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Section 7.2 Integrating Trig Functions

Recall the following:

  • \(\displaystyle \dfrac{d}{dx}( \sin(x)) = \cos(x) \)

  • \(\displaystyle \dfrac{d}{dx}(\cos(x)) = -\sin(x) \)

  • \(\displaystyle \dfrac{d}{dx}(\tan(x)) = \sec^2(x) \)

  • \(\displaystyle \dfrac{d}{dx}(\csc(x)) = - \csc(x) \cot(x) \)

  • \(\displaystyle \dfrac{d}{dx}(\sec(x)) = \sec(x)\tan(x) \)

  • \(\displaystyle \dfrac{d}{dx}(\cot(x)) = -csc^2(x) \)

With our current methods we are able to easily solve the integrals of the right hand side of the functions above (ex: \(\int \sec^2 (x) dx = \tan(x) + c \)), but we do not currently have methods for most of the left hand side (ex: \(\int \sec(x) dx\)). The only exception currently is how to find the integral of \(\tan(x) \mbox{ and } \cot(x) \text{.}\)

\begin{align*} \int \cot(x) dx \amp = \int \dfrac{\cos(x)}{\sin(x)} dx \\ \amp = \int \dfrac{1}{u} du \\ \amp = \ln|u| + c \\ \amp = \ln| \sin(x)| + c \end{align*}

Table 7.2.1.
\(u = \sin(x) \)
\(du = \cos(x)dx \)

Subsection 7.2.1 Using Trig Identities

So, how would we find something like \(\int \tan^2(x)dx\text{?}\)

\begin{align*} \int \tan^2(x) dx \amp = \int \sec^2(x) - 1 dx \\ \amp = \tan(x) - x + c \end{align*}

Recall: \(\sin^2(x) + \cos^2(x) = 1\text{.}\) If we divide both sides by \(cos^2(x)\) we get \(\dfrac{\sin^2(x) + \cos^2(x)}{\cos^2(x)} = \dfrac{1}{\cos^2(x)}\text{.}\) Simplified we get \(\tan^2(x) + 1 = \sec^2(x)\text{.}\)

Another example of using trig identities for would be \(\int \cos^2(x)dx\text{.}\)

\begin{align*} \int \cos^2(x)dx \amp = \int \dfrac{1}{2} (1 + \cos(2x)) dx \\ \amp = \dfrac{1}{2} \int 1 + \cos(2x) dx \\ \amp = \dfrac{1}{2} (x + \dfrac{1}{2} sin(2x)) + c \\ \amp = \dfrac{1}{2}x + \dfrac{1}{4}\sin(2x) + c \end{align*}

Recall the following identities:

  • \(\displaystyle sin^2(x) = \dfrac{1}{2} (1 - \cos(2x)) \)

  • \(\displaystyle cos^2(x) = \dfrac{1}{2} (1 + \cos(2x)) \)

Subsection 7.2.2 Multiplying by One

For some trig functions integrals, you have to multiply by a wierd arrangement of 1, such as \(\int \sec (x) dx\text{.}\)

\begin{align*} \int \sec (x) dx \amp = \int \sec(x) \cdot \left( \dfrac{\sec(x) + \tan(x)}{\sec(x) + \tan(x)} \right) dx\\ \amp = \int \dfrac{\sec^2(x) + \sec(x)\tan(x)}{\sec(x) + \tan(x)} dx \\ \amp = \int \dfrac{1}{u} du \\ \amp = \ln | u | + c \\ \amp = \ln | \sec(x) + \tan(x) | + c \end{align*}

Table 7.2.2.
\(u = \sec(x) + \tan(x)\)
\(du = \sec(x)\tan(x) + sec^2(x) dx\)

Subsection 7.2.3 Trig Functions with Higher Powers

For higher powers of sine and cosine, there a few different methods of exaluating. One of these is by using one of the following reduction formulas:

  • \(\displaystyle \int \sin^n(x)dx = -\dfrac{1}{n} \sin^{n-1}(x) cos(x) + \dfrac{n-1}{n} \int \sin^{n-2}(x)dx \)

  • \(\displaystyle \int \cos^n(x)dx = \dfrac{1}{n} \cos^{n-1}(x) sin(x) + \dfrac{n-1}{n} \int \cos^{n-2}(x)dx \)

When you use this method, sometimes you have to apply this method multiple times.

Example 7.2.1.

