Exercise 7.2 · Q28: $\int\frac{\cos x}{\sqrt{1+\sin x}}\,dx$

Basics you need

Solution

1. Where is the integrand defined?

   The root is in the denominator, so we need $1 + \sin x > 0$ (not just $\ge 0$⁠: a zero denominator is not allowed). Since $\sin x \ge -1$ always, $1 + \sin x \ge 0$ always, and it is zero only when $\sin x = -1$⁠:

   $$1 + \sin x = 0 \iff x = -\tfrac{\pi}{2} + 2n\pi.$$

   So the integrand is defined for every real $x$ except these points, that is, on each open interval $\left(-\frac{\pi}{2} + 2n\pi,\ \frac{3\pi}{2} + 2n\pi\right)$⁠.

   The answer can only be claimed where the integrand exists (Note, Section 7.2). Here only isolated points are lost, one in every $2\pi$⁠.
   $\sin x$ reaches its least value $-1$ at $x = -\frac{\pi}{2}$⁠, and repeats every $2\pi$⁠. Basics: trigonometric facts

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

   Put

   $$t = 1 + \sin x.$$

   Differentiate with respect to $x$⁠: the derivative of the constant 1 is 0 and of $\sin x$ is $\cos x$⁠, so

   $$\frac{dt}{dx} = \cos x, \quad\text{that is,}\quad dt = \cos x\,dx.$$

   On the domain of Step 1, $t > 0$ (indeed $0 < t \le 2$⁠).

   The function under the root, $1 + \sin x$⁠, has its derivative $\cos x$ sitting in the numerator. That is the signal for this substitution (Section 7.3.1). Including the "1" in $t$ is what makes the root become simply $\sqrt t$⁠.
   $\frac{d}{dx}\sin x = \cos x$ (standard formula (ii), Section 7.2, read as a derivative). Lesson: integration by substitution

3. Rewrite the integral in $t$

   $$\begin{aligned} &\int \frac{\cos x}{\sqrt{1 + \sin x}}\,dx \\ &= \int \frac{1}{\sqrt{1 + \sin x}}\cdot\cos x\,dx \\ &= \int \frac{1}{\sqrt t}\,dt = \int t^{-1/2}\,dt. \end{aligned}$$
   We separate the integrand into "a function of $1 + \sin x$" times "$\cos x\,dx$", then replace each piece. No $x$ is left, so the substitution is complete. Writing $\frac{1}{\sqrt t}$ as $t^{-1/2}$ puts it in power-rule form.
   $\frac{1}{\sqrt t} = \frac{1}{t^{1/2}} = t^{-1/2}$⁠. Basics: powers

4. Integrate by the power rule

   With $n = -\frac12$⁠, $n + 1 = \frac12$⁠:

   $$\int t^{-1/2}\,dt = \frac{t^{1/2}}{1/2} + C = 2\sqrt t + C.$$
   The power rule needs $n \ne -1$⁠; here $n = -\frac12$⁠, so it applies. Dividing by $\frac12$ doubles.
   Standard formula (i), Section 7.2.

5. Return to $x$

   $$\int \frac{\cos x}{\sqrt{1 + \sin x}}\,dx = 2\sqrt{1 + \sin x} + C,$$

   valid wherever $\sin x \ne -1$ (on each interval of Step 1).

   Replacing $t$ by $1 + \sin x$ gives the answer in the original variable, as the question requires.

6. Check: differentiate the answer back

   Let $F(x) = 2(1 + \sin x)^{1/2}$⁠. Chain rule, outer $2u^{1/2}$⁠, inner $u = 1 + \sin x$⁠:

   $$\begin{aligned} F'(x) &= 2\cdot\tfrac12(1 + \sin x)^{-1/2}\cdot\cos x \\ &= \frac{\cos x}{\sqrt{1 + \sin x}}. \end{aligned}$$

   This is the integrand. ✓

   An antiderivative is checked by differentiating it (Property (I), Section 7.2.1). The derivative of $u^{1/2}$ needs $u > 0$⁠, which is exactly the domain from Step 1.
   Chain rule, Theorem 4, Section 5.3.1. Basics: integration undoes differentiation

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