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Mathematics LibreTexts

2.3: Powers of Trig Functions

  • Page ID
    520
  • [ "article:topic", "authorname:green" ]

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    Example 1: Powers of Sine and Cosine

    Evaluate

    \[ \int \sin^5 x \, dx.\]

    Solution

    We write

    \[\begin{align} \int \sin^5 x \, dx &= \int \sin^4 x \, \sin x \, dx \\ &= \int (\sin^2 x)^2 \sin x \,dx  \\ &= \int (1-\cos^2x)^2 \sin x \, dx \\  &\text{substitute } u = \cos x \text{ and } du = -\sin x\, dx \\  &= -\int (1 - u^2)^2 du \\ & =  -\int [1 - 2u^2 + u^4] \; du \\ &=  -u + \dfrac{2}{3} u^3 - \dfrac{1}{5} u^5 + C \\ &=  -\cos x + \dfrac{2}{3} \cos^3 x - \dfrac{1}{5} \cos^5x + C \end{align}. \]

    Exercise

    Evaluate \( \int \cos^7 x \, dx\).

    Example 2

    Evaluate \(\int \sin^4 x\, dx\).

    Solution

    We write  

    \[\begin{align} \int \sin^4 x \; dx &= \int (\sin^2 x)^2 dx \\ &= \int \Big(\dfrac{1}{2} - \dfrac{1}{2} \cos(2x)\Big)^2 \; dx \\ &= \int \Big[ \dfrac{1}{4}-\dfrac{1}{2} \cos(2x) + \dfrac{1}{4} \cos^2(2x)  \Big] dx \end{align}\]

    For the second integral let \( u = 2x\)  and  \(dx = \dfrac{1}{2} du\).

    The integral becomes

    \[ \dfrac{1}{4} x - \dfrac{1}{4} \sin(2x) + \dfrac{1}{4}\int \Big[ \dfrac{1}{2}+\dfrac{1}{2} \cos(4x) \Big] dx \]

    \[\text{Let } u = 4x, dx = \dfrac{1}{4} \; du \]

    \[\dfrac{1}{4}x - \dfrac{1}{4}\sin(2x) +\dfrac{1}{8}x +\dfrac{1}{32}\sin(4x) +C. \]

    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

    \[ \sin^2x  =  \dfrac{1}{2} - \dfrac{1}{2} \cos (2x) \]

    and

    \[ \cos^2 x  =  \dfrac{1}{2} + \dfrac{1}{2} \cos (2x). \]

    This method will always work and is always long and tedious.

    Example 3

    Evaluate

    \[ \sin^2 x \cos^3 x \; dx.\]

    Solution

    We pull out a \(\cos x\) and convert the \(\cos^2 x\) to \(1 - \sin^2 x\).

    \[ \sin^2 x (1-\sin^2 x) \cos x \; dx \; \;\; \; \text{Let } u=\sin x, \; du=\cos x \; dx :\]

    We have

    \[\begin{align} \int u^2 (1-u^2) \; du &= \int u^2-u^4 \; du \\ &= \dfrac{1}{3}u^5 + C \\ &= \dfrac{1}{3} \sin^3 x -\dfrac{1}{5} \sin^5 x + C \end{align}\]

    Exercise

    \[ \int \sin^5 x \cos^2 x \, dx \]

    Powers of Tangents and Secants

    To integrate powers of tangents and secants we use the formula

    \[ \tan^2 x  =  \sec^2 x - 1.\]

    Example 4

    Evaluate

    \[ \int \tan^4 x\, dx. \]

    Solution

    We write

    \[\begin{align} \int \tan^2 x \tan^2 x \, dx &=  \int \tan^2 x (\sec^2 x - 1) \\ &=  \int \tan^2 x \sec^2 x\, dx  - \int \tan^2 x\, dx \\ &=  \int \tan^2 x \sec^2 x \, dx  - \int (\sec^2 x - 1) \,dx   \end{align}.\]

    For the first integral let

    \[  u = \tan x,\;\; du = \sec^2 x\, dx. \]

    We have

    \[\begin{align}  \int u^2 \,du - \tan x + x &=  \dfrac{1}{3} u^3 - \tan x + x + C \\ &=  \dfrac{1}{3} \tan^3 x - \tan x + x + C   \end{align}.\]

    Exercise

    \[ \int \sec^5 x \tan^3 x \, dx.\]

    Mixed Angles

    We have the following formulas:

    \[ \sin(m \theta) \sin(n \theta) = \dfrac{1}{2} \left[ \cos[(m - n) \theta] - \cos[(m + n) \theta] \right]\]

    \[ \sin(m \theta) \cos(n \theta) = \dfrac{1}{2} \left[ \sin[(m - n) \theta] + \sin[(m + n) \theta] \right] \]

    \[ \cos(m \theta) \cos(n \theta) = \dfrac{1}{2} \left[ \cos[(m - n) \theta] + \cos[(m + n) \theta] \right]. \]

    Example 5

    \[ \int \sin{(3x)} \sin{(4x)}\, dx \]

    \[  = \dfrac{1}{2} \int \left[ \cos{(-x)} - \cos{(7x)} \right]\, dx  \]

    We integrate the first by letting \(u = -x\) and the second by letting \(u = 7x\) to get

    \[ \dfrac{1}{2} -\sin{(-x)} + \dfrac{1}{7} \sin{(7x)} + C.\]

    Larry Green (Lake Tahoe Community College)

    • Integrated by Justin Marshall.