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Last edited on 10. November 2016

Trigonometric Integrals and Identities

In this article you will see how to easily calculate trigonometric integrals like \begin{align} \int_a^b \sin^2{x} \cdot \cos{x} \D{x} \end{align} without the help of trigonometric identities (which - in my experience - turn out to be difficult to remember) or finding sophisticated substitutions. The only thing you need to know is the Euler formula \begin{align} e^{ix} = \cos{x} + i \sin{x} \end{align} which relates the complex exponential function to the trigonometric functions.[1]

By applying basic methods of complex analysis this identiy can be used to express the trigonometric functions in terms of the complex exponential function only:[2] \begin{align} \sin{x} = \text{Im} \left( e^{ix} \right) = \frac{e^{ix} - e^{-ix}}{2i} \\ \cos{x} = \text{Re} \left( e^{ix} \right) = \frac{e^{ix} + e^{-ix}}{2} \end{align} Note that $e^{-ix}$ is the complex conjugate of $e^{ix}$. Now one proceeds as follows:

  1. express all trigonometric functions in terms of the complex exponential function
  2. perform eventual multiplications
  3. rewrite the complex expressions as basic trigonometric functions.

This method is universially applicable and enables you to evaluate complex trigonometric integrals without the help of external reference.

Examples

sin(x) cos(x)

\begin{align} \int_a^b \sin{x} \cdot \cos{x} \D{x} &= \int_a^b \frac{e^{ix} - e^{-ix}}{2i} \cdot \frac{e^{ix} + e^{-ix}}{2} \D{x} \nonumber \\ &= \frac{1}{2} \cdot \int_a^b \frac{e^{2ix} \cancel{-1} \cancel{+1} - e^{-i2x}}{2i} \D{x} \nonumber \\ &= \frac{1}{2} \cdot \int_a^b \sin{\left( 2x \right)} \D{x} \nonumber \\[2ex] &= \left. - \frac{ \cos{ \left( 2x \right) } }{4} \right|_a^b \end{align}

sin3(x) cos(x)

\begin{align} \int_a^b & \sin^3{x} \cdot \cos{x} \D{x} = \int_a^b \left( \frac{e^{ix} - e^{-ix}}{2i} \right)^3 \cdot \frac{e^{ix} + e^{-ix}}{2} \D{x} \nonumber \\ &= - \frac{1}{16i} \cdot \int_a^b \left( e^{i 3x} - 3e^{i x} + 3e^{-ix} + e^{-i3x} \right) \cdot \left( e^{ix} + e^{-ix} \right) \D{x} \nonumber \\ &= -\frac{1}{16i} \cdot \int_a^b e^{i4x} - 3e^{i2x} \cancel{+3} - e^{-i2x} + e^{i2x} - 3 + 3 e^{-i2x} - e^{-i4x} \D{x} \nonumber \\ &= -\frac{1}{8} \cdot \int_a^b \underbrace{ \frac{ e^{i4x} - e^{-i4x} }{2i} }_{\sin{(4x)}} - 2 \cdot \underbrace{ \frac{ e^{i2x} - e^{-i2x} }{2i} }_{\sin{(2x)}} \D{x} \nonumber \\ &= \frac{1}{8} \cdot \int_a^b 2 \cdot \sin{\left( 2x \right)} - \sin{\left( 4x \right)} \D{x} \nonumber \\[2ex] &= \left. \frac{1}{8} \cdot \left[ - \cos{(2x)} + \frac{\cos{(4x)}}{4} \right] \right|_a^b \end{align}

sin(x) cos2(x)

\begin{align} \int_a^b & \sin{x} \cdot \cos^2{x} \D{x} = \int_a^b \frac{ e^{ix} - e^{-ix} }{2i} \cdot \left( \frac{ e^{ix} + e^{-ix} }{2} \right)^2 \D{x} \nonumber \\ &= \frac{1}{8i} \cdot \int_a^b \left( e^{ix} - e^{-ix} \right) \cdot \left( e^{i2x} + 2 + e^{-i2x} \right) \D{x} \nonumber \\ &= \frac{1}{8i} \cdot \int_a^b e^{i3x} + 2 e^{ix} + e^{-ix} - e^{ix} - 2 e^{-ix} - e^{-i3x} \D{x} \nonumber \\ &= \frac{1}{4} \cdot \int_a^b \underbrace{ \frac{e^{i3x} - e^{-i3x} }{2i} }_{\sin{(3x)}} + \underbrace{ \frac{ e^{ix} - e^{-ix} }{2i} }_{\sin{(x)}} \D{x} \nonumber \\ &= \frac{1}{4} \cdot \int_a^b \sin{\left( 3x \right)} + \sin{x} \D{x} \nonumber \\[2ex] &= \left. - \frac{1}{4} \cdot \left[ \frac{ \cos{(3x)} }{3} + \cos{x} \right] \right|_a^b \end{align}

Trigonometric Identities

The same logic can be used to find trigonometric identities:

sin(a)cos(b) + sin(b)cos(a)

\begin{align} \sin(a) &\cos(b) + \sin(b)\cos(a) \nonumber\\ &= \frac{e^{ia}-e^{-ia}}{2i} \frac{e^{ib} + e^{-ib}}{2} + \frac{e^{ib}-e^{-ib}}{2i} \frac{e^{ib}+e^{-ib}}{2} \nonumber\\ &= \frac{1}{4} \, [ e^{i(a+b)}-\cancel{e^{i(-a+b)}}+\cancel{e^{i( a-b)}}-e^{i(-a-b)}+ \nonumber\\ &\qquad\quad e^{i(a+b)}-\cancel{e^{i( a-b)}}+\cancel{e^{i(-a+b)}}-e^{i(-a-b)} ] \nonumber\\ &= \frac{2}{2} \, \frac{e^{i(a+b)}-e^{-i(a+b)}}{2i} = \sin(a+b) \end{align}

References

[1] B. Kusse, E. Westwig Mathematical Physics Wiley 1998 (ch. 6.1.3)
[2] B. Kusse, E. Westwig Mathematical Physics Wiley 1998 (ch. 6.1.2)