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Chapter 7 · Integrals

Exercise 7.10 · Q3: ∫0π/2sin⁡3/2xsin⁡3/2x+cos⁡3/2x dx\int_0^{\pi/2}\frac{\sin^{3/2}x}{\sin^{3/2}x+\cos^{3/2}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.

3. ∫0π2sin⁡32x dxsin⁡32x+cos⁡32x\displaystyle\int_0^{\frac{\pi}{2}}\frac{\sin^{\frac32}x\,dx}{\sin^{\frac32}x + \cos^{\frac32}x}

Exam answer

Let

I=∫0π2sin⁡32xsin⁡32x+cos⁡32x dx…(i)\begin{aligned} I &= \int_0^{\frac{\pi}{2}}\frac{\sin^{\frac32}x}{\sin^{\frac32}x + \cos^{\frac32}x}\,dx \\ &\qquad\ldots(\text{i}) \end{aligned}

Using ∫0af(x) dx=∫0af(a−x) dx\displaystyle\int_0^a f(x)\,dx = \int_0^a f(a - x)\,dx with sin⁡(π2−x)=cos⁡x\sin\left(\frac{\pi}{2} - x\right) = \cos x and cos⁡(π2−x)=sin⁡x\cos\left(\frac{\pi}{2} - x\right) = \sin x⁠,

I=∫0π2cos⁡32xcos⁡32x+sin⁡32x dx…(ii)\begin{aligned} I &= \int_0^{\frac{\pi}{2}}\frac{\cos^{\frac32}x}{\cos^{\frac32}x + \sin^{\frac32}x}\,dx \\ &\qquad\ldots(\text{ii}) \end{aligned}

Adding (i) and (ii),

2I=∫0π2sin⁡32x+cos⁡32xsin⁡32x+cos⁡32x dx=∫0π21 dx=[x]0π2=π2\begin{aligned} 2I &= \int_0^{\frac{\pi}{2}}\frac{\sin^{\frac32}x + \cos^{\frac32}x}{\sin^{\frac32}x + \cos^{\frac32}x}\,dx \\ &= \int_0^{\frac{\pi}{2}}1\,dx = \Big[x\Big]_0^{\frac{\pi}{2}} = \frac{\pi}{2} \end{aligned}

∴ I=π4\therefore\ I = \dfrac{\pi}{4}⁠.

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

The one idea this problem is built on

This is the same shape as Question 2 with the power 32\frac32 in place of 12\frac12⁠. Property P4 with a=π2a = \frac{\pi}{2} swaps sin⁡x\sin x and cos⁡x\cos x⁠, so II is also the integral of the fraction with cos⁡3/2x\cos^{3/2}x on top. Adding the two forms, the numerators add up to the common denominator and we integrate 1.

Given, to find, and the plan
Given
I=∫0π/2sin⁡3/2xsin⁡3/2x+cos⁡3/2x dxI = \displaystyle\int_0^{\pi/2}\frac{\sin^{3/2}x}{\sin^{3/2}x + \cos^{3/2}x}\,dx⁠, where sin⁡3/2x\sin^{3/2}x means (sin⁡x)3/2(\sin x)^{3/2}⁠.
To find
The exact value of II⁠, and its decimal value.
Plan
Check the powers are defined and the denominator is never 0 on [0,π2]\left[0, \frac{\pi}{2}\right]⁠; apply P4; add the two forms; solve for II⁠.

Basics you need

1Property P4 (Section 7.10)

∫0af(x) dx=∫0af(a−x) dx\int_0^a f(x)\,dx = \int_0^a f(a - x)\,dx⁠. Lesson: properties of definite integrals

2Complementary angles

sin⁡(π2−x)=cos⁡x\sin\left(\frac{\pi}{2} - x\right) = \cos x and cos⁡(π2−x)=sin⁡x\cos\left(\frac{\pi}{2} - x\right) = \sin x⁠. Basics: trigonometric identities

3Fractional powers

For u≥0u \ge 0⁠, u3/2=uuu^{3/2} = u\sqrt{u}⁠, the square root of u3u^3⁠. It is not defined (as a real number) for u<0u < 0⁠. Basics: powers of x

4Adding integrals over the same interval

∫abf dx+∫abg dx=∫ab(f+g) dx\int_a^b f\,dx + \int_a^b g\,dx = \int_a^b (f + g)\,dx⁠. Basics: symmetry and mirror images

