Chapter 7 · Integrals

Exercise 7.10 · Q4: $\int_0^{\pi/2}\frac{\cos^5x}{\sin^5x+\cos^5x}\,dx$

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.

4. $\displaystyle\int_0^{\frac{\pi}{2}}\frac{\cos^5 x\,dx}{\sin^5 x + \cos^5 x}$

Exam answer

Let

I = \int_0^{\frac{\pi}{2}}\frac{\cos^5 x}{\sin^5 x + \cos^5 x}\,dx\quad\ldots(\text{i})

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

I = \int_0^{\frac{\pi}{2}}\frac{\sin^5 x}{\cos^5 x + \sin^5 x}\,dx\quad\ldots(\text{ii})

Adding (i) and (ii),

\begin{aligned} 2I &= \int_0^{\frac{\pi}{2}}\frac{\cos^5 x + \sin^5 x}{\sin^5 x + \cos^5 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

Property P4 with $a = \frac{\pi}{2}$ replaces $x$ by $\frac{\pi}{2} - x$ ⁠, which swaps $\sin$ and $\cos$ ⁠. So the same number $I$ is also the integral of $\frac{\sin^5 x}{\cos^5 x + \sin^5 x}$ ⁠. The two fractions add up to 1, so $2I$ is the length of the interval.

Given, to find, and the plan

Given: $I = \displaystyle\int_0^{\pi/2}\frac{\cos^5 x}{\sin^5 x + \cos^5 x}\,dx$ ⁠.
To find: The exact value of $I$ ⁠, and its decimal value.
Plan: Check the denominator is never 0 on $\left[0, \frac{\pi}{2}\right]$ ⁠; apply P4; add the two forms; solve for $I$ ⁠.

Basics you need

1Property P4 (Section 7.10)

$\int_0^a f(x)\,dx = \int_0^a f(a - x)\,dx$ ⁠. Lesson: properties of definite integrals

2Complementary angles

$\sin\left(\frac{\pi}{2} - x\right) = \cos x$ and $\cos\left(\frac{\pi}{2} - x\right) = \sin x$ ⁠; so $\cos^5\left(\frac{\pi}{2} - x\right) = \sin^5 x$ ⁠. Basics: trigonometric identities

3Adding 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 denominator is positive on $\left[0, \frac{\pi}{2}\right]$

For $0 \le x \le \frac{\pi}{2}$ ⁠, $\sin x \ge 0$ and $\cos x \ge 0$ ⁠, and they are never both 0 (because $\sin^2 x + \cos^2 x = 1$ ⁠). So $\sin^5 x + \cos^5 x$ is a sum of two non-negative numbers, at least one of them positive:
$\sin^5 x + \cos^5 x > 0 \quad \text{on } \left[0, \tfrac{\pi}{2}\right].$
(Its smallest value there is $\frac{1}{2\sqrt2} \approx 0.3535534$ ⁠, at $x = \frac{\pi}{4}$ ⁠: its derivative, $5\sin x\cos x$ times $\sin^3 x - \cos^3 x$ ⁠, is negative before $\frac{\pi}{4}$ and positive after it, and the end values are 1.) So the integrand is continuous on the closed interval. Label
$I = \int_0^{\frac{\pi}{2}}\frac{\cos^5 x}{\sin^5 x + \cos^5 x}\,dx \qquad (1)$
This check is not a formality. On the whole real line the denominator can be 0: at $x = \frac{3\pi}{4}$ ⁠, $\sin x = \frac{1}{\sqrt2}$ and $\cos x = -\frac{1}{\sqrt2}$ ⁠, and the fifth powers cancel. That point lies outside our interval, so on $\left[0, \frac{\pi}{2}\right]$ all is well, and $I$ is a finite number.

At $x = \frac{\pi}{4}$ ⁠: $\sin^5 x + \cos^5 x = 2\left(\frac{1}{\sqrt2}\right)^5$ ⁠, which is $\frac{2}{4\sqrt2} = \frac{1}{2\sqrt2}$ ⁠.
Apply P4 with $a = \frac{\pi}{2}$

Replace $x$ by $\frac{\pi}{2} - x$ everywhere. Since $\cos^5\left(\frac{\pi}{2} - x\right) = \sin^5 x$ and $\sin^5\left(\frac{\pi}{2} - x\right) = \cos^5 x$ ⁠,
$f\left(\tfrac{\pi}{2} - x\right) = \frac{\sin^5 x}{\cos^5 x + \sin^5 x} .$
So, by P4,
$I = \int_0^{\frac{\pi}{2}}\frac{\sin^5 x}{\cos^5 x + \sin^5 x}\,dx \qquad (2)$
P4 is valid for every function continuous on $[0, a]$ ⁠, which Step 1 established. The new form has the same denominator (in the other order) and the "other" term on top, which is exactly what we need to add.

