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2.3: Powers of Trig Functions

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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.