Chapter 7 · Integrals

Exercise 7.10 · Q2: $\int_0^{\pi/2}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx$

Exercise 7.10Properties of definite integrals (P4) · Section 7.10

The question

By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

2. $\displaystyle\int_0^{\frac{\pi}{2}}\frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}}\,dx$

Exam answer

Let

\begin{aligned} I &= \int_0^{\frac{\pi}{2}}\frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}}\,dx \\ &\qquad\ldots(\text{i}) \end{aligned}

Using $\displaystyle\int_0^a f(x)\,dx = \int_0^a f(a - x)\,dx$ with $\sin\left(\frac{\pi}{2} - x\right) = \cos x$ and $\cos\left(\frac{\pi}{2} - x\right) = \sin x$ ⁠,

\begin{aligned} I &= \int_0^{\frac{\pi}{2}}\frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}}\,dx \\ &\qquad\ldots(\text{ii}) \end{aligned}

Adding (i) and (ii),

\begin{aligned} 2I &= \int_0^{\frac{\pi}{2}}\frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}}\,dx \\ &= \int_0^{\frac{\pi}{2}}1\,dx = \Big[x\Big]_0^{\frac{\pi}{2}} = \frac{\pi}{2} \end{aligned}

$\therefore\ I = \dfrac{\pi}{4}$ ⁠.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

Finding an antiderivative of this integrand would be long and hard (none of the standard formulas of this chapter fits it), so we do not look for one. Property P4 replaces $x$ by $\frac{\pi}{2} - x$ ⁠, which swaps $\sin x$ and $\cos x$ and so gives the same integral $I$ with $\sqrt{\cos x}$ on top. The two fractions have the same denominator and their numerators add up to it, so together they integrate the constant 1.

Given, to find, and the plan

Given: $I = \displaystyle\int_0^{\pi/2}\frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}}\,dx$ ⁠.
To find: The exact value of $I$ ⁠, and its decimal value.
Plan: Check that the square roots are defined and the denominator is never 0 on $\left[0, \frac{\pi}{2}\right]$ ⁠; apply P4 with $a = \frac{\pi}{2}$ ⁠; add the two forms; solve for $I$ ⁠.

Basics you need

1Property P4 (Section 7.10)

$\int_0^a f(x)\,dx = \int_0^a f(a - x)\,dx$ ⁠: running through $[0, a]$ backwards does not change the integral. Lesson: properties of definite integrals

2Complementary angles

$\sin\left(\frac{\pi}{2} - x\right) = \cos x$ and $\cos\left(\frac{\pi}{2} - x\right) = \sin x$ ⁠. Basics: trigonometric identities

3Square roots

$\sqrt{u}$ is defined (as a real number) only for $u \ge 0$ ⁠, and it is continuous there.

4Adding integrals over the same interval

$\int_a^b f\,dx + \int_a^b g\,dx = \int_a^b (f + g)\,dx$ ⁠. Basics: symmetry and mirror images

Solution

The integrand is defined and continuous on $\left[0, \frac{\pi}{2}\right]$

For $0 \le x \le \frac{\pi}{2}$ ⁠, both $\sin x \ge 0$ and $\cos x \ge 0$ ⁠, so both square roots are defined and continuous. The denominator is never 0: $\sin x$ and $\cos x$ are never 0 together (since $\sin^2 x + \cos^2 x = 1$ ⁠), so at least one root is positive. In fact, for $0 \le s \le 1$ we have $\sqrt{s} \ge s^2$ ⁠, so
$\sqrt{\sin x} + \sqrt{\cos x} \ \ge\ \sin^2 x + \cos^2 x = 1 .$
So the integrand is continuous on the closed interval, with values between 0 and 1. Label
$I = \int_0^{\frac{\pi}{2}}\frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}}\,dx \qquad (1)$
The properties of Section 7.10 are used for continuous integrands, and the method below adds $I$ to itself and divides by 2, which needs $I$ to be a finite number. Outside $\left[0, \frac{\pi}{2}\right]$ one of $\sin x$ ⁠, $\cos x$ can be negative and the integrand would not even be defined, so the interval matters.

For $0 \le s \le 1$ ⁠: $s^2 \le s \le \sqrt{s}$ (squaring a number between 0 and 1 makes it smaller, taking its square root makes it larger).
Apply P4 with $a = \frac{\pi}{2}$

With $f(x)$ the integrand, compute $f\left(\frac{\pi}{2} - x\right)$ by replacing every $x$ ⁠. The two pieces become
$\begin{aligned}\sqrt{\sin\left(\tfrac{\pi}{2} - x\right)} &= \sqrt{\cos x}, \\ \sqrt{\cos\left(\tfrac{\pi}{2} - x\right)} &= \sqrt{\sin x},\end{aligned}$
so
$f\left(\tfrac{\pi}{2} - x\right) = \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} .$
So, by P4,
$I = \int_0^{\frac{\pi}{2}}\frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}}\,dx \qquad (2)$
P4 holds for every function continuous on $[0, a]$ ⁠; its only "condition" is that, which Step 1 checked. Its use lies in what $f(a - x)$ becomes: here the numerator changes from $\sqrt{\sin x}$ to $\sqrt{\cos x}$ while the denominator stays the same (only its two terms change places).