P4, Section 7.10. Raising both sides of $\cos\left(\frac{\pi}{2} - x\right) = \sin x$ to the fifth power. Lesson: properties of definite integrals
Add (1) and (2)
$\begin{aligned}2I &= \int_0^{\frac{\pi}{2}}\frac{\cos^5 x + \sin^5 x}{\sin^5 x + \cos^5 x}\,dx \\ &= \int_0^{\frac{\pi}{2}}1\,dx = \big[x\big]_0^{\frac{\pi}{2}} = \frac{\pi}{2}.\end{aligned}$
Same interval and same non-zero denominator, so the two integrands add to 1 at every point. The integral of 1 is the interval length.

$\int dx = x + C$ (Section 7.2); Theorem 2, Section 7.8.3. Lesson: the fundamental theorem
Solve for $I$
$I = \frac{\pi}{4} \approx 0.7853982 .$
Divide $2I = \frac{\pi}{2}$ by 2. Here $\cos^5 x$ is on top rather than $\sin^5 x$ ⁠, but by Step 2 the two versions have the same integral, so it makes no difference.

At $x = \frac{\pi}{4}$ the integrand equals $\frac12$ ⁠: numerator and the two denominator terms are all equal.
The picture

The blue curve $y = \frac{\cos^5 x}{\sin^5 x + \cos^5 x}$ falls from 1 to 0, staying close to 1 at first and close to 0 at the end; the region under it (area $I$ ⁠) is shaded. The dashed gold curve, with $\sin^5 x$ on top, is its mirror image in $x = \frac{\pi}{4}$ ⁠. The heights add to 1, so the two equal areas fill the rectangle of area $\frac{\pi}{2}$ ⁠.

The high power 5 makes the switch from 1 to 0 sharper than in Questions 2 and 3, but the half-turn symmetry about $\left(\frac{\pi}{4}, \frac12\right)$ is the same, so the area is still half the rectangle.

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

Result

$\int_0^{\frac{\pi}{2}}\frac{\cos^5 x}{\sin^5 x + \cos^5 x}\,dx = \frac{\pi}{4} \approx 0.7853982$ (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 0.9971756 and its P4 partner is about 0.0028244; they add to 1.

Common mistakes

Skipping the denominator check. $\sin^5 x + \cos^5 x$ is 0 at $x = \frac{3\pi}{4}$ ⁠; it is safe only because that point is outside $\left[0, \frac{\pi}{2}\right]$ ⁠.
Writing $\cos^5\left(\frac{\pi}{2} - x\right) = -\sin^5 x$ ⁠. There is no sign change: $\cos\left(\frac{\pi}{2} - x\right) = +\sin x$ ⁠.
Expanding fifth powers. No expansion is needed; P4 and one addition do all the work.

What to take away

For any power $p > 0$ ⁠, whether $\sin^p x$ or $\cos^p x$ is on top, $\int_0^{\pi/2}$ of such a fraction is $\frac{\pi}{4}$ ⁠.
P4 with $a = \frac{\pi}{2}$ swaps $\sin$ and $\cos$ ⁠; then add the two forms.
Check that a denominator has no zeros on the interval; zeros elsewhere do not matter.

Every fact used, and where it comes from

Fact	Used in	Source
$\sin x, \cos x \ge 0$ on $\left[0, \frac{\pi}{2}\right]$ ⁠, never both 0	Step 1	Trigonometry (Basics)
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)
$\int dx = x + C$ ⁠; $\int_a^b f = F(b) - F(a)$	Step 3	Section 7.2; Theorem 2, Section 7.8.3 (Lesson)
Same method	Steps 2 to 4	Example 32, Exercise 7.10 Q3

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Exercise 7.10 · Q4: ∫0π/2cos⁡5xsin⁡5x+cos⁡5x dx\int_0^{\pi/2}\frac{\cos^5x}{\sin^5x+\cos^5x}\,dx∫0π/2​sin5x+cos5xcos5x​dx

Basics you need

Solution

The denominator is positive 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 · Q4: $\int_0^{\pi/2}\frac{\cos^5x}{\sin^5x+\cos^5x}\,dx$

The denominator is positive on $\left[0, \frac{\pi}{2}\right]$

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

Solve for $I$