Miscellaneous · Q26: $\int_0^{\pi/2}\frac{\cos^2x\,dx}{\cos^2x+4\sin^2x}$

Basics you need

Solution

1. Continuity on $[0, \frac{\pi}{2}]$

   $$\cos^2x + 4\sin^2x = 1 + 3\sin^2x \ \ge\ 1 .$$

   The denominator never vanishes, so $g$ is continuous on $[0, \frac{\pi}{2}]$⁠.

   Theorem 2 and the properties of Section 7.10 are used for continuous integrands on closed intervals. Writing $\cos^2x = 1 - \sin^2x$ shows the denominator is at least 1.
   Section 7.8.3. Lesson: definite integrals

2. Write the integrand as a constant plus a simpler fraction

   Let $D = \cos^2x + 4\sin^2x = 4 - 3\cos^2x$⁠. Then $\cos^2x = \frac{4 - D}{3}$⁠, so

   $$g(x) = \frac{4 - D}{3D} = \frac43\cdot\frac1D - \frac13 .$$

   Hence, with $J = \int_0^{\pi/2}\frac{dx}{\cos^2x + 4\sin^2x}$⁠,

   $$I = \frac43J - \frac13\cdot\frac{\pi}{2} = \frac43J - \frac{\pi}{6} .$$
   Why this method: $\cos^2x$ in the numerator does not pair with any differential, but a constant numerator does: $\frac{1}{\cos^2x + 4\sin^2x}$ becomes $\frac{\sec^2x}{1 + 4\tan^2x}$ after dividing by $\cos^2x$⁠. So we move the $\cos^2x$ into the denominator's own terms. The $-\frac13$ integrates to $-\frac13$ times the length $\frac{\pi}{2}$ of the interval.
   Basics: splitting fractions

3. Split $J$ at $\frac{\pi}{4}$ (P2)

   $$J = \underbrace{\int_0^{\pi/4}\frac{dx}{\cos^2x + 4\sin^2x}}_{J_1} + \underbrace{\int_{\pi/4}^{\pi/2}\frac{dx}{\cos^2x + 4\sin^2x}}_{J_2} .$$
   The natural substitution $t = \tan x$ breaks down at $x = \frac{\pi}{2}$⁠, where $\tan x$ does not exist; using it on all of $[0, \frac{\pi}{2}]$ would need the upper limit "$t = \infty$", an improper integral that this chapter does not cover. On $[\frac{\pi}{4}, \frac{\pi}{2}]$ we use $\cot x$ instead, which is finite there. This is the device of Example 42.
   P2, Section 7.10; Example 42 (Miscellaneous Examples). Lesson: properties of definite integrals

4. $J_1$ with $t = \tan x$

   On $[0, \frac{\pi}{4}]$⁠, $\cos x \ne 0$⁠; divide by $\cos^2x$⁠:

   $$J_1 = \int_0^{\pi/4}\frac{\sec^2x\,dx}{1 + 4\tan^2x} .$$

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

   $$\begin{aligned}J_1 &= \int_0^1\frac{dt}{1 + 4t^2} = \frac14\int_0^1\frac{dt}{t^2 + \left(\frac12\right)^2}\\ &= \frac14\cdot 2\left[\tan^{-1}2t\right]_0^1 = \frac12\tan^{-1}2 .\end{aligned}$$
   $\tan x$ is differentiable and increasing on $[0, \frac{\pi}{4}]$⁠, so the substitution is valid with new limits 0 and 1 (Section 7.9). Formula (3) with $a = \frac12$ gives $\frac{1}{a}\tan^{-1}\frac{t}{a} = 2\tan^{-1}2t$⁠. Numerically $J_1 \approx 0.5535743589$⁠.
   Section 7.4, formula (3); Section 7.9. Lesson: integrals of particular functions

5. $J_2$ with $u = \cot x$

   On $[\frac{\pi}{4}, \frac{\pi}{2}]$⁠, $\sin x \ne 0$⁠; divide by $\sin^2x$⁠:

   $$J_2 = \int_{\pi/4}^{\pi/2}\frac{\operatorname{cosec}^2x\,dx}{\cot^2x + 4} .$$

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

   $$\begin{aligned}J_2 &= -\int_1^0\frac{du}{u^2 + 4} = \int_0^1\frac{du}{u^2 + 2^2}\\ &= \frac12\left[\tan^{-1}\frac{u}{2}\right]_0^1 = \frac12\tan^{-1}\frac12 .\end{aligned}$$
   $\cot x$ is differentiable and decreasing on $[\frac{\pi}{4}, \frac{\pi}{2}]$⁠, with $\cot\frac{\pi}{2} = 0$⁠, so the new limits are 1 and 0. The minus sign from $du$ and the reversed limits cancel by P1. Numerically $J_2 \approx 0.2318238045$⁠.
   P1, Section 7.10; Section 7.4, formula (3). Lesson: properties of definite integrals

6. Add: $J = \frac{\pi}{4}$

   $$J = J_1 + J_2 = \frac12\left(\tan^{-1}2 + \tan^{-1}\frac12\right).$$

   Let $\phi = \tan^{-1}2$⁠, so $0 < \phi < \frac{\pi}{2}$⁠. Then $\frac{\pi}{2} - \phi$ also lies in $(0, \frac{\pi}{2})$ and $\tan\left(\frac{\pi}{2} - \phi\right) = \cot\phi = \frac12$⁠. Since $\frac{\pi}{2} - \phi$ is in the principal range of $\tan^{-1}$⁠, $\tan^{-1}\frac12 = \frac{\pi}{2} - \phi$⁠. Therefore

   $$J = \frac12\cdot\frac{\pi}{2} = \frac{\pi}{4} \approx 0.7853981634 .$$
   $\tan^{-1}y$ is the unique angle in $(-\frac{\pi}{2}, \frac{\pi}{2})$ whose tangent is $y$⁠, so finding one such angle identifies it. This is the special case $a = 1$⁠, $b = 2$ of Example 42's inner result $\int_0^{\pi/2}\frac{dx}{a^2\cos^2x + b^2\sin^2x} = \frac{\pi}{2ab}$⁠.
   Principal range of $\tan^{-1}$ (Chapter 2).

7. Assemble $I$

   $$I = \frac43\cdot\frac{\pi}{4} - \frac{\pi}{6} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6} \approx 0.5235987756 .$$
   Substituting $J = \frac{\pi}{4}$ into Step 2. The answer is less than the length $\frac{\pi}{2}$ of the interval, as it must be, since $0 \le g(x) \le 1$⁠.
   Basics: area as a sum of thin strips

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