Chapter 7 · Integrals

Example 32: $\int_0^{\pi/2}\frac{\sin^4x}{\sin^4x+\cos^4x}\,dx$

ExamplesProperties of definite integrals · Section 7.10

The question

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

Exam answer

Let

I = \int_0^{\frac{\pi}{2}}\frac{\sin^4 x\,dx}{\sin^4 x + \cos^4 x}\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{\cos^4 x\,dx}{\cos^4 x + \sin^4 x}\quad\ldots(\text{ii})

Adding (i) and (ii),

\begin{aligned} 2I &= \int_0^{\frac{\pi}{2}}\frac{\sin^4 x + \cos^4 x}{\sin^4 x + \cos^4 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

On $\left[0, \frac{\pi}{2}\right]$ ⁠, replacing $x$ by $\frac{\pi}{2} - x$ swaps $\sin$ and $\cos$ ⁠. Property P4 says this does not change the integral. So the same number $I$ is also the integral of $\frac{\cos^4 x}{\cos^4 x + \sin^4 x}$ ⁠. The two integrands add up to exactly 1, so $2I = \int_0^{\pi/2}1\,dx = \frac{\pi}{2}$ ⁠.

Given, to find, and the plan

Given: $I = \int_0^{\pi/2}\frac{\sin^4 x}{\sin^4 x + \cos^4 x}\,dx$ ⁠.
To find: The exact value of $I$ ⁠.
Plan: Check the denominator never vanishes; apply P4 with $a = \frac{\pi}{2}$ to get a second form (2); add (1) and (2); solve for $I$ ⁠.

Basics you need

1Property P4

$\int_0^a f(x)\,dx = \int_0^a f(a - x)\,dx$ (Section 7.10): reading the interval from the other end does not change the area. 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

3Adding integrals over the same interval

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

Solution

The denominator is never 0

Using $\sin^2 x + \cos^2 x = 1$ ⁠:
$\begin{aligned}\sin^4 x + \cos^4 x &= (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x\cos^2 x \\ &= 1 - \tfrac12\sin^2 2x \ \ge\ \tfrac12.\end{aligned}$
So the integrand is continuous on $\left[0, \frac{\pi}{2}\right]$ (in fact everywhere), and its values lie between 0 and 1.

P4 and the addition below need integrals of continuous functions. Since $\sin^2 2x \le 1$ ⁠, the denominator is at least $\frac12$ ⁠, never 0. (It is also clear directly: $\sin x$ and $\cos x$ are never 0 at the same time.)

$2\sin x\cos x = \sin 2x$ ⁠, so $2\sin^2 x\cos^2 x = \frac12\sin^2 2x$ ⁠. Basics: trigonometric identities
Label the integral (1) and apply P4
$I = \int_0^{\frac{\pi}{2}}\frac{\sin^4 x}{\sin^4 x + \cos^4 x}\,dx \qquad (1)$
By P4 with $a = \frac{\pi}{2}$ ⁠, replace $x$ by $\frac{\pi}{2} - x$ ⁠:
$\begin{aligned}I &= \int_0^{\frac{\pi}{2}}\frac{\sin^4\left(\frac{\pi}{2} - x\right)}{\sin^4\left(\frac{\pi}{2} - x\right) + \cos^4\left(\frac{\pi}{2} - x\right)}\,dx \\ &= \int_0^{\frac{\pi}{2}}\frac{\cos^4 x}{\cos^4 x + \sin^4 x}\,dx \qquad (2)\end{aligned}$
The complementary-angle identities turn every $\sin$ into $\cos$ and every $\cos$ into $\sin$ ⁠. The denominator $\sin^4 + \cos^4$ is unchanged (only its order), while the numerator becomes $\cos^4 x$ ⁠. We now have two different-looking integrands with the same integral.

P4, Section 7.10, with $a = \frac{\pi}{2}$ ⁠. Lesson: properties of definite integrals
Add (1) and (2)
$\begin{aligned}2I &= \int_0^{\frac{\pi}{2}}\frac{\sin^4 x + \cos^4 x}{\sin^4 x + \cos^4 x}\,dx \\ &= \int_0^{\frac{\pi}{2}}1\,dx = \big[x\big]_0^{\frac{\pi}{2}} = \frac{\pi}{2}.\end{aligned}$
Hence $I = \frac{\pi}{4} \approx 0.7853982$ ⁠.

