Miscellaneous · Q27: $\int_{\pi/6}^{\pi/3}\frac{\sin x+\cos x}{\sqrt{\sin 2x}}\,dx$

Basics you need

Solution

1. The integrand is continuous on $[\frac{\pi}{6}, \frac{\pi}{3}]$

   For $\frac{\pi}{6} \le x \le \frac{\pi}{3}$ we have $\frac{\pi}{3} \le 2x \le \frac{2\pi}{3}$⁠, so

   $$\frac{\sqrt3}{2} \le \sin 2x \le 1 .$$

   So $\sin 2x$ is positive, its square root is defined and at least $\sqrt{\sqrt3/2} \approx 0.9306$⁠, and the denominator is never zero. The numerator is continuous, so the whole integrand is continuous on the closed interval.

   Theorem 2 of Section 7.8.3 (the second fundamental theorem) needs a function continuous on $[a, b]$⁠; Remark (iv) after it warns that the integrand must be well defined there. A square root of $\sin 2x$ would fail wherever $\sin 2x < 0$⁠, and the fraction would blow up where $\sin 2x = 0$⁠. Neither happens here.
   On $[\frac{\pi}{3}, \frac{2\pi}{3}]$ the sine is smallest at the ends, where it equals $\frac{\sqrt3}{2}$⁠. Lesson: definite integrals and the fundamental theorem

2. Choose the substitution $t = \sin x - \cos x$

   Differentiate the candidate:

   $$\frac{d}{dx}(\sin x - \cos x) = \cos x + \sin x .$$

   So with $t = \sin x - \cos x$⁠,

   $$dt = (\sin x + \cos x)\,dx ,$$

   which is exactly the numerator times $dx$⁠.

   Substitution works when the integrand contains a function together with its derivative. The derivative of $\sin x - \cos x$ is $\sin x + \cos x$⁠, the numerator. The other natural guess, $t = \sin x + \cos x$⁠, has derivative $\cos x - \sin x$⁠, which is not in the integrand, so it would not simplify anything.
   $\frac{d}{dx}\sin x = \cos x$⁠, $\frac{d}{dx}\cos x = -\sin x$⁠. Basics: substitution

3. Write $\sin 2x$ in terms of $t$

   Square $t$ and use the two identities:

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

   Hence $\sin 2x = 1 - t^2$ and $\sqrt{\sin 2x} = \sqrt{1 - t^2}$⁠.

   This is the second job of the same substitution. The cross term $-2\sin x\cos x$ in the square is $-\sin 2x$⁠, and the squares add up to 1. Because $\sin 2x > 0$ on our interval (Step 1), $1 - t^2 > 0$ as well, so the square root is the ordinary positive root.
   $(p - q)^2 = p^2 - 2pq + q^2$⁠; $\sin^2 x + \cos^2 x = 1$⁠; $2\sin x\cos x = \sin 2x$⁠. Basics: trigonometric identities

4. Change the limits

   At $x = \frac{\pi}{6}$⁠: $t = \frac12 - \frac{\sqrt3}{2} = -\frac{\sqrt3 - 1}{2}$⁠.

   At $x = \frac{\pi}{3}$⁠: $t = \frac{\sqrt3}{2} - \frac12 = \frac{\sqrt3 - 1}{2}$⁠.

   Write $k = \frac{\sqrt3 - 1}{2} \approx 0.366025$⁠. Then

   $$I = \int_{-k}^{k} \frac{dt}{\sqrt{1 - t^2}} .$$
   The substitution is valid on the whole interval: $t = \sin x - \cos x$ is differentiable, and $\frac{dt}{dx} = \sin x + \cos x > 0$ there (both terms are positive in the first quadrant), so $t$ increases steadily from $-k$ to $k$ without turning back. Every $t$ in $[-k, k]$ has $|t| \le 0.37 < 1$⁠, so $\frac{1}{\sqrt{1-t^2}}$ is continuous on the new interval too.
   $\sin\frac{\pi}{6} = \cos\frac{\pi}{3} = \frac12$⁠, $\cos\frac{\pi}{6} = \sin\frac{\pi}{3} = \frac{\sqrt3}{2}$⁠. Lesson: integration by substitution

5. Integrate with standard formula (viii)

   $$\begin{aligned} I &= \Big[\sin^{-1} t\Big]_{-k}^{k} \\ &= \sin^{-1} k - \sin^{-1}(-k) \\ &= 2\sin^{-1} k . \end{aligned}$$

   So

   $$I = 2\sin^{-1}\left(\frac{\sqrt3 - 1}{2}\right).$$
   The last line uses that $\sin^{-1}$ is odd (Basic 4). Another way to see the factor 2: $\frac{1}{\sqrt{1-t^2}}$ is an even function of $t$⁠, so P7 (i) of Section 7.10 gives $\int_{-k}^{k} = 2\int_0^{k}$⁠, which is again $2\sin^{-1} k$⁠. The value $\frac{\sqrt3 - 1}{2}$ is not the sine of any standard angle, so the answer stays in this form.
   Theorem 2, Section 7.8.3: $\int_a^b f = F(b) - F(a)$ for an antiderivative $F$⁠. P7 (i): $\int_{-a}^{a} f = 2\int_0^a f$ for even $f$⁠. Lesson: properties of definite integrals

6. Check: differentiate the antiderivative in $x$

   Going back to $x$⁠, the antiderivative is $F(x) = \sin^{-1}(\sin x - \cos x)$⁠. By the chain rule (outer $\sin^{-1}$⁠, inner $\sin x - \cos x$⁠):

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

   This is the integrand ✓, valid wherever $\sin 2x > 0$⁠, which includes all of $[\frac{\pi}{6}, \frac{\pi}{3}]$⁠. Then $F(\frac{\pi}{3}) - F(\frac{\pi}{6}) = \sin^{-1}k - \sin^{-1}(-k)$⁠, the same as Step 5.

   Differentiating back is the surest test that no factor was lost in the substitution. The second line is Step 3 read backwards: $1 - (\sin x - \cos x)^2 = \sin 2x$⁠.
   $\frac{d}{du}\sin^{-1} u = \frac{1}{\sqrt{1-u^2}}$ for $|u| < 1$⁠. Basics: integration undoes differentiation

7. The decimal value

   $k = \frac{\sqrt3 - 1}{2} \approx 0.3660254$⁠, $\sin^{-1} k \approx 0.3747344$⁠, so

   $$I = 2\sin^{-1}\left(\frac{\sqrt3 - 1}{2}\right) \approx 0.749469 .$$

   It is positive, as it must be: the integrand is positive on the whole interval.

   A rough size check: the integrand stays between about 1.41 and 1.47 on the interval (at the ends it is $\frac{(1+\sqrt3)/2}{\sqrt{\sqrt3/2}} \approx 1.468$⁠; at the middle point $\frac{\pi}{4}$ it is $\sqrt2 \approx 1.414$⁠, its smallest value), and the interval has length $\frac{\pi}{6} \approx 0.5236$⁠. So the integral must lie between $1.414 \times 0.5236 \approx 0.740$ and $1.468 \times 0.5236 \approx 0.769$⁠. It does.

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