Miscellaneous · Q39 (MCQ): $\int\frac{\cos 2x}{(\sin x+\cos x)^2}\,dx$

Basics you need

Solution

1. Where is the integrand defined?

   We need $\sin x + \cos x \ne 0$⁠, that is $\tan x \ne -1$⁠, so

   $$x \ne -\frac{\pi}{4} + n\pi, \quad n \text{ an integer}.$$

   We work on any interval between consecutive such points, for example $\left(-\frac{\pi}{4}, \frac{3\pi}{4}\right)$⁠.

   $\sin x + \cos x = \sqrt2\sin\left(x + \frac{\pi}{4}\right)$⁠, which is zero exactly when $x + \frac{\pi}{4}$ is a multiple of $\pi$⁠. At those points the integrand is undefined, and so is $\log|\sin x + \cos x|$⁠.

2. Factorise $\cos 2x$ and cancel

   $$\begin{aligned} &\frac{\cos 2x}{(\sin x + \cos x)^2} \\ &= \frac{(\cos x - \sin x)(\cos x + \sin x)}{(\sin x + \cos x)^2} \\ &= \frac{\cos x - \sin x}{\sin x + \cos x} . \end{aligned}$$
   $\cos 2x = \cos^2 x - \sin^2 x$ is a difference of squares. Of the three forms of $\cos 2x$⁠, this is the one that shares a factor with the denominator. The cancellation is allowed because $\sin x + \cos x \ne 0$ on the domain (Step 1).
   $\cos 2x = \cos^2 x - \sin^2 x$⁠. Basics: trigonometric identities

3. Substitute $t = \sin x + \cos x$

   Then $dt = (\cos x - \sin x)\,dx$⁠, exactly the numerator, and

   $$\begin{aligned} \int\frac{\cos x - \sin x}{\sin x + \cos x}\,dx &= \int\frac{dt}{t} \\ &= \log|t| + C . \end{aligned}$$

   So

   $$\begin{aligned} &\int\frac{\cos 2x}{(\sin x + \cos x)^2}\,dx \\ &= \log|\sin x + \cos x| + C . \end{aligned}$$
   The numerator is the derivative of the denominator, the pattern $\int\frac{g'}{g} = \log|g|$⁠. The absolute value is needed because $\sin x + \cos x$ is negative on some intervals of the domain, for example on $\left(\frac{3\pi}{4}, \frac{7\pi}{4}\right)$⁠.
   Section 7.3.1; standard formula (xii), Section 7.2. Lesson: integration by substitution

4. Check: differentiate the answer back

   $$\begin{aligned} &\frac{d}{dx}\log|\sin x + \cos x| \\ &= \frac{\cos x - \sin x}{\sin x + \cos x} \\ &=\frac{(\cos x - \sin x)(\cos x + \sin x)}{(\sin x + \cos x)^2} \\ &= \frac{\cos 2x}{(\sin x + \cos x)^2} \ ✓ \end{aligned}$$
   $\frac{d}{dx}\log|u| = \frac{u'}{u}$⁠. The second line reverses Step 2 by multiplying top and bottom by $\sin x + \cos x$⁠.
   Basics: integration undoes differentiation

5. Examine every option

   Use the general point $x = \frac{\pi}{6}$⁠: $s = \sin x + \cos x = \frac{1 + \sqrt3}{2}$⁠, $\cos x - \sin x = \frac{\sqrt3 - 1}{2}$⁠, and $s^2 = 1 + \sin 2x = \frac{2 + \sqrt3}{2}$⁠. The integrand there is $\frac{\sqrt3 - 1}{\sqrt3 + 1} = 2 - \sqrt3 \approx 0.267949$⁠.

   • (A) $\frac{d}{dx}\left(\frac{-1}{s}\right) = \frac{s'}{s^2} = \frac{\cos x - \sin x}{(\sin x + \cos x)^2}$⁠. At $\frac{\pi}{6}$⁠: $\frac{\sqrt3 - 1}{2 + \sqrt3}$ $= 3\sqrt3 - 5$ $\approx 0.196152$ $\ne 0.267949$⁠. Wrong. (At $x = 0$ both happen to equal 1, a coincidence: that is why a special point is not enough.)
   • (B) Derivative equals the integrand everywhere on the domain (Step 4); at $\frac{\pi}{6}$ it is $2 - \sqrt3$ ✓. Correct.
   • (C) $\frac{d}{dx}\log|\sin x - \cos x| = \frac{\cos x + \sin x}{\sin x - \cos x}$⁠. At $\frac{\pi}{6}$⁠: $\frac{(1 + \sqrt3)/2}{(1 - \sqrt3)/2}$ $= -(2 + \sqrt3)$ $\approx -3.732051$⁠. Wrong.
   • (D) $\frac{d}{dx}s^{-2}$ $= -2s^{-3}s'$ $= \frac{-2(\cos x - \sin x)}{(\sin x + \cos x)^3}$⁠. At $\frac{\pi}{6}$⁠: $\approx -0.287187$⁠. Wrong. (It also lacks the constant $C$⁠, so it could not be "the" indefinite integral anyway.)
   Option (D) is just the denominator's reciprocal, a reminder that integrating is not "dropping the numerator". Option (C) has the wrong sign inside: $\sin x - \cos x$ is not the expression whose derivative is the numerator.

Every fact used, and where it comes from