Solution

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

    For 0≤x≤π20 \le x \le \frac{\pi}{2}⁠, sin⁡x≥0\sin x \ge 0 and cos⁡x≥0\cos x \ge 0⁠, so sin⁡3/2x=sin⁡xsin⁡x\sin^{3/2}x = \sin x\sqrt{\sin x} and cos⁡3/2x=cos⁡xcos⁡x\cos^{3/2}x = \cos x\sqrt{\cos x} are defined and continuous. For 0≤s≤10 \le s \le 1⁠, s3/2≥s2s^{3/2} \ge s^2⁠, so the denominator satisfies

    sin⁡32x+cos⁡32x ≥ sin⁡2x+cos⁡2x=1.\sin^{\frac32}x + \cos^{\frac32}x \ \ge\ \sin^2 x + \cos^2 x = 1 .

    It is never 0, and the integrand is continuous on the closed interval. Label

    I=∫0π2sin⁡32xsin⁡32x+cos⁡32x dx(1)I = \int_0^{\frac{\pi}{2}}\frac{\sin^{\frac32}x}{\sin^{\frac32}x + \cos^{\frac32}x}\,dx \qquad (1)
    P4 and the addition below need a continuous integrand, so that II is a definite finite number. The fractional power is why the interval matters: for xx slightly beyond π2\frac{\pi}{2}⁠, cos⁡x<0\cos x < 0 and cos⁡3/2x\cos^{3/2}x would not be defined.
    For 0≤s≤10 \le s \le 1⁠: s2=s3/2⋅s1/2≤s3/2s^2 = s^{3/2}\cdot s^{1/2} \le s^{3/2}⁠, because s1/2≤1s^{1/2} \le 1⁠. Basics: powers of x

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

    Replacing xx by π2−x\frac{\pi}{2} - x turns sin⁡x\sin x into cos⁡x\cos x and cos⁡x\cos x into sin⁡x\sin x⁠:

    sin⁡32(π2−x)=cos⁡32x,cos⁡32(π2−x)=sin⁡32x.\begin{aligned}\sin^{\frac32}\left(\tfrac{\pi}{2} - x\right) &= \cos^{\frac32}x, \\ \cos^{\frac32}\left(\tfrac{\pi}{2} - x\right) &= \sin^{\frac32}x .\end{aligned}

    Making both replacements in numerator and denominator,

    f(π2−x)=cos⁡32xcos⁡32x+sin⁡32x.f\left(\tfrac{\pi}{2} - x\right) = \frac{\cos^{\frac32}x}{\cos^{\frac32}x + \sin^{\frac32}x} .

    So, by P4,

    I=∫0π2cos⁡32xcos⁡32x+sin⁡32x dx(2)I = \int_0^{\frac{\pi}{2}}\frac{\cos^{\frac32}x}{\cos^{\frac32}x + \sin^{\frac32}x}\,dx \qquad (2)
    P4 holds for every function continuous on [0,a][0, a] (checked in Step 1). The point of using it is that f(π2−x)f\left(\frac{\pi}{2} - x\right) has the same denominator as f(x)f(x) but the other term on top.
    P4, Section 7.10, with a=π2a = \frac{\pi}{2}⁠. Lesson: properties of definite integrals

  3. Add (1) and (2)

    2I=∫0π2sin⁡32x+cos⁡32xsin⁡32x+cos⁡32x dx=∫0π21 dx=π2.\begin{aligned}2I &= \int_0^{\frac{\pi}{2}}\frac{\sin^{\frac32}x + \cos^{\frac32}x}{\sin^{\frac32}x + \cos^{\frac32}x}\,dx \\ &= \int_0^{\frac{\pi}{2}}1\,dx = \frac{\pi}{2}.\end{aligned}
    The two fractions share a denominator that is never 0 (Step 1), and their numerators add to it, so the sum is 1 at every point. Integrating 1 over [0,π2]\left[0, \frac{\pi}{2}\right] gives the length π2\frac{\pi}{2}⁠.
    ∫dx=x+C\int dx = x + C (Section 7.2), then [x]0π/2=π2\big[x\big]_0^{\pi/2} = \frac{\pi}{2} by Theorem 2, Section 7.8.3. Lesson: the fundamental theorem

  4. Solve for II

    I=π4≈0.7853982.I = \frac{\pi}{4} \approx 0.7853982 .
    Divide 2I=π22I = \frac{\pi}{2} by 2. The power 32\frac32 changes the shape of the curve but not the answer: the argument never used the value of the power.
    At x=π4x = \frac{\pi}{4} the integrand is 12\frac12⁠, and the curve is symmetric about that point (see the figure).

