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Chapter 7 · Integrals

Example 7: ∫cos⁡2x dx\int\cos^2x\,dx⁠, ∫sin⁡2xcos⁡3x dx\int\sin 2x\cos 3x\,dx⁠, ∫sin⁡3x dx\int\sin^3x\,dx

ExamplesIntegration using trigonometric identities · Section 7.3.2
The question

Find (i) ∫cos⁡2x dx\displaystyle\int\cos^2x\,dx    (ii) ∫sin⁡2xcos⁡3x dx\displaystyle\int\sin 2x\cos 3x\,dx    (iii) ∫sin⁡3x dx\displaystyle\int\sin^3x\,dx

Exam answer

(i) Let I=∫cos⁡2x dxI = \displaystyle\int\cos^2x\,dx⁠. Using cos⁡2x=1+cos⁡2x2\cos^2x = \frac{1 + \cos 2x}{2}⁠,

I=12∫dx+12∫cos⁡2x dxI = \frac12\int dx + \frac12\int\cos 2x\,dx

∴ I=x2+14sin⁡2x+C\therefore\ I = \frac x2 + \frac14\sin 2x + C⁠.

(ii) Let I=∫sin⁡2xcos⁡3x dxI = \displaystyle\int\sin 2x\cos 3x\,dx⁠.

Using 2sin⁡Acos⁡B2\sin A\cos B =sin⁡(A+B)+sin⁡(A−B)= \sin(A + B) + \sin(A - B)⁠,

I=12∫(sin⁡5x+sin⁡(−x)) dx=12∫(sin⁡5x−sin⁡x) dx=12(−cos⁡5x5+cos⁡x)+C\begin{aligned} I &= \frac12\int\big(\sin 5x + \sin(-x)\big)\,dx \\ &= \frac12\int(\sin 5x - \sin x)\,dx \\ &= \frac12\left(-\frac{\cos 5x}{5} + \cos x\right) + C \end{aligned}

∴ I=−110cos⁡5x+12cos⁡x\therefore\ I = -\frac{1}{10}\cos 5x + \frac12\cos x +C+ C⁠.

(iii) Let I=∫sin⁡3x dxI = \displaystyle\int\sin^3x\,dx⁠.

Using sin⁡3x=3sin⁡x−4sin⁡3x\sin 3x = 3\sin x - 4\sin^3x⁠,

I=∫3sin⁡x−sin⁡3x4 dx=34∫sin⁡x dx−14∫sin⁡3x dx\begin{aligned} I &= \int\frac{3\sin x - \sin 3x}{4}\,dx \\ &= \frac34\int\sin x\,dx \\ &\quad - \frac14\int\sin 3x\,dx \end{aligned}

∴ I=−34cos⁡x+112cos⁡3x\therefore\ I = -\dfrac34\cos x + \dfrac{1}{12}\cos 3x +C+ C⁠.

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

The one idea this problem is built on

Powers and products of sines and cosines are not on the list of standard integrals, but single sines and cosines of multiples of xx are: ∫cos⁡kx dx=1ksin⁡kx\int\cos kx\,dx = \frac1k\sin kx and ∫sin⁡kx dx=−1kcos⁡kx\int\sin kx\,dx = -\frac1k\cos kx⁠. Trigonometric identities convert squares, cubes and products into sums of single sines and cosines, which can then be integrated term by term.

Given, to find, and the plan
Given
cos⁡2x\cos^2x⁠, sin⁡2xcos⁡3x\sin 2x\cos 3x and sin⁡3x\sin^3x⁠, each defined for all real xx⁠.
To find
Their indefinite integrals.
Plan
(i) cos⁡2x=1+cos⁡2x2\cos^2x = \frac{1 + \cos 2x}{2}⁠. (ii) sin⁡Acos⁡B\sin A\cos B =12[sin⁡(A+B)+sin⁡(A−B)]= \frac12[\sin(A + B) + \sin(A - B)]⁠. (iii) sin⁡3x=3sin⁡x−sin⁡3x4\sin^3x = \frac{3\sin x - \sin 3x}{4}⁠; also the alternative t=cos⁡xt = \cos x⁠, and compare the two answers. Differentiate every answer back.

Basics you need

1Integrals of sin⁡kx\sin kx and cos⁡kx\cos kx

For a constant k≠0k \ne 0⁠: ∫cos⁡kx dx=1ksin⁡kx+C\int\cos kx\,dx = \frac1k\sin kx + C⁠, ∫sin⁡kx dx=−1kcos⁡kx+C\int\sin kx\,dx = -\frac1k\cos kx + C (Example 5(i)). Basics: substitution, the chain rule backwards

2Double and triple angles

cos⁡2x=2cos⁡2x−1\cos 2x = 2\cos^2x - 1⁠; sin⁡3x=3sin⁡x−4sin⁡3x\sin 3x = 3\sin x - 4\sin^3x⁠; cos⁡3x=4cos⁡3x−3cos⁡x\cos 3x = 4\cos^3x - 3\cos x⁠. Basics: trigonometric identities for integrals

3Compound angles

sin⁡(A±B)\sin(A \pm B) =sin⁡Acos⁡B±cos⁡Asin⁡B= \sin A\cos B \pm \cos A\sin B⁠, and sin⁡(−θ)=−sin⁡θ\sin(-\theta) = -\sin\theta⁠. Basics: trigonometric identities

