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Evaluate the integral of 1 / (2 sin x – cos x + 5)

by RoRi

Compute the following integral.

    \[ \int \frac{dx}{2 \sin x - \cos x + 5}. \]

Following Example 1 in Apostol’s text (p. 264-265) we make the substitution u = \tan \frac{x}{2}. This gives us

    \[ x = 2 \arctan u, \qquad dx = \frac{2}{1+u^2} \, du. \]

We also have the following expressions for \sin x and \cos x,

    \begin{align*} \sin x &= 2 \sin \frac{x}{2} \cos \frac{x}{2} = \frac{2 \tan \frac{x}{2}}{\sec^2 \frac{x}{2}} = \frac{2u}{1+u^2},\\[10pt] \cos x &= 2 \cos^2 \frac{x}{2} - 1 = \frac{2}{\sec^2 \frac{x}{2}} - 1 = \frac{2}{1+u^2}-1 = \frac{1-u^2}{1+u^2}. \end{align*}

Therefore,

    \begin{align*} \int \frac{dx}{2 \sin x - \cos x + 5} &= \int \left(\frac{1+u^2}{4u-1+u^2+5+5u^2} \cdot \frac{2}{1+u^2} \right) \, du \\[9pt] &= \int \frac{du}{3u^2 + 2u + 2} \\[9pt] &= \int \frac{3 \, du}{(3u+1)^2 + 5}. \end{align*}

Now we make the substitution 3u + 1 = t \sqrt{5} which implies du = \frac{\sqrt{5}}{3} \, dt. We then have

    \begin{align*} \int \frac{3 \, du}{(3u+1)^2 + 5} &= \int \frac{\sqrt{5}}{5t^2 + 5}\, dt \\ &= \frac{1}{\sqrt{5}} \int \frac{dt}{t^2+1} \\ &= \frac{1}{\sqrt{5}} \arctan t + C \\ &= \frac{1}{\sqrt{5}} \arctan \left( \frac{3u+1}{\sqrt{5}} \right) + C \\ &+ \frac{1}{\sqrt{5}} \arctan \left( \frac{3 \tan \left(\frac{x}{2} \right) + 1}{\sqrt{5}} \right) + C. \end{align*}

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