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Evaluate the integral of 1 / (a sin x + b cos x)2

by RoRi

Compute the following integral.

    \[ \int \frac{dx}{(a \sin x + b \cos x)^2} \qquad \text{for} \quad a \neq 0. \]

First, we divide the numerator and denominator by \cos^2 x to obtain

    \[ \int \frac{dx}{(a \sin x + b \cos x)^2} = \int \frac{\sec^2 x \, dx}{(a \tan x + b)^2}. \]

Then we make the substitution u = a \tan x + b which implies du = a \sec^2 x \, dx. Therefore,

    \begin{align*} \int \frac{dx}{(a \sin x + b \cos x)^2} &= \frac{1}{a} \int \frac{du}{u^2} \\[9pt] &= -\frac{1}{au} + C \\[9pt] &= \frac{-1}{a (a \tan x + b)} + C \\[9pt] &= \frac{- \cos x}{a (a \sin x + b \cos )} + C. \end{align*}

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