2.3: Powers of Trig Functions
- Page ID
- 520
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We write
\[\begin{align} \int \sin^4 x \; dx &= \int (\sin^2 x)^2 dx \\ &= \int \Big(\dfrac{1}{2} - \dfrac{1}{2} \cos(2x)\Big)^2 \; dx \\ &= \int \Big[ \dfrac{1}{4}-\dfrac{1}{2} \cos(2x) + \dfrac{1}{4} \cos^2(2x) \Big] dx \end{align} \nonumber \]
For the second integral let \( u = 2x\) and \(dx = \dfrac{1}{2} du\).
The integral becomes
\[ \dfrac{1}{4} x - \dfrac{1}{4} \sin(2x) + \dfrac{1}{4}\int \Big[ \dfrac{1}{2}+\dfrac{1}{2} \cos(4x) \Big] dx \nonumber \]
\[\text{Let } u = 4x, dx = \dfrac{1}{4} \; du \nonumber \]
\[\dfrac{1}{4}x - \dfrac{1}{4}\sin(2x) +\dfrac{1}{8}x +\dfrac{1}{32}\sin(4x) +C. \nonumber \]
We see that if the power is odd we can pull out one of the sin functions and convert the other to an expression involving the cos function only. Then use \( u = \cos x. \)
If the power is even, we must use the trig identities
\[ \sin^2x = \dfrac{1}{2} - \dfrac{1}{2} \cos (2x) \nonumber \]
and
\[ \cos^2 x = \dfrac{1}{2} + \dfrac{1}{2} \cos (2x). \nonumber \]
This method will always work and is always long and tedious.
Powers of Tangents and Secants
To integrate powers of tangents and secants we use the formula
\[ \tan^2 x = \sec^2 x - 1. \nonumber \]
Mixed Angles
We have the following formulas:
\[ \sin(m \theta) \sin(n \theta) = \dfrac{1}{2} \left[ \cos[(m - n) \theta] - \cos[(m + n) \theta] \right] \nonumber \]
\[ \sin(m \theta) \cos(n \theta) = \dfrac{1}{2} \left[ \sin[(m - n) \theta] + \sin[(m + n) \theta] \right] \nonumber \]
\[ \cos(m \theta) \cos(n \theta) = \dfrac{1}{2} \left[ \cos[(m - n) \theta] + \cos[(m + n) \theta] \right]. \nonumber \]
Larry Green (Lake Tahoe Community College)
Integrated by Justin Marshall.