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Chapter 7 · Integrals

Exercise 7.8 · Q4: ∫0π/4sin⁡2x dx\int_0^{\pi/4}\sin 2x\,dx

Exercise 7.8Fundamental theorem of calculus · Section 7.8
The question

Evaluate the definite integral

∫0π4sin⁡2x dx.\int_0^{\frac{\pi}{4}}\sin 2x\,dx.
Exam answer

Let I=∫0π4sin⁡2x dxI = \displaystyle\int_0^{\frac{\pi}{4}}\sin 2x\,dx⁠.

I=[−cos⁡2x2]0π4=−12cos⁡π2+12cos⁡0=−12⋅0+12⋅1\begin{aligned} I &= \left[-\frac{\cos 2x}{2}\right]_0^{\frac{\pi}{4}} \\ &= -\frac12\cos\frac{\pi}{2} + \frac12\cos 0 \\ &= -\frac12\cdot 0 + \frac12\cdot 1 \end{aligned}

∴ I=12\therefore\ I = \dfrac12⁠.

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

The one idea this problem is built on

The second fundamental theorem (Theorem 2, Section 7.8.3): ∫abf(x) dx=F(b)−F(a)\int_a^b f(x)\,dx = F(b) - F(a)⁠. The antiderivative of sin⁡2x\sin 2x is −12cos⁡2x-\frac12\cos 2x⁠: the 12\frac12 undoes the factor 2 that the chain rule produces when cos⁡2x\cos 2x is differentiated. Then we evaluate at π4\frac{\pi}{4} and at 0.

Given, to find, and the plan
Given
f(x)=sin⁡2xf(x) = \sin 2x on [0,π4]\left[0, \frac{\pi}{4}\right]⁠.
To find
The exact value of ∫0π/4sin⁡2x dx\int_0^{\pi/4}\sin 2x\,dx⁠.
Plan
(1) Continuity. (2) Antiderivative by the substitution u=2xu = 2x⁠, then differentiate it back. (3) F(π4)−F(0)F\left(\frac{\pi}{4}\right) - F(0)⁠. (4) Size check against the graph.

Basics you need

1Second fundamental theorem

∫abf(x) dx=[F(x)]ab=F(b)−F(a)\int_a^b f(x)\,dx = \big[F(x)\big]_a^b = F(b) - F(a) for ff continuous on [a,b][a, b] and F′=fF' = f⁠. Lesson: definite integrals and the fundamental theorem

2Integral of sin⁡\sin

∫sin⁡u du=−cos⁡u+C\int\sin u\,du = -\cos u + C (standard formula (iii), Section 7.2). Basics: trigonometric identities for integrals

3Substitution

If u=g(x)u = g(x)⁠, then du=g′(x) dxdu = g'(x)\,dx⁠; rewrite the whole integrand in uu (Section 7.3.1). Basics: substitution (the chain rule backwards)

4Exact values

cos⁡0=1\cos 0 = 1⁠, cos⁡π2=0\cos\frac{\pi}{2} = 0⁠. Basics: trigonometric values

Solution

  1. Check continuity on [0,π4]\left[0, \frac{\pi}{4}\right]

    sin⁡2x\sin 2x is the composition of the continuous functions x↦2xx \mapsto 2x and sin⁡\sin⁠, so it is continuous for every real xx⁠, in particular on [0,π4]\left[0, \frac{\pi}{4}\right]⁠.

    Theorem 2 requires a continuous integrand on the closed interval (Remark (iv) after Theorem 2). Sine has no breaks anywhere.
    A composition of continuous functions is continuous (Chapter 5, Theorem 2, Section 5.2.1). Lesson: the fundamental theorem

  2. Antiderivative by the substitution u=2xu = 2x

    Let u=2xu = 2x⁠. Then dudx=2\frac{du}{dx} = 2⁠, so du=2 dxdu = 2\,dx and dx=12 dudx = \frac12\,du⁠. Hence

    ∫sin⁡2x dx=∫sin⁡u⋅12 du=12(−cos⁡u)+C=−12cos⁡2x+C.\begin{aligned}\int\sin 2x\,dx &= \int\sin u\cdot\tfrac12\,du = \tfrac12\left(-\cos u\right) + C \\ &= -\tfrac12\cos 2x + C.\end{aligned}

    Take F(x)=−12cos⁡2xF(x) = -\frac12\cos 2x⁠.

    The inside function 2x2x has derivative 2, a constant, so the substitution only brings in the factor 12\frac12⁠. Writing −cos⁡2x-\cos 2x alone would be off by a factor 2.
    Standard formula (iii), Section 7.2, and substitution, Section 7.3.1. Lesson: integration by substitution

  3. Check: differentiate FF back

    By the chain rule (outer cos⁡\cos⁠, inner 2x2x⁠):

    F′(x)=−12⋅(−sin⁡2x)⋅2=sin⁡2x.F'(x) = -\tfrac12\cdot\left(-\sin 2x\right)\cdot 2 = \sin 2x.
    The two minus signs cancel and the 12\frac12 cancels the 2 from the inner derivative, returning exactly the integrand.
    ddxcos⁡u=−sin⁡u⋅dudx\frac{d}{dx}\cos u = -\sin u\cdot\frac{du}{dx}⁠. Basics: the chain rule

