Miscellaneous · Q25: $\int_0^{\pi/4}\frac{\sin x\cos x}{\cos^4x+\sin^4x}\,dx$

Basics you need

Solution

1. Continuity on $[0, \frac{\pi}{4}]$

   $$\begin{aligned}\cos^4x + \sin^4x &= (\cos^2x + \sin^2x)^2 - 2\sin^2x\cos^2x\\ &= 1 - \tfrac12\sin^22x \ \ge\ \tfrac12 .\end{aligned}$$

   The denominator is never 0, so $g$ is continuous on $[0, \frac{\pi}{4}]$ (indeed on all of $\mathbb{R}$⁠).

   Theorem 2 (Section 7.8.3) needs a continuous integrand on the closed interval. Writing $a^2 + b^2 = (a + b)^2 - 2ab$ with $a = \cos^2x$⁠, $b = \sin^2x$ shows the denominator is at least $\frac12$⁠, because $\sin^22x \le 1$⁠.
   Basics: trigonometric identities

2. Divide numerator and denominator by $\cos^4x$

   On $[0, \frac{\pi}{4}]$⁠, $\cos x \ge \frac{1}{\sqrt2} > 0$⁠, so this is allowed:

   $$\frac{\sin x\cos x}{\cos^4x + \sin^4x} = \frac{\frac{\sin x}{\cos x}\cdot\frac{1}{\cos^2x}}{1 + \frac{\sin^4x}{\cos^4x}} = \frac{\tan x\sec^2x}{1 + \tan^4x} .$$
   Why this method: $\sec^2x\,dx$ is the differential of $\tan x$⁠, so after the division the whole integrand is a function of $\tan x$ times something close to its differential. The numerator $\sin x\cos x$ divided by $\cos^4x$ is $\frac{\sin x}{\cos^3x}$⁠, split as $\frac{\sin x}{\cos x}\cdot\frac{1}{\cos^2x}$⁠.
   Basics: trigonometric identities

3. Substitute $t = \tan^2x$ and change the limits

   $$t = \tan^2x ,\qquad dt = 2\tan x\sec^2x\,dx .$$

   Limits: $x = 0$ gives $t = 0$⁠; $x = \frac{\pi}{4}$ gives $t = \tan^2\frac{\pi}{4} = 1$⁠. And $\tan^4x = t^2$⁠. So

   $$\int_0^{\pi/4}\frac{\tan x\sec^2x\,dx}{1 + \tan^4x} = \frac12\int_0^1\frac{dt}{1 + t^2} .$$
   Why $\tan^2x$ and not $\tan x$⁠: with $u = \tan x$ we would get $\int\frac{u\,du}{1 + u^4}$⁠, which still needs a second substitution $t = u^2$⁠; taking $t = \tan^2x$ does both at once. The substitution is valid on the whole interval: $\tan^2x$ is differentiable on $[0, \frac{\pi}{4}]$ (where $\cos x \ne 0$⁠) and increases steadily from 0 to 1, so the new limits are 0 and 1 (Section 7.9).
   Section 7.9. Lesson: integration by substitution

4. Integrate and evaluate

   $$\frac12\int_0^1\frac{dt}{1+t^2} = \frac12\left[\tan^{-1}t\right]_0^1 = \frac12\left(\frac{\pi}{4} - 0\right) = \frac{\pi}{8} .$$

   So

   $$\int_0^{\pi/4}\frac{\sin x\cos x}{\cos^4x + \sin^4x}\,dx = \frac{\pi}{8} \approx 0.3926990817 .$$
   Standard formula (x), then Theorem 2 in the variable $t$⁠; $\tan^{-1}1 = \frac{\pi}{4}$ and $\tan^{-1}0 = 0$ are principal values. Because the limits were changed, there is no need to go back to $x$⁠.
   Theorem 2, Section 7.8.3. Basics: area as a sum of thin strips

5. Check: an antiderivative in $x$⁠, differentiated

   Steps 3 and 4 say $F(x) = \frac12\tan^{-1}(\tan^2x)$ is an antiderivative on $[0, \frac{\pi}{4}]$⁠. Differentiating with the chain rule,

   $$\begin{aligned}F'(x) &= \frac12\cdot\frac{1}{1 + \tan^4x}\cdot 2\tan x\sec^2x\\ &= \frac{\tan x\sec^2x}{1 + \tan^4x} = g(x) ,\end{aligned}$$

   and $F(\frac{\pi}{4}) - F(0) = \frac12\cdot\frac{\pi}{4} - 0 = \frac{\pi}{8}$⁠, the same value.

   The last equality is Step 2. Evaluating in $x$ with the old limits and in $t$ with the new limits must give the same number; this confirms the limits were changed correctly.
   Basics: integration undoes differentiation

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