Exercise 7.2 · Q27: $\int\sqrt{\sin 2x}\,\cos 2x\,dx$

Basics you need

Solution

1. Where is the integrand defined?

   A square root of a real number needs the number to be $\ge 0$⁠, so we need $\sin 2x \ge 0$⁠. The sine of an angle is $\ge 0$ when the angle lies in $[2n\pi,\ 2n\pi + \pi]$⁠, $n$ an integer. With angle $2x$⁠, halve everything:

   $$\sin 2x \ge 0 \iff n\pi \le x \le n\pi + \tfrac{\pi}{2}.$$

   So the integrand lives on the intervals $[0, \frac{\pi}{2}]$⁠, $[\pi, \frac{3\pi}{2}]$⁠, $[-\pi, -\frac{\pi}{2}]$⁠, and so on. We integrate on any one of these intervals.

   The book's Note in Section 7.2 says we usually do not mention the interval, "however, in any specific problem one has to keep it in mind". Here it matters: between these intervals $\sin 2x$ is negative and the integrand simply does not exist, so the answer can only be claimed where the integrand exists.
   $\sin\theta \ge 0$ for $0 \le \theta \le \pi$⁠, and sine repeats every $2\pi$⁠. Basics: trigonometric facts

2. Choose the substitution $t = \sin 2x$ and write out $dt$

   Put

   $$t = \sin 2x.$$

   Differentiate both sides with respect to $x$ (chain rule, inner function $2x$⁠, outer function $\sin$⁠):

   $$\frac{dt}{dx} = 2\cos 2x, \quad\text{so}\quad dt = 2\cos 2x\,dx.$$

   Divide by 2:

   $$\cos 2x\,dx = \tfrac12\,dt.$$

   On the intervals of Step 1, $t = \sin 2x$ runs over $0 \le t \le 1$⁠.

   The book's rule of thumb (Section 7.3.1): "we make a substitution for a function whose derivative also occurs in the integrand". The function $\sin 2x$ sits inside the root, and its derivative $2\cos 2x$ appears outside, apart from the constant 2. A missing constant is never a problem: we just divide by it.
   If $t = g(x)$⁠, then $dt = g'(x)\,dx$⁠. Lesson: integration by substitution

3. Rewrite the integral in $t$

   Group the integrand as (root) times ($\cos 2x\,dx$⁠) and replace each part:

   $$\begin{aligned} \int \sqrt{\sin 2x}\,\cos 2x\,dx &= \int \sqrt{t}\cdot\tfrac12\,dt \\ &= \tfrac12\int t^{1/2}\,dt. \end{aligned}$$
   After a substitution, no $x$ may be left over: every piece must be in $t$⁠. Here $\sqrt{\sin 2x}$ became $\sqrt t$ and $\cos 2x\,dx$ became $\frac12\,dt$⁠, so nothing in $x$ remains. The constant $\frac12$ comes out in front by Property (IV) of Section 7.2.1, $\int k f = k\int f$⁠.
   $\sqrt t = t^{1/2}$ for $t \ge 0$⁠. Basics: powers

4. Integrate the power of $t$

   Power rule with $n = \frac12$⁠, so $n + 1 = \frac32$⁠:

   $$\begin{aligned} \tfrac12\int t^{1/2}\,dt &= \tfrac12\cdot\frac{t^{3/2}}{3/2} + C \\ &= \tfrac12\cdot\tfrac23\,t^{3/2} + C \\ &= \tfrac13\,t^{3/2} + C. \end{aligned}$$
   Dividing by $\frac32$ is the same as multiplying by $\frac23$⁠. Then $\frac12\cdot\frac23 = \frac13$⁠. The power rule applies because $n = \frac12 \ne -1$⁠.
   Standard formula (i), Section 7.2: $\int t^n\,dt = \frac{t^{n+1}}{n+1} + C$⁠, $n \ne -1$⁠.

5. Return to $x$

   Replace $t$ by $\sin 2x$⁠:

   $$\int \sqrt{\sin 2x}\,\cos 2x\,dx = \tfrac13(\sin 2x)^{3/2} + C.$$

   This holds on each interval $n\pi \le x \le n\pi + \frac{\pi}{2}$ where the integrand is defined.

   The question was asked in $x$⁠, so the answer must be in $x$⁠. The power $\frac32$ means $(\sin 2x)\sqrt{\sin 2x}$⁠, which makes sense exactly because $\sin 2x \ge 0$ there.

6. Check: differentiate the answer back

   Let $F(x) = \frac13(\sin 2x)^{3/2}$⁠. Chain rule twice: outer $u^{3/2}$⁠, middle $\sin$⁠, inner $2x$⁠:

   $$\begin{aligned} F'(x) &= \tfrac13\cdot\tfrac32(\sin 2x)^{1/2}\cdot\cos 2x\cdot 2 \\ &= \sqrt{\sin 2x}\,\cos 2x. \end{aligned}$$

   This is the integrand, so the answer is right. ✓

   This check is the definition of the indefinite integral (Property (I), Section 7.2.1): $F$ is an antiderivative of $f$ exactly when $F' = f$⁠. At an endpoint of an interval, where $\sin 2x = 0$⁠, the derivative is one-sided; it is $0$ there, which again equals the integrand.
   Chain rule, Theorem 4, Section 5.3.1. Basics: integration undoes differentiation

Every fact used, and where it comes from