The two fractions have the same denominator, so their numerators add to the denominator itself and the quotient is 1 (allowed because the denominator is never 0, Step 1). Integrating 1 over an interval gives its length, $\frac{\pi}{2}$ ⁠.

$\int 1\,dx = x + C$ (standard formula (i) with $n = 0$ ⁠). Lesson: the fundamental theorem
The picture

The blue curve $y = \frac{\sin^4 x}{\sin^4 x + \cos^4 x}$ rises from 0 to 1 through $\left(\frac{\pi}{4}, \frac12\right)$ ⁠; the region under it is shaded. The dashed gold curve $\frac{\cos^4 x}{\sin^4 x + \cos^4 x}$ is its mirror image in $x = \frac{\pi}{4}$ and has the same area I. At every $x$ the two heights add to 1, so the two areas fill the rectangle of area $\frac{\pi}{2}$ ⁠.

The blue curve is symmetric about the point $\left(\frac{\pi}{4}, \frac12\right)$ ⁠: the unshaded part of the rectangle above it is a half-turn copy of the shaded part. So the shaded region is exactly half of the $\frac{\pi}{2}$ by 1 rectangle.

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

Result

$\int_0^{\frac{\pi}{2}}\frac{\sin^4 x}{\sin^4 x + \cos^4 x}\,dx = \frac{\pi}{4} \approx 0.7853982,$ the same as the book's answer.

Check it with numbers

A numerical integration over $\left[0, \frac{\pi}{2}\right]$ gives 0.7853982, which is $\frac{\pi}{4}$ ⁠. At the sample point $x = 0.3$ ⁠, the integrand is about 0.0090733 and its partner $\frac{\cos^4}{\sin^4 + \cos^4}$ is about 0.9909267: they add to 1.

Common mistakes

Trying to find an antiderivative. It is possible but long; P4 gives the value in three lines.
Changing only the numerator. P4 replaces $x$ by $\frac{\pi}{2} - x$ everywhere, in the denominator too.
Writing $2I = \frac{\pi}{2}$ ⁠, so $I = \pi$ ⁠. Divide by 2: $I = \frac{\pi}{4}$ ⁠.

What to take away

For $\int_0^{\pi/2}\frac{f(\sin x)}{f(\sin x) + f(\cos x)}\,dx$ the answer is $\frac{\pi}{4}$ by the same three lines, whenever $f$ is continuous and the denominator is never 0 on the interval.
P4 with $a = \frac{\pi}{2}$ swaps $\sin$ and $\cos$ ⁠; adding the two forms often cancels the difficult part.
Label the two forms (1) and (2) so that "adding (1) and (2)" is clear.

Every fact used, and where it comes from

Fact	Used in	Source
$\sin^4 x + \cos^4 x = 1 - \frac12\sin^2 2x$	Step 1	Identities $\sin^2 + \cos^2 = 1$ ⁠, $\sin 2x = 2\sin x\cos x$
P4: $\int_0^a f(x)\,dx = \int_0^a f(a - x)\,dx$	Step 2	Section 7.10 (Lesson)
$\sin(\frac{\pi}{2} - x) = \cos x$ ⁠, $\cos(\frac{\pi}{2} - x) = \sin x$	Step 2	Trigonometry (Basics)
$\int_a^b f + \int_a^b g = \int_a^b(f + g)$	Step 3	Linearity of integrals
$\int_a^b f = F(b) - F(a)$	Step 3	Theorem 2, Section 7.8.3 (Lesson)

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Example 32: ∫0π/2sin⁡4xsin⁡4x+cos⁡4x dx\int_0^{\pi/2}\frac{\sin^4x}{\sin^4x+\cos^4x}\,dx∫0π/2​sin4x+cos4xsin4x​dx

Basics you need

Solution

The denominator is never 0

Label the integral (1) and apply P4

Add (1) and (2)

The picture

Every fact used, and where it comes from

Example 32: $\int_0^{\pi/2}\frac{\sin^4x}{\sin^4x+\cos^4x}\,dx$