Exercise 7.2 · Q30: $\int\frac{\sin x}{1+\cos x}\,dx$

Basics you need

Solution

1. Where is the integrand defined?

   The denominator must not be zero: $1 + \cos x \ne 0$⁠, that is, $\cos x \ne -1$⁠, that is,

   $$x \ne (2n+1)\pi, \quad n \text{ an integer}.$$

   Since $\cos x \ge -1$ always, at every other point $1 + \cos x > 0$⁠. So on the domain the denominator is positive.

   This positivity will matter in Step 4: it lets us drop the modulus in $\log|1 + \cos x|$⁠. Knowing the sign of what goes into a logarithm is always worth a line.
   $\cos\pi = -1$⁠, and cosine repeats every $2\pi$⁠. Basics: trigonometric facts

2. Substitute $t = 1 + \cos x$

   Put $t = 1 + \cos x$⁠. Then

   $$\frac{dt}{dx} = -\sin x, \quad\text{so}\quad \sin x\,dx = -dt.$$

   The integral becomes

   $$\int \frac{\sin x}{1 + \cos x}\,dx = \int \frac{-dt}{t} = -\int \frac{dt}{t}.$$
   The numerator $\sin x\,dx$ is, apart from the sign, the differential of the denominator. The minus sign comes from $\frac{d}{dx}\cos x = -\sin x$⁠; it is easy to lose, so we write $\sin x\,dx = -dt$ explicitly before replacing anything.
   This is the same idea the book uses for $\int\tan x\,dx$ in Section 7.3.1 (put $\cos x = t$⁠, $\sin x\,dx = -dt$⁠). Lesson: integration by substitution

3. Integrate $\frac1t$

   $$-\int \frac{dt}{t} = -\log|t| + C.$$
   The power rule fails for $n = -1$⁠; the integral of $\frac1t$ is the logarithm of $|t|$⁠. The modulus makes the formula valid for negative $t$ too.
   Standard formula (xii), Section 7.2. Basics: logarithms and 1/x

4. Return to $x$⁠; the modulus is harmless

   $$\int \frac{\sin x}{1 + \cos x}\,dx = -\log|1 + \cos x| + C.$$

   By Step 1, $1 + \cos x > 0$ on the whole domain, so $|1 + \cos x| = 1 + \cos x$⁠, and the answer may equally be written $-\log(1 + \cos x) + C$⁠.

   The textbook keeps the modulus, which is never wrong. Here it changes nothing, because the quantity inside is already positive wherever the integrand exists.

5. Check: differentiate the answer back

   Let $F(x) = -\log(1 + \cos x)$⁠. Chain rule, outer $-\log u$⁠, inner $u = 1 + \cos x$⁠:

   $$F'(x) = -\frac{1}{1 + \cos x}\cdot(-\sin x) = \frac{\sin x}{1 + \cos x}.$$

   This is the integrand. ✓

   Differentiating back is the definition of an antiderivative (Property (I), Section 7.2.1). The two minus signs cancel, which confirms that the sign in Step 2 was handled correctly.
   Basics: integration undoes differentiation

6. A second route through $\tan\frac{x}{2}$⁠, and why it agrees

   Using the half-angle identities,

   $$\begin{aligned} \frac{\sin x}{1 + \cos x} &= \frac{2\sin\frac{x}{2}\cos\frac{x}{2}}{2\cos^2\frac{x}{2}} \\ &= \tan\tfrac{x}{2}. \end{aligned}$$

   With $u = \frac{x}{2}$⁠, $dx = 2\,du$⁠, and $\int\tan u\,du = -\log|\cos u|$ (Section 7.3.1):

   $$\int\tan\tfrac{x}{2}\,dx = -2\log\left|\cos\tfrac{x}{2}\right| + C.$$

   This equals our answer up to a constant:

   $$\begin{aligned} -2\log\left|\cos\tfrac{x}{2}\right| &= -\log\cos^2\tfrac{x}{2} \\ &= -\log\frac{1 + \cos x}{2} \\ &= -\log(1 + \cos x) + \log 2. \end{aligned}$$
   Two correct methods can give answers that look different; they must then differ by a constant (Property (II), Section 7.2.1). Here the difference is $\log 2$⁠, which is absorbed into $C$⁠. Seeing that both routes agree is a strong check.
   $2\log|a| = \log a^2$ and $1 + \cos x = 2\cos^2\frac{x}{2}$⁠. Basics: trigonometric identities

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