Evaluate the integral of 1 / (a2sin2 x + b2cos2 x)

Compute the following integral.

$\int \frac{dx}{a^2 \sin^2 x + b^2 \cos^2 x} \qquad \text{for} \quad ab \neq 0.$

Since $ab \neq 0$ implies $a \neq 0$ and $b \neq 0$ we know $\frac{1}{a}$ and $\frac{1}{b}$ exist, so we can divide by $a$ and $b$ whenever we want. First, we divide the numerator and denominator by $\cos^2 x$ to obtain

$\int \frac{dx}{a^2 \sin^2 x + b^2 \cos^2 x} = \int \frac{\sec^2 x \, dx}{a^2 \tan^2 x + b^2} = \frac{1}{a^2} \int \frac{\sec^2 x \, dx}{\tan^2 x + \left( \frac{b}{a} \right)^2}.$

Then we make the substitution $u = \tan x$ so $du = \sec^2 x \, dx$ . Therefore,

$\begin{align*} \int \frac{dx}{a^2 \sin^2 x + b^2 \cos^2 x} &= \frac{1}{a^2} \int \frac{\sec^2 x \, dx}{\tan^2 x + \left( \frac{b}{a} \right)^2} \\[9pt] &= \frac{1}{a^2} \int \frac{du}{u^2 + \left( \frac{b}{a} \right)^2} \\[9pt] &= \frac{1}{b^2} \int \frac{du}{\left( \frac{a}{b} \right)^2 u^2 + 1}. \end{align*}$

Next, we make another substitution. This time let $t = \frac{a}{b} u$ which implies $dt = \frac{a}{b} \, du$ . Therefore,

$\begin{align*} \frac{1}{b^2} \int \frac{du}{\left( \frac{a}{b} \right)^2 u^2 + 1} &= \frac{1}{ab} \int \frac{dt}{1+t^2} \\[9pt] &= \frac{1}{ab} \arctan t + C \\[9pt] &= \frac{1}{ab} \arctan \left( \frac{a}{b} u \right) + C \\[9pt] &= \frac{1}{ab} \arctan \left( \frac{a}{b} \tan x \right) + C. \end{align*}$

Stumbling Robot

Evaluate the integral of 1 / (a²sin² x + b²cos² x)

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