Miscellaneous · Q29: $\int_0^{\pi/4}\frac{\sin x+\cos x}{9+16\sin 2x}\,dx$

Basics you need

Solution

1. The integrand is continuous on $[0, \frac{\pi}{4}]$

   For $0 \le x \le \frac{\pi}{4}$⁠, $0 \le 2x \le \frac{\pi}{2}$⁠, so $0 \le \sin 2x \le 1$ and

   $$9 \le 9 + 16\sin 2x \le 25 .$$

   The denominator is never zero, so the integrand is continuous on the closed interval.

   Theorem 2 of Section 7.8.3 needs continuity on $[a, b]$⁠. A denominator of the form $9 + 16\sin 2x$ would vanish where $\sin 2x = -\frac{9}{16}$⁠, but on our interval $\sin 2x$ is never negative.
   Lesson: definite integrals and the fundamental theorem

2. Substitute $t = \sin x - \cos x$

   Then

   $$dt = (\cos x + \sin x)\,dx ,$$

   which is exactly the numerator times $dx$⁠. Squaring,

   $$\begin{aligned} t^2 &= \sin^2 x + \cos^2 x - 2\sin x\cos x \\ &= 1 - \sin 2x , \end{aligned}$$

   so $\sin 2x = 1 - t^2$⁠.

   We need a new variable whose differential is the numerator. $\sin x - \cos x$ is that variable. The choice $t = \sin x + \cos x$ also turns $\sin 2x$ into $t^2 - 1$⁠, but its differential is $(\cos x - \sin x)\,dx$⁠, which is not in the integrand. So only $t = \sin x - \cos x$ fits.
   $\frac{d}{dx}\sin x = \cos x$⁠, $\frac{d}{dx}(-\cos x) = \sin x$⁠. Basics: substitution

3. Rewrite the denominator and change the limits

   $$\begin{aligned} 9 + 16\sin 2x &= 9 + 16(1 - t^2) \\ &= 25 - 16t^2 . \end{aligned}$$

   Limits: at $x = 0$⁠, $t = 0 - 1 = -1$⁠; at $x = \frac{\pi}{4}$⁠, $t = \frac{1}{\sqrt2} - \frac{1}{\sqrt2} = 0$⁠. So

   $$I = \int_{-1}^{0}\frac{dt}{25 - 16t^2} .$$
   The substitution is valid on the whole interval: $\frac{dt}{dx} = \sin x + \cos x > 0$ for $0 \le x \le \frac{\pi}{4}$ (since $\cos x \ge \frac{1}{\sqrt2}$ there), so $t$ runs steadily from $-1$ up to 0. On $[-1, 0]$⁠, $25 - 16t^2$ lies between 9 and 25, so the new integrand is continuous too.
   $\sin\frac{\pi}{4} = \cos\frac{\pi}{4} = \frac{1}{\sqrt2}$⁠. Lesson: integration by substitution

4. Bring it to the form $\frac{1}{a^2 - t^2}$ and integrate

   Take out the 16:

   $$\frac{1}{25 - 16t^2} = \frac{1}{16}\cdot\frac{1}{\left(\frac54\right)^2 - t^2} .$$

   Formula 7.4 (2) with $a = \frac54$⁠, so $\frac{1}{2a} = \frac25$⁠:

   $$\begin{aligned} \int\frac{dt}{25 - 16t^2} &= \frac{1}{16}\cdot\frac25\log\left|\frac{\frac54 + t}{\frac54 - t}\right| \\ &= \frac{1}{40}\log\left|\frac{5 + 4t}{5 - 4t}\right| + C . \end{aligned}$$
   Formula 7.4 (2) needs the coefficient of $t^2$ to be 1, so the 16 must come out first; then $25 \div 16 = \left(\frac54\right)^2$ gives $a$⁠. In the last line we multiplied the top and bottom of the fraction inside the log by 4, which does not change it. The two constants $\frac{1}{16}$ and $\frac25$ multiply to $\frac{1}{40}$⁠.
   Section 7.4, formula (2). Lesson: integrals of particular functions

5. Check: differentiate the antiderivative

   Let $G(t) = \frac{1}{40}\big(\log|5 + 4t| - \log|5 - 4t|\big)$⁠. Then

   $$\begin{aligned} G'(t) &= \frac{1}{40}\left(\frac{4}{5 + 4t} + \frac{4}{5 - 4t}\right) \\ &= \frac{1}{40}\cdot\frac{4(5 - 4t) + 4(5 + 4t)}{25 - 16t^2} \\ &= \frac{1}{40}\cdot\frac{40}{25 - 16t^2} = \frac{1}{25 - 16t^2} \ ✓ \end{aligned}$$

   valid for $t \ne \pm\frac54$⁠. In terms of $x$⁠, the chain rule multiplies by $\frac{dt}{dx} = \sin x + \cos x$⁠, and $25 - 16t^2 = 9 + 16\sin 2x$⁠, which gives back the original integrand.

   The derivative of $\log|5 - 4t|$ is $\frac{-4}{5 - 4t}$⁠, and the minus sign in front of it turns that into $+\frac{4}{5 - 4t}$⁠. Getting this sign right is what makes the two fractions add rather than cancel.
   $\frac{d}{dt}\log|u| = \frac{u'}{u}$⁠. Basics: logarithms and 1/x

6. Evaluate between $t = -1$ and $t = 0$

   At $t = 0$⁠: $\frac{1}{40}\log\frac55 = \frac{1}{40}\log 1 = 0$⁠.

   At $t = -1$⁠: $\frac{1}{40}\log\left|\frac{5 - 4}{5 + 4}\right| = \frac{1}{40}\log\frac19 = -\frac{1}{40}\log 9$⁠.

   $$I = 0 - \left(-\frac{1}{40}\log 9\right) = \frac{1}{40}\log 9 .$$

   Since $\log 9 = 2\log 3$⁠, this is also $\frac{1}{20}\log 3$⁠. As a decimal, $I \approx 0.054931$⁠.

   Theorem 2: upper value minus lower value. Both values of $\frac{5 + 4t}{5 - 4t}$ are positive on $[-1, 0]$⁠, so the absolute value bars change nothing. Size check: the integrand falls from $\frac19 \approx 0.111$ at $x = 0$ to $\frac{\sqrt2}{25} \approx 0.057$ at $x = \frac{\pi}{4}$⁠, and the interval has length $\frac{\pi}{4} \approx 0.785$⁠, so $I$ lies between $0.057 \times 0.785 \approx 0.044$ and $0.111 \times 0.785 \approx 0.087$⁠. It does.
   $\log 1 = 0$⁠, $\log\frac19 = -\log 9$⁠, $\log 9 = \log 3^2 = 2\log 3$⁠. Basics: logarithms

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