  4. Evaluate F(π4)−F(0)F\left(\frac{\pi}{4}\right) - F(0)

    At the upper limit, 2x=π22x = \frac{\pi}{2}⁠; at the lower limit, 2x=02x = 0⁠:

    F(π4)=−12cos⁡π2=−12⋅0=0,F(0)=−12cos⁡0=−12⋅1=−12.\begin{aligned}F\left(\tfrac{\pi}{4}\right) &= -\tfrac12\cos\tfrac{\pi}{2} = -\tfrac12\cdot 0 = 0, \\ F(0) &= -\tfrac12\cos 0 = -\tfrac12\cdot 1 = -\tfrac12.\end{aligned} ∫0π4sin⁡2x dx=0−(−12)=12=0.5.\int_0^{\frac{\pi}{4}}\sin 2x\,dx = 0 - \left(-\tfrac12\right) = \tfrac12 = 0.5.
    Upper minus lower. The value at the lower limit is negative, so subtracting it adds 12\frac12⁠.
    cos⁡π2=0\cos\frac{\pi}{2} = 0⁠, cos⁡0=1\cos 0 = 1⁠. Basics: trigonometric values

  5. Does the size make sense?

    For xx in [0,π4]\left[0, \frac{\pi}{4}\right]⁠, 2x2x is in [0,π2]\left[0, \frac{\pi}{2}\right]⁠, so 0≤sin⁡2x≤10 \le \sin 2x \le 1 and the integral is the shaded area. The arch is bent outwards (it lies above the chord from (0,0)(0, 0) to (π4,1)\left(\frac{\pi}{4}, 1\right)⁠), so the area is more than that triangle's and less than the rectangle of height 1:

    π8≈0.393<0.5<π4≈0.785.\frac{\pi}{8} \approx 0.393 < 0.5 < \frac{\pi}{4} \approx 0.785.
    π/4 π/2 0.5 1 x y y = sin 2x area 1/2 chord (π/4, 1)
    One arch of y=sin⁡2xy = \sin 2x runs from 0 to π2\frac{\pi}{2}⁠. The shaded first half, from 0 to π4\frac{\pi}{4}⁠, has area 12\frac12⁠; by symmetry the whole arch has area 1.
    The triangle with vertices (0,0)(0, 0)⁠, (π4,0)\left(\frac{\pi}{4}, 0\right)⁠, (π4,1)\left(\frac{\pi}{4}, 1\right) has area 12⋅π4⋅1=π8\frac12\cdot\frac{\pi}{4}\cdot 1 = \frac{\pi}{8}⁠; the rectangle on the same base with height 1 has area π4\frac{\pi}{4}⁠. The answer 0.5 sits between them.
    Definite integral as area, Section 7.8.1. Basics: area as a sum of thin strips
Result ∫0π4sin⁡2x dx=12.\int_0^{\frac{\pi}{4}}\sin 2x\,dx = \frac12.

(Textbook answer: 12\frac12⁠, the same.)

Check it with numbers

Numerical integration of sin⁡2x\sin 2x from 0 to π4≈0.7853982\frac{\pi}{4} \approx 0.7853982 (thin strips, no antiderivative used) gives 0.50000000.5000000⁠.

Common mistakes
  1. Missing the 12\frac12⁠. ∫sin⁡2x dx=−12cos⁡2x\int\sin 2x\,dx = -\frac12\cos 2x⁠, not −cos⁡2x-\cos 2x (that would give 1).
  2. Sign of the antiderivative. ∫sin⁡=−cos⁡\int\sin = -\cos⁠; with +cos⁡+\cos the answer comes out as −12-\frac12⁠.
  3. Evaluating cos⁡π4\cos\frac{\pi}{4} instead of cos⁡π2\cos\frac{\pi}{2}⁠. The angle inside is 2x2x⁠, so at x=π4x = \frac{\pi}{4} it is π2\frac{\pi}{2}⁠.
What to take away
  1. ∫sin⁡kx dx=−1kcos⁡kx+C\int\sin kx\,dx = -\frac1k\cos kx + C for a constant k≠0k \ne 0⁠.
  2. In F(b)F(b)⁠, substitute the limit into the whole inside expression (here 2x2x⁠).
  3. An arch-shaped area lies between the inscribed triangle and the enclosing rectangle.

Every fact used, and where it comes from

FactUsed inSource
∫abf=F(b)−F(a)\int_a^b f = F(b) - F(a) for continuous ffSteps 1, 4Theorem 2, Section 7.8.3 (Lesson)
Composition of continuous functions is continuousStep 1Chapter 5, Theorem 2, Section 5.2.1
∫sin⁡u du=−cos⁡u+C\int\sin u\,du = -\cos u + CStep 2Standard formula (iii), Section 7.2
Substitution u=2xu = 2xStep 2Section 7.3.1 (Lesson)
Chain ruleStep 3Chapter 5, Section 5.3.1 (Basics)
cos⁡0=1\cos 0 = 1⁠, cos⁡π2=0\cos\frac{\pi}{2} = 0Step 4Trigonometric values (Basics)
Definite integral as areaStep 5Section 7.8.1 (Basics)
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