Solve the Following using the reduction formula:

  1. \(\displaystyle\int \cos^5 (x) dx\)

    Answer

    \begin{align*} \int \cos^5 (x) dx \amp = -\dfrac{1}{5} \cos^4(x)\sin(x) + \dfrac{4}{5} \int \cos^3 (x) dx\\ \amp = -\dfrac{1}{5} \cos^4(x)\sin(x) + \dfrac{4}{5} \left( \dfrac{1}{3} \cos^2(x) \sin(x) + \dfrac{2}{3} \int \cos(x) dx \right) \\ \amp = -\dfrac{1}{5} \cos^4(x)\sin(x) + \dfrac{4}{15} \cos^2(x) \sin(x) + \dfrac{8}{15} \sin(x) + c \end{align*}
    Note 7.2.3.
    Notice we had to use the reduction formula twice.

  2. \(\displaystyle\int \sin^4(x) dx \)

    Answer

    \begin{align*} \int \sin^4(x) dx \amp = -\dfrac{1}{4}\sin^3(x)\cos(x) + \dfrac{3}{4} \int \sin^2(x)dx\\ \amp = -\dfrac{1}{4}\sin^3(x)\cos(x) + \dfrac{3}{4} \left( -\dfrac{1}{2} \sin(x)\cos(x) + \dfrac{1}{2}\int \sin^0(x)dx \right)\\ \amp = -\dfrac{1}{4}\sin^3(x)\cos(x) - \dfrac{3}{8} \sin(x)\cos(x) + \dfrac{3}{8} x + c \end{align*}

U-substitution can also be used in specific situations:

\begin{align*} \int \sin^4(x) \cos(x) dx \amp = \int u^4 du\\ \amp = \dfrac{u^5}{5} + c\\ \amp \dfrac{1}{5}\sin^5(x) + c \end{align*}

Table 7.2.4.
\(u = \sin(x) \)
\(du = \cos(x)dx\)

Here is a procedure to evaluate functions of the form \(\displaystyle\int\sin^m(x)\cos^n(x)dx\text{:}\)

  • n is odd

    split off a factor of \(\cos(x)\text{,}\) apply relevant identity, make u-sub where \(u = \sin(x)\)

    \(\cos^2(x) = 1 - sin^2(x)\)

  • m is odd

    split off a factor of \(\sin(x)\text{,}\) apply relevant identity, make u-sub where \(u=\cos(x)\)

    \(\sin^2(x) = 1 - \cos^2(x)\)>

  • \(\begin{cases} \text{m is even} \\ \text{n is even} \end{cases} \)

    use the relevant identities to reduce the powers on \(sin(x) \mbox{ and } \cos(x) \)

    \(\begin{cases} \sin^2(x) = \dfrac{1}{2}(1 - \cos(2x)) \\ \cos^2(x) = \dfrac{1}{2}(1 + \cos(2x)) \end{cases}\)

Subsection 7.2.4 Using Angular Trig Identities

How do we evaluate integrals of the form: \(\displaystyle \int \sin(mx) \cos(nx) dx \mbox{ , }\) \(\displaystyle \int \sin(mx) \sin(nx) dx \mbox{ , or }\) \(\displaystyle \int \cos(mx) \cos(nx) dx \text{?}\) We can use a trig identity and express them as sums.

Recall:

  • \(\displaystyle \sin\alpha\cos\beta = \dfrac{1}{2}\left( \sin(\alpha - \beta) + \sin( \alpha + \beta)\right) dx \)

  • \(\displaystyle \sin\alpha\sin\beta = \dfrac{1}{2}\left( \cos(\alpha - \beta) - \cos( \alpha + \beta)\right) dx \)

  • \(\displaystyle \cos\alpha\cos\beta = \dfrac{1}{2}\left( \cos(\alpha - \beta) + \cos( \alpha + \beta)\right) dx \)

Example 7.2.2.

Evaluate \(\displaystyle \int \sin(5x)\cos(3x) dx\text{:}\)

Answer

\begin{align*} \int \sin(5x)\cos(3x) dx \amp = \int \dfrac{1}{2} ( \sin(2x) + \sin(8x)) dx \\ \amp = \dfrac{1}{2} \int \sin(2x) + \sin(8x) dx\\ \amp = \dfrac{1}{2} \left( - \dfrac{1}{2} \cos(2x) - \dfrac{1}{8} \cos(8x) \right) + c \\ \amp = - \dfrac{1}{4} \cos(2x) - \dfrac{1}{16} \cos(8x) + c \end{align*}