Exercise 7.10 · Q12: $\int_0^{\pi}\frac{x}{1+\sin x}\,dx$

Basics you need

Solution

1. The integrand is continuous on $[0, \pi]$

   For $0 \le x \le \pi$ we have $\sin x \ge 0$⁠, so

   $$1 + \sin x \ge 1 > 0 .$$

   The integrand $\frac{x}{1+\sin x}$ is a quotient of continuous functions whose denominator never vanishes, so it is continuous on the whole closed interval. $I$ is an ordinary definite integral and the properties of Section 7.10 apply.

   Every property P0 to P7 and Theorem 2 are stated for functions continuous on the closed interval (Remark (iv) after Theorem 2). Checking this first means nothing later can go wrong because of a zero denominator.
   $\sin x \ge 0$ on $[0, \pi]$⁠: the angle is in the first or second quadrant (or on the axis).

2. P4 with $a = \pi$⁠: replace $x$ by $\pi - x$

   P4 says $\int_0^{\pi} f(x)\,dx = \int_0^{\pi} f(\pi - x)\,dx$⁠. With $f(x) = \frac{x}{1+\sin x}$⁠:

   $$f(\pi - x) = \frac{\pi - x}{1 + \sin(\pi - x)} = \frac{\pi - x}{1 + \sin x} .$$

   So

   $$I = \int_0^{\pi}\frac{\pi - x}{1+\sin x}\,dx .$$
   P4 has no extra condition: it holds for any continuous $f$ (Step 1). We use it because it changes the numerator $x$ into $\pi - x$ while leaving the denominator exactly as it was, thanks to $\sin(\pi - x) = \sin x$⁠. That is what makes the next step work.
   $\sin(\pi - x) = \sin x$⁠: the angles $x$ and $\pi - x$ are mirror images in the $y$-axis of the unit circle, so they have the same height. Basics: symmetry and mirror images

3. Add the two forms of $I$

   Add the original integral and the result of Step 2. Both run over the same interval $[0, \pi]$⁠, so we may simply add the integrands:

   $$\begin{aligned}2I &= \int_0^{\pi}\frac{x + (\pi - x)}{1+\sin x}\,dx \\ &= \pi\int_0^{\pi}\frac{dx}{1+\sin x} .\end{aligned}$$

   Write $J = \int_0^{\pi}\frac{dx}{1+\sin x}$⁠. Then $2I = \pi J$⁠, that is, $I = \frac{\pi}{2}J$⁠.

   The sum of two integrals over the same interval is the integral of the sum (Properties (III) and (IV) of Section 7.2.1, which carry over to definite integrals through Theorem 2). The $x$ in the numerator was the only obstacle; $x + (\pi - x) = \pi$ is a constant, which comes out of the integral.
   $\int_a^b (f + g)\,dx = \int_a^b f\,dx + \int_a^b g\,dx$ and $\int_a^b k f\,dx = k\int_a^b f\,dx$⁠. Basics: area as a sum of thin strips

4. An antiderivative of $\frac{1}{1+\sin x}$

   Multiply top and bottom by $1 - \sin x$ (allowed where $\sin x \ne 1$⁠, that is $x \ne \frac{\pi}{2}$ on our interval):

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

   Both terms are standard derivatives, so an antiderivative is $\tan x - \sec x$⁠. This form is undefined at $x = \frac{\pi}{2}$⁠, so simplify it:

   $$\begin{aligned}\tan x - \sec x &= \frac{\sin x - 1}{\cos x} \\ &= \frac{(\sin x - 1)(1 + \sin x)}{\cos x\,(1+\sin x)} \\ &= \frac{-\cos^2 x}{\cos x\,(1 + \sin x)} = \frac{-\cos x}{1 + \sin x}.\end{aligned}$$

   So we take

   $$F(x) = -\frac{\cos x}{1 + \sin x} ,$$

   which is defined and differentiable on all of $[0, \pi]$⁠, including $x = \frac{\pi}{2}$⁠, because its denominator is at least 1 (Step 1).

   Theorem 2 needs an antiderivative on the whole closed interval $[0, \pi]$⁠. The form $\tan x - \sec x$ has a gap at $\frac{\pi}{2}$⁠, right in the middle of the interval. The simplified form $F$ agrees with it everywhere else and fills the gap, so it is the one to use.
   Multiplying by $\frac{1 - \sin x}{1 - \sin x}$ is the "conjugate" trick: $(1+\sin x)(1-\sin x)$ $= 1 - \sin^2 x = \cos^2 x$⁠. Basics: trigonometric identities

5. Check: differentiate $F$ back

   Quotient rule with $u = -\cos x$ ($u' = \sin x$⁠) and $v = 1 + \sin x$ ($v' = \cos x$⁠):

   $$\begin{aligned}F'(x) &= \frac{\sin x\,(1 + \sin x) + \cos x\cos x}{(1+\sin x)^2} \\ &= \frac{\sin x + \sin^2 x + \cos^2 x}{(1+\sin x)^2} \\ &= \frac{1 + \sin x}{(1+\sin x)^2} = \frac{1}{1+\sin x} .\end{aligned}$$

   This is the integrand of $J$ at every point of $[0, \pi]$⁠.

   Differentiating back is the surest test of an antiderivative. It also confirms that $F$ works at $x = \frac{\pi}{2}$⁠, where the intermediate form of Step 4 did not exist.
   Here $u'v = \sin x\,(1 + \sin x)$ and $uv' = -\cos^2 x$⁠, so $u'v - uv'$ has $+\cos^2 x$⁠; then $\sin^2 x + \cos^2 x = 1$⁠. Basics: integration undoes differentiation

6. Evaluate $J$ and then $I$

   By Theorem 2 with $F$ from Step 4:

   $$\begin{aligned}J &= F(\pi) - F(0) \\ &= -\frac{\cos\pi}{1 + \sin\pi} + \frac{\cos 0}{1 + \sin 0} \\ &= -\frac{-1}{1} + \frac{1}{1} = 2 .\end{aligned}$$

   Then, from Step 3,

   $$I = \frac{\pi}{2}\,J = \frac{\pi}{2}\cdot 2 = \pi .$$
   $\cos\pi = -1$⁠, $\sin\pi = 0$⁠, $\cos 0 = 1$⁠, $\sin 0 = 0$⁠. The answer is positive, as it must be: the integrand is $\ge 0$ on $[0, \pi]$⁠, so its integral is the (positive) shaded area in the figure.
   Theorem 2, Section 7.8.3: value of the antiderivative at the upper limit minus its value at the lower limit. Lesson: definite integrals

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