Part (i): ∫cos⁡2x dx\int\cos^2x\,dx

  1. Lower the power with the double-angle formula

    From cos⁡2x=2cos⁡2x−1\cos 2x = 2\cos^2x - 1⁠, solve for cos⁡2x\cos^2x⁠:

    cos⁡2x=1+cos⁡2x2.\cos^2x = \frac{1 + \cos 2x}{2}.
    cos⁡2x\cos^2x is a square, which we cannot integrate directly. The double-angle formula trades the square for a first power of cos⁡2x\cos 2x⁠, which we can integrate.
    cos⁡2x=2cos⁡2x−1\cos 2x = 2\cos^2x - 1 (Class XI). Basics: trigonometric identities

  2. Integrate term by term

    ∫cos⁡2x dx=12∫dx+12∫cos⁡2x dx=x2+14sin⁡2x+C.\begin{aligned} \int\cos^2x\,dx &= \frac12\int dx + \frac12\int\cos 2x\,dx \\ &= \frac x2 + \frac14\sin 2x + C. \end{aligned}
    ∫dx=x\int dx = x⁠. ∫cos⁡2x dx=12sin⁡2x\int\cos 2x\,dx = \frac12\sin 2x (Example 1(i)), and 12⋅12=14\frac12\cdot\frac12 = \frac14⁠.
    Section 7.3.2. Lesson: integration by substitution

  3. Check: differentiate the answer

    ddx(x2+14sin⁡2x)=12+12cos⁡2x=1+cos⁡2x2=cos⁡2x. ✓\begin{aligned} \frac{d}{dx}\left(\frac x2 + \frac14\sin 2x\right) &= \frac12 + \frac12\cos 2x \\ &= \frac{1 + \cos 2x}{2} = \cos^2x. \ \checkmark \end{aligned}
    The chain rule gives ddxsin⁡2x=2cos⁡2x\frac{d}{dx}\sin 2x = 2\cos 2x⁠, and 14⋅2=12\frac14\cdot 2 = \frac12⁠. Part (i), Step 1 read backwards returns cos⁡2x\cos^2x⁠.

Part (ii): ∫sin⁡2xcos⁡3x dx\int\sin 2x\cos 3x\,dx

  1. Why sin⁡Acos⁡B\sin A\cos B =12[sin⁡(A+B)+sin⁡(A−B)]= \frac12[\sin(A + B) + \sin(A - B)]

    The book asks "(Why?)". Write the two compound-angle formulas and add them:

    sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡Bsin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\begin{aligned} \sin(A + B) &= \sin A\cos B + \cos A\sin B \\ \sin(A - B) &= \sin A\cos B - \cos A\sin B \end{aligned}

    Adding, the cos⁡Asin⁡B\cos A\sin B terms cancel:

    sin⁡(A+B)+sin⁡(A−B)=2sin⁡Acos⁡B.\sin(A + B) + \sin(A - B) = 2\sin A\cos B.

    Divide by 2 to get the product-to-sum formula.

    A product of two trigonometric functions is not integrable term by term, but a sum is. This identity converts the product into a sum of two single sines.
    Compound-angle formulas (Class XI). Basics: trigonometric identities

  2. Apply it with A=2xA = 2x⁠, B=3xB = 3x

    sin⁡2xcos⁡3x=12[sin⁡5x+sin⁡(−x)]=12[sin⁡5x−sin⁡x].\begin{aligned} \sin 2x\cos 3x &= \tfrac12[\sin 5x + \sin(-x)] \\ &= \tfrac12[\sin 5x - \sin x]. \end{aligned}
    A+B=5xA + B = 5x and A−B=−xA - B = -x⁠. Sine is an odd function, sin⁡(−x)=−sin⁡x\sin(-x) = -\sin x⁠, which is where the minus sign in the book's line comes from.
    sin⁡(−θ)=−sin⁡θ\sin(-\theta) = -\sin\theta⁠. Basics: trigonometric identities

  3. Integrate

    ∫sin⁡2xcos⁡3x dx=12[∫sin⁡5x dx−∫sin⁡x dx]=12[−15cos⁡5x+cos⁡x]+C=−110cos⁡5x+12cos⁡x+C.\begin{aligned} &\int\sin 2x\cos 3x\,dx \\ &= \frac12\left[\int\sin 5x\,dx - \int\sin x\,dx\right] \\ &= \frac12\left[-\frac15\cos 5x + \cos x\right] + C \\ &= -\frac{1}{10}\cos 5x + \frac12\cos x + C. \end{aligned}
    ∫sin⁡5x dx=−15cos⁡5x\int\sin 5x\,dx = -\frac15\cos 5x (the 1k\frac1k rule with k=5k = 5⁠), and −∫sin⁡x dx=−(−cos⁡x)=+cos⁡x-\int\sin x\,dx = -(-\cos x) = +\cos x⁠.
    Example 5(i) with m=5m = 5⁠. Lesson: integration by substitution

  4. Check: differentiate the answer

    ddx(−110cos⁡5x+12cos⁡x)=12sin⁡5x−12sin⁡x=sin⁡2xcos⁡3x. ✓\begin{aligned} &\frac{d}{dx}\left(-\frac{1}{10}\cos 5x + \frac12\cos x\right) \\ &= \frac12\sin 5x - \frac12\sin x \\ &= \sin 2x\cos 3x. \ \checkmark \end{aligned}
    ddxcos⁡5x=−5sin⁡5x\frac{d}{dx}\cos 5x = -5\sin 5x⁠, and −110⋅(−5)=12-\frac{1}{10}\cdot(-5) = \frac12⁠. The last equality is Part (ii), Step 2.

Part (iii): ∫sin⁡3x dx\int