  5. The picture

    x y π/4 π/2 1/2 1 (π/4, 1/2) y = sin^(3/2)x / (sin^(3/2)x + cos^(3/2)x) dashed: mirror area I
    The blue curve is the integrand, rising from 0 to 1; the region under it (area II⁠) is shaded. The dashed gold curve, cos⁡3/2xsin⁡3/2x+cos⁡3/2x\frac{\cos^{3/2}x}{\sin^{3/2}x + \cos^{3/2}x}⁠, is its mirror image in x=π4x = \frac{\pi}{4} and has the same area. Their heights add to 1, so the two areas fill the rectangle of area π2\frac{\pi}{2}⁠.
    Reflecting in the line x=π4x = \frac{\pi}{4} is exactly the change x→π2−xx \to \frac{\pi}{2} - x of P4, and reflection does not change area.
    Mirror images have equal areas. Basics: symmetry and mirror images
Result

∫0π2sin⁡32xsin⁡32x+cos⁡32x dx=π4≈0.7853982\int_0^{\frac{\pi}{2}}\frac{\sin^{\frac32}x}{\sin^{\frac32}x + \cos^{\frac32}x}\,dx = \frac{\pi}{4} \approx 0.7853982 (Textbook answer: π4\frac{\pi}{4}⁠. Ours is the same.)

Check it with numbers

A numerical integration over [0,π2]\left[0, \frac{\pi}{2}\right] gives 0.7853982, which is π4\frac{\pi}{4}⁠. At x=0.3x = 0.3 the integrand is about 0.1467916 and its P4 partner is about 0.8532084; they add to 1.

Common mistakes
  1. Reading sin⁡3/2x\sin^{3/2}x as sin⁡(x3/2)\sin\left(x^{3/2}\right)⁠. The power belongs to sin⁡x\sin x⁠: (sin⁡x)3/2(\sin x)^{3/2}⁠.
  2. Changing sin⁡\sin to cos⁡\cos in one place only. P4 replaces xx everywhere, so the denominator is rewritten too (it only changes order).
  3. Expecting the answer to depend on the power. Any positive power gives π4\frac{\pi}{4}⁠; compare Questions 2 and 4.
What to take away
  1. ∫0π/2sin⁡pxsin⁡px+cos⁡px dx=π4\int_0^{\pi/2}\frac{\sin^p x}{\sin^p x + \cos^p x}\,dx = \frac{\pi}{4} for every p>0p > 0⁠.
  2. P4 on [0,π2]\left[0, \frac{\pi}{2}\right] swaps sin⁡\sin and cos⁡\cos⁠; add the two forms when they share a denominator.
  3. Fractional powers of sin⁡x\sin x and cos⁡x\cos x need sin⁡x≥0\sin x \ge 0 and cos⁡x≥0\cos x \ge 0⁠: check the interval.

Every fact used, and where it comes from

FactUsed inSource
u3/2=uuu^{3/2} = u\sqrt{u} for u≥0u \ge 0⁠; s3/2≥s2s^{3/2} \ge s^2 on [0,1][0, 1]Step 1Laws of indices (Basics)
sin⁡x,cos⁡x≥0\sin x, \cos x \ge 0 on [0,π2]\left[0, \frac{\pi}{2}\right]⁠; sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1Step 1Trigonometry (Basics)
P4: ∫0af(x) dx=∫0af(a−x) dx\int_0^a f(x)\,dx = \int_0^a f(a - x)\,dxStep 2Section 7.10 (Lesson)
sin⁡(π2−x)=cos⁡x\sin\left(\frac{\pi}{2} - x\right) = \cos x⁠, cos⁡(π2−x)=sin⁡x\cos\left(\frac{\pi}{2} - x\right) = \sin xStep 2Trigonometry (Basics)
∫dx=x+C\int dx = x + C⁠; ∫abf=F(b)−F(a)\int_a^b f = F(b) - F(a)Step 3Section 7.2; Theorem 2, Section 7.8.3 (Lesson)
Same methodSteps 2 to 4Example 32, Exercise 7.10 Q2
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