P4, Section 7.10; $\sin\left(\frac{\pi}{2} - x\right) = \cos x$ ⁠, $\cos\left(\frac{\pi}{2} - x\right) = \sin x$ ⁠. Lesson: properties of definite integrals
Add (1) and (2)
$\begin{aligned}2I &= \int_0^{\frac{\pi}{2}}\frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}}\,dx \\ &= \int_0^{\frac{\pi}{2}}1\,dx = \big[x\big]_0^{\frac{\pi}{2}} = \frac{\pi}{2}.\end{aligned}$
Same interval, same denominator: the numerators add to the denominator, and the fraction is 1 (allowed because the denominator is never 0, Step 1). The integral of 1 over an interval is its length.

$\int dx = x + C$ (Section 7.2) and Theorem 2, Section 7.8.3. Lesson: the fundamental theorem
Solve for $I$
$I = \frac{\pi}{4} \approx 0.7853982 .$
Dividing $2I = \frac{\pi}{2}$ by 2. A quick sense check: the integrand lies between 0 and 1 and equals $\frac12$ at the middle point $x = \frac{\pi}{4}$ ⁠, so an average height of $\frac12$ over a length $\frac{\pi}{2}$ is believable.

At $x = \frac{\pi}{4}$ ⁠, $\sin x = \cos x$ ⁠, so the fraction is $\frac{\sqrt{\sin x}}{2\sqrt{\sin x}} = \frac12$ ⁠.
The picture

The blue curve is the integrand; the region under it (area $I$ ⁠) is shaded. The dashed gold curve, $\frac{\sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}}$ ⁠, is its mirror image in the line $x = \frac{\pi}{4}$ ⁠, so it has the same area $I$ (this is P4). The two heights add to 1 at every $x$ ⁠, so the two areas fill the dashed rectangle of area $\frac{\pi}{2}$ ⁠.

The blue curve is symmetric under a half turn about $\left(\frac{\pi}{4}, \frac12\right)$ ⁠: the unshaded part of the rectangle is a copy of the shaded part. So the shaded part is exactly half the rectangle.

Mirror images have equal areas. Basics: symmetry and mirror images

Result

$\begin{aligned}&\int_0^{\frac{\pi}{2}}\frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}}\,dx \\ &= \frac{\pi}{4} \approx 0.7853982\end{aligned}$ (Textbook answer: $\frac{\pi}{4}$ ⁠. Ours is the same.)

Check it with numbers

A numerical integration over $\left[0, \frac{\pi}{2}\right]$ gives 0.7853982, which is $\frac{\pi}{4}$ ⁠. At $x = 0.3$ the integrand is about 0.3574008 and its P4 partner (form (2)) is about 0.6425992; they add to 1.

Common mistakes

Replacing $x$ by $\frac{\pi}{2} - x$ in the numerator only. P4 changes the whole function, denominator included.
Thinking $\sqrt{\sin x} + \sqrt{\cos x}$ could be 0. On $\left[0, \frac{\pi}{2}\right]$ it is at least 1 (Step 1).
Trying $\sqrt{\sin x + \cos x}$ in the denominator. The denominator is a sum of two separate roots; $\sqrt{a} + \sqrt{b} \ne \sqrt{a + b}$ ⁠.

What to take away

For any power $p > 0$ ⁠, $\int_0^{\pi/2}\frac{\sin^p x}{\sin^p x + \cos^p x}\,dx = \frac{\pi}{4}$ by exactly these steps (Questions 3 and 4 are the cases $p = \frac32$ and $p = 5$ ⁠).
When an integrand on $\left[0, \frac{\pi}{2}\right]$ is a fraction whose P4 partner has the same denominator, add the two forms.
Always check the integrand is defined and continuous on the whole closed interval first.

Every fact used, and where it comes from

Fact	Used in	Source
$\sin x, \cos x \ge 0$ on $\left[0, \frac{\pi}{2}\right]$ ⁠; $\sin^2 x + \cos^2 x = 1$	Step 1	Trigonometry (Basics)
$\sqrt{s} \ge s^2$ for $0 \le s \le 1$	Step 1	Algebra of numbers between 0 and 1
P4: $\int_0^a f(x)\,dx = \int_0^a f(a - x)\,dx$	Step 2	Section 7.10 (Lesson)
$\sin\left(\frac{\pi}{2} - x\right) = \cos x$ ⁠, $\cos\left(\frac{\pi}{2} - x\right) = \sin x$	Step 2	Trigonometry (Basics)
$\int_a^b f + \int_a^b g = \int_a^b (f + g)$	Step 3	Linearity of integrals
$\int dx = x + C$ ⁠; $\int_a^b f = F(b) - F(a)$	Step 3	Section 7.2; Theorem 2, Section 7.8.3 (Lesson)
Same method	Steps 2 to 4	Example 32, Exercise 7.10 Q1

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Exercise 7.10 · Q2: ∫0π/2sin⁡xsin⁡x+cos⁡x dx\int_0^{\pi/2}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx∫0π/2​sinx​+cosx​sinx​​dx

Basics you need

Solution

The integrand is defined and continuous on [0,π2]\left[0, \frac{\pi}{2}\right][0,2π​]

Apply P4 with a=π2a = \frac{\pi}{2}a=2π​

Add (1) and (2)

Solve for III

The picture

Every fact used, and where it comes from

Exercise 7.10 · Q2: $\int_0^{\pi/2}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx$

The integrand is defined and continuous on $\left[0, \frac{\pi}{2}\right]$

Apply P4 with $a = \frac{\pi}{2}$

Solve for $I$