Example 35: $\int\cos 6x\sqrt{1+\sin 6x}\,dx$

Basics you need

Solution

1. Choose the substitution $t = 1 + \sin 6x$

   Put $t = 1 + \sin 6x$⁠. Differentiating with the chain rule,

   $$\frac{dt}{dx} = 6\cos 6x, \qquad\text{so}\qquad dt = 6\cos 6x\,dx$$

   and therefore $\cos 6x\,dx = \frac{1}{6}\,dt$⁠.

   We substitute for the expression whose derivative is sitting in the integrand. The $\cos 6x\,dx$ is then completely absorbed into $dt$⁠, and the only thing left is the root, which becomes $\sqrt t$⁠. Choosing $t = \sin 6x$ would also work (then $\sqrt{1 + t}$ remains); taking the whole inside of the root is one step shorter.
   The derivative of the constant 1 is 0, so $t = 1 + \sin 6x$ and $t = \sin 6x$ have the same differential. Lesson: integration by substitution

2. Rewrite the integral in $t$

   $$\int \cos 6x\sqrt{1+\sin 6x}\,dx = \int \sqrt{t}\cdot\frac{1}{6}\,dt = \frac{1}{6}\int t^{1/2}\,dt .$$
   Every $x$ has disappeared: $\sqrt{1 + \sin 6x}$ became $\sqrt t$ and $\cos 6x\,dx$ became $\frac16\,dt$⁠. A substitution is successful exactly when no $x$ is left over. The constant $\frac16$ comes out of the integral (Property IV, Section 7.2.1). Since $t = 1 + \sin 6x$ lies in $[0, 2]$⁠, $\sqrt t$ is always defined.
   $\sqrt t = t^{1/2}$ for $t \ge 0$⁠. Basics: powers

3. Integrate with the power rule

   $$\begin{aligned}\frac{1}{6}\int t^{1/2}\,dt &= \frac{1}{6}\cdot\frac{t^{3/2}}{3/2} + C\\ &= \frac{1}{6}\cdot\frac{2}{3}\,t^{3/2} + C = \frac{1}{9}\,t^{3/2} + C .\end{aligned}$$
   The power rule raises the exponent by 1 ($\frac12 + 1 = \frac32$⁠) and divides by the new exponent; dividing by $\frac32$ is multiplying by $\frac23$⁠. Then $\frac16 \cdot \frac23 = \frac{2}{18} = \frac19$⁠.
   Standard formula (i) of Section 7.2: $\int x^n\,dx = \frac{x^{n+1}}{n+1} + C$⁠, $n \ne -1$⁠.

4. Substitute back

   $$\begin{aligned}&\int \cos 6x\sqrt{1+\sin 6x}\,dx\\ &= \frac{1}{9}\,(1 + \sin 6x)^{3/2} + C .\end{aligned}$$
   The question is in $x$⁠, so the answer must be in $x$⁠. Replace $t$ by $1 + \sin 6x$⁠.

5. Check: $\frac{d}{dx}$ of our answer

   Let $F(x) = \frac19(1 + \sin 6x)^{3/2}$⁠. By the chain rule (outer: $u^{3/2}$⁠; inner: $u = 1 + \sin 6x$⁠, with $u' = 6\cos 6x$⁠):

   $$\begin{aligned}F'(x) &= \frac19\cdot\frac32(1+\sin 6x)^{1/2}\cdot 6\cos 6x\\ &= \frac{18}{18}\cos 6x\sqrt{1+\sin 6x}\\ &= \cos 6x\sqrt{1+\sin 6x} .\end{aligned}$$

   This is the integrand, so the answer is right.

   Differentiation undoes integration, so differentiating the answer must return the integrand exactly. The constants $\frac19 \cdot \frac32 \cdot 6 = 1$ cancel perfectly. The check holds for every real $x$⁠; at the points where $\sin 6x = -1$ both sides are 0 ($u^{3/2}$ is flat where $u = 0$⁠, and $u$ never goes negative), so the answer is valid on the whole real line.
   Basics: integration undoes differentiation

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