) \] \[ = \dfrac{1}{2}(3 - 4\cos(2x) + \cos(4x)) + \dfrac{1}{2}(1 - \cos(2x)) \] \[ = \dfrac{1}{2}[3 - 4\cos(2x) + \cos(4x) + 1 - \cos(2x)] = \dfrac{1}{2}[4 - 5\cos(2x) + \cos(4x)] \] \[ = 2 - \dfrac{5}{2}\cos(2x) + \dfrac{1}{2}\cos(4x) \]Integrating term by term yields:
\[ 2x - \dfrac{5}{4}\sin(2x) + \dfrac{1}{8}\sin(4x) + C \]