Example 42: $\int_0^{\pi}\frac{x\,dx}{a^2\cos^2x+b^2\sin^2x}$

Basics you need

Solution

1. The integrand is continuous on $[0, \pi]$

   Write $D(x) = a^2\cos^2 x + b^2\sin^2 x$⁠. Since $\cos^2 x + \sin^2 x = 1$⁠,

   $$D(x) \ge \min(a^2, b^2)\,(\cos^2 x + \sin^2 x) = \min(a^2, b^2) > 0 .$$

   So $D$ never vanishes, and $\frac{x}{D(x)}$ and $\frac{1}{D(x)}$ are continuous on all of $[0, \pi]$⁠.

   This is why $a$ and $b$ must both be non-zero: if, say, $b = 0$⁠, then $D(x) = a^2\cos^2 x$ is 0 at $x = \frac{\pi}{2}$ and the integral is not a proper one. With continuity, all the properties and Theorem 2 apply.
   Each of $a^2\cos^2 x$ and $b^2\sin^2 x$ is at least $\min(a^2, b^2)$ times $\cos^2 x$ or $\sin^2 x$ respectively.

2. P4 removes the $x$ from the numerator

   P4 with $c = \pi$ replaces $x$ by $\pi - x$⁠. The denominator does not change, because $\cos^2(\pi - x) = \cos^2 x$ and $\sin^2(\pi - x) = \sin^2 x$⁠:

   $$\begin{aligned}I &= \int_0^{\pi}\frac{(\pi - x)\,dx}{a^2\cos^2 x + b^2\sin^2 x}\\ &= \pi\int_0^{\pi}\frac{dx}{D(x)} - \int_0^{\pi}\frac{x\,dx}{D(x)}\\ &= \pi\int_0^{\pi}\frac{dx}{D(x)} - I .\end{aligned}$$

   Adding $I$ to both sides and dividing by 2:

   $$I = \frac{\pi}{2}\int_0^{\pi}\frac{dx}{a^2\cos^2 x + b^2\sin^2 x} .$$
   The second integral is $I$ itself, so we can move it to the left: $2I = \pi\int_0^\pi\frac{dx}{D}$⁠. The figure in Step 7 shows the idea: $\frac{x}{D}$ and $\frac{\pi - x}{D}$ are mirror images with equal areas, and they add up to $\frac{\pi}{D}$⁠, so each has half the area under $\frac{\pi}{D}$⁠.
   P4 works because the interval is $[0, c]$ and $x \mapsto c - x$ maps it onto itself. Basics: symmetry and mirror images

3. P6 halves the interval

   Let $g(x) = \frac{1}{D(x)}$⁠. With $2c = \pi$⁠, $c = \frac{\pi}{2}$⁠, the condition of P6 holds, since $g(\pi - x) = g(x)$ (the same reflection facts as in Step 2). So

   $$\int_0^{\pi}g(x)\,dx = 2\int_0^{\pi/2}g(x)\,dx ,$$

   and therefore

   $$I = \pi\int_0^{\pi/2}\frac{dx}{a^2\cos^2 x + b^2\sin^2 x} .$$
   The graph of $g$ on $[0, \pi]$ is symmetric about $x = \frac{\pi}{2}$⁠, so its two halves have equal areas. Working on $[0, \frac{\pi}{2}]$ is needed because the substitutions ahead use $\tan x$ and $\cot x$⁠, which behave well only inside one quadrant.
   $\frac{\pi}{2}\cdot 2 = \pi$⁠.

4. Split at $\frac{\pi}{4}$ by P2

   $$I = \pi\left[\int_0^{\pi/4}\frac{dx}{D(x)} + \int_{\pi/4}^{\pi/2}\frac{dx}{D(x)}\right] .$$
   Next we want to divide by $\cos^2 x$ to bring in $\tan x$⁠. But $\cos x = 0$ at $\frac{\pi}{2}$⁠, so that division is not allowed at the upper end. Dividing by $\sin^2 x$ instead (to bring in $\cot x$⁠) fails at 0. Splitting at $\frac{\pi}{4}$ lets us use each division where it is safe: on $[0, \frac{\pi}{4}]$⁠, $\cos x \ge \frac{1}{\sqrt2}$⁠; on $[\frac{\pi}{4}, \frac{\pi}{2}]$⁠, $\sin x \ge \frac{1}{\sqrt2}$⁠. Any split point strictly between 0 and $\frac{\pi}{2}$ would work; $\frac{\pi}{4}$ gives the neat limit 1 for both $\tan$ and $\cot$⁠.

5. The first piece: divide by $\cos^2 x$⁠, put $t = \tan x$

   Dividing top and bottom by $\cos^2 x$⁠:

   $$\begin{aligned}&\int_0^{\pi/4}\frac{dx}{a^2\cos^2 x + b^2\sin^2 x}\\ &= \int_0^{\pi/4}\frac{\sec^2 x\,dx}{a^2 + b^2\tan^2 x} .\end{aligned}$$

   Put $t = \tan x$⁠, $dt = \sec^2 x\,dx$⁠; $x = 0 \Rightarrow t = 0$⁠, $x = \frac{\pi}{4} \Rightarrow t = 1$⁠:

   $$\begin{aligned}\int_0^1\frac{dt}{a^2 + b^2t^2} &= \frac{1}{b^2}\int_0^1\frac{dt}{t^2 + (a/b)^2}\\ &= \frac{1}{b^2}\cdot\frac{b}{a}\Big[\tan^{-1}\frac{bt}{a}\Big]_0^1 = \frac{1}{ab}\tan^{-1}\frac{b}{a} .\end{aligned}$$
   $\tan x$ is increasing and differentiable on $[0, \frac{\pi}{4}]$⁠, so the substitution is valid on the whole piece. Section 7.4, formula (3) with $k = \frac{a}{b}$⁠: $\frac{1}{k}\tan^{-1}\frac{t}{k} = \frac{b}{a}\tan^{-1}\frac{bt}{a}$⁠. At $t = 0$⁠, $\tan^{-1}0 = 0$⁠.
   $\frac{\cos^2 x}{\cos^2 x} = 1$⁠, $\frac{\sin^2 x}{\cos^2 x} = \tan^2 x$⁠, $\frac{1}{\cos^2 x} = \sec^2 x$⁠. Basics: substitution

6. The second piece: divide by $\sin^2 x$⁠, put $u = \cot x$

   Dividing top and bottom by $\sin^2 x$⁠:

   $$\begin{aligned}&\int_{\pi/4}^{\pi/2}\frac{dx}{a^2\cos^2 x + b^2\sin^2 x}\\ &= \int_{\pi/4}^{\pi/2}\frac{\operatorname{cosec}^2 x\,dx}{a^2\cot^2 x + b^2} .\end{aligned}$$

   Put $u = \cot x$⁠, $du = -\operatorname{cosec}^2 x\,dx$⁠; $x = \frac{\pi}{4} \Rightarrow u = 1$⁠, $x = \frac{\pi}{2} \Rightarrow u = 0$⁠:

   $$\begin{aligned}-\int_1^0\frac{du}{a^2u^2 + b^2} &= \int_0^1\frac{du}{a^2u^2 + b^2}\\ &= \frac{1}{ab}\Big[\tan^{-1}\frac{au}{b}\Big]_0^1 = \frac{1}{ab}\tan^{-1}\frac{a}{b} .\end{aligned}$$
   The minus sign from $du$ is absorbed by swapping the limits (P1). The integral is the same as in Step 5 with $a$ and $b$ interchanged, so the answer is too.
   $\frac{d}{dx}\cot x = -\operatorname{cosec}^2 x$⁠; $\cot\frac{\pi}{4} = 1$⁠, $\cot\frac{\pi}{2} = 0$⁠.

7. Add and simplify

   $$I = \pi\cdot\frac{1}{ab}\left[\tan^{-1}\frac{b}{a} + \tan^{-1}\frac{a}{b}\right] = \frac{\pi}{ab}\cdot\frac{\pi}{2} = \frac{\pi^2}{2ab} .$$
   Since $a, b > 0$⁠, $p = \frac{b}{a}$ is positive and $\frac{a}{b} = \frac1p$⁠, so Basic 4 gives $\tan^{-1}\frac{b}{a} + \tan^{-1}\frac{a}{b} = \frac{\pi}{2}$⁠. (If $p$ were negative, the sum would be $-\frac{\pi}{2}$⁠.)

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