3.7 Rational Functions of Trigonometric Functions
The integration of any rational function of the trigonometric functions \(\sin x\) and \(\cos x\) can be transformed into integration of rational functions of \(z\).
Let \(\hspace {0.2cm} z = \tan \dfrac {x}{2}\hspace {0.2cm}\) then \(\hspace {0.2cm} dz = \dfrac {1}{2}\sec ^2 \Big (\dfrac {x}{2}\Big )dx\)
\[\implies \hspace {0.4cm} dz = \dfrac {1}{2}\Bigg ( \tan ^2\Big (\dfrac {x}{2}\Big ) + 1 \Bigg ) dx\]
\[\implies \hspace {0.4cm} dz = \dfrac {z^2 + 1}{2}\hspace {0.1cm} dx \hspace {0.5cm} \text {or}\hspace {0.5cm} dx = \dfrac {2}{z^2 + 1}dz\]
\(\sin x\) and \(\cos x\) can also be transformed as rational functions of \(z\) as follows: \begin {align*} \cos x & = 2 \cos ^2\Big (\dfrac {x}{2}\Big ) - 1 = \dfrac {2}{\sec ^2\Big (\dfrac {x}{2}\Big )} - 1\\ & = \dfrac {2}{1 + \tan ^2\Big (\dfrac {x}{2}\Big )} - 1 \\ & = \dfrac {2}{1 + z^2} - 1\\\\ & = \dfrac {1 - z^2}{1 + z^2}\\ \end {align*}
\[\text {i.e}\hspace {0.5cm} \cos x = \dfrac {1 - z^2}{1 + z^2}\hspace {1cm} \sin x = \dfrac {2z}{1 + z^2}\]
Solution.
Let \(\hspace {0.2cm} z = \tan \dfrac {x}{2}\hspace {0.2cm}\), then \(\hspace {0.2cm} dx = \dfrac {2}{1 + z^2}dz\hspace {0.2cm}\) and \(\hspace {0.2cm} \cos x = \dfrac {1 - z^2}{1 + z^2}\hspace {0.2cm}\), thus \begin {align*} \int \dfrac {dx}{2 + \cos x}dx & = \int \frac {\dfrac {2}{1 + z^2}dz}{2 + \Bigg (\dfrac {1 - z^2}{1 + z^2}\Bigg )}\\ & = 2 \int \dfrac {1}{z^2 + 3}dz\\\\ & = \dfrac {2}{\sqrt {3}}\tan ^{-1}\dfrac {z}{\sqrt {3}} + c\\\\ & = \dfrac {2}{\sqrt {3}}\tan ^{-1}\Bigg (\dfrac {\tan (x/2)}{\sqrt {3}}\Bigg ) + c\\\\ \end {align*}
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