Example 33: $\int_{\pi/6}^{\pi/3}\frac{dx}{1+\sqrt{\tan x}}$

Basics you need

Solution

1. The integrand is continuous on $[\pi/6, \pi/3]$

   For $\frac{\pi}{6} \le x \le \frac{\pi}{3}$ the angle is in the first quadrant, so $\sin x > 0$⁠, $\cos x > 0$ and $\tan x > 0$⁠. In fact $\tan x$ runs from $\tan\frac{\pi}{6} = \frac{1}{\sqrt3}$ up to $\tan\frac{\pi}{3} = \sqrt3$⁠. So $\sqrt{\tan x}$ is defined and continuous, and the denominator $1 + \sqrt{\tan x}$ is at least 1, never 0. The integrand is therefore continuous on the whole closed interval.

   A definite integral of a continuous function on a closed interval is an ordinary finite number (Remark (iv) after Theorem 2, Section 7.8.3), and all the properties P0 to P7 apply to it. Checking this first means nothing later can go wrong at an endpoint.
   $\tan x$ is increasing on $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$⁠, so its smallest and largest values on $[\pi/6, \pi/3]$ are at the two ends.

2. Rewrite the integrand with $\sqrt{\sin x}$ and $\sqrt{\cos x}$

   Since $\sin x > 0$ and $\cos x > 0$ here, $\sqrt{\tan x} = \frac{\sqrt{\sin x}}{\sqrt{\cos x}}$⁠. Then multiply the top and bottom by $\sqrt{\cos x}$⁠:

   $$\frac{1}{1+\sqrt{\tan x}} = \frac{1}{1+\dfrac{\sqrt{\sin x}}{\sqrt{\cos x}}} = \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}} .$$

   So

   $$I = \int_{\pi/6}^{\pi/3} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}\,dx \qquad (1)$$
   This form treats $\sin$ and $\cos$ on an equal footing: the denominator is symmetric in them. That is exactly what makes the swap $\sin \leftrightarrow \cos$ in the next step useful. Splitting the square root of the quotient is allowed only because $\sin x$ and $\cos x$ are positive on our interval (Step 1).
   Multiplying the numerator and the denominator of a fraction by the same non-zero quantity ($\sqrt{\cos x} > 0$⁠) does not change its value. Basics: working with fractions

3. Apply P3 with $a = \pi/6$⁠, $b = \pi/3$

   Here $a + b = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2}$⁠, so P3 says we may replace $x$ by $\frac{\pi}{2} - x$ in the integrand of (1):

   $$I = \int_{\pi/6}^{\pi/3} \frac{\sqrt{\cos\left(\frac{\pi}{2}-x\right)}}{\sqrt{\cos\left(\frac{\pi}{2}-x\right)}+\sqrt{\sin\left(\frac{\pi}{2}-x\right)}}\,dx .$$

   Using $\cos\left(\frac{\pi}{2}-x\right) = \sin x$ and $\sin\left(\frac{\pi}{2}-x\right) = \cos x$⁠:

   $$I = \int_{\pi/6}^{\pi/3} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx \qquad (2)$$
   P3 holds for any integrand continuous on $[a, b]$ (Step 1). As $x$ runs over $[\pi/6, \pi/3]$⁠, the new point $\frac{\pi}{2} - x$ runs over the same interval backwards, from $\pi/3$ down to $\pi/6$⁠. So it adds up the same values of the integrand in the opposite order, and the total cannot change. The point of choosing $a + b = \pi/2$ is that this reflection is the complementary-angle swap.
   Proof of P3 (Section 7.10): put $t = a + b - x$⁠, so $dt = -dx$⁠; $x = a$ gives $t = b$ and $x = b$ gives $t = a$⁠; then use P1 to swap the limits back and P0 to rename $t$ as $x$⁠. Basics: symmetry and mirror images

4. Add (1) and (2)

   Both integrals have the same limits, so they can be combined into one integral (the sum property of Section 7.2.1 holds for definite integrals too):

   $$\begin{aligned}2I &= \int_{\pi/6}^{\pi/3} \frac{\sqrt{\cos x}+\sqrt{\sin x}}{\sqrt{\cos x}+\sqrt{\sin x}}\,dx\\ &= \int_{\pi/6}^{\pi/3} 1\,dx .\end{aligned}$$
   The two integrands have the same denominator, and their numerators add up to that denominator. So the sum is exactly 1 at every point of the interval. This is the whole trick: neither integral is easy on its own, but their sum is trivial.
   For continuous $f$ and $g$⁠: $$\int_a^b [f(x) + g(x)]\,dx = \int_a^b f(x)\,dx + \int_a^b g(x)\,dx .$$

5. Evaluate and solve for $I$

   $$2I = \Big[x\Big]_{\pi/6}^{\pi/3} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6} .$$

   Dividing by 2,

   $$I = \frac{\pi}{12} \approx 0.261799 .$$
   An antiderivative of 1 is $x$⁠, so Theorem 2 gives upper limit minus lower limit, which is the length of the interval. Halving is legitimate because $I$ is a finite number (Step 1).
   $\frac{\pi}{3} - \frac{\pi}{6} = \frac{2\pi - \pi}{6} = \frac{\pi}{6}$⁠.

6. See it in a picture

   Write $f(x) = \frac{1}{1+\sqrt{\tan x}}$⁠. Steps 3 and 4 say $f(x) + f\left(\frac{\pi}{2}-x\right) = 1$⁠. So at every $x$⁠, the gap from the curve up to the line $y = 1$ is $1 - f(x) = f\left(\frac{\pi}{2} - x\right)$⁠: the region above the curve (inside the rectangle) is the region under the curve turned half a turn about the centre point $\left(\frac{\pi}{4}, \frac12\right)$⁠.

   The algebra and the picture say the same thing. The curve passes through the centre $\left(\frac{\pi}{4}, \frac12\right)$ because $\tan\frac{\pi}{4} = 1$ gives $f\left(\frac{\pi}{4}\right) = \frac{1}{2}$⁠, and it is symmetric about that point. Any curve with this point symmetry cuts the rectangle into two equal halves.
   Basics: area as a sum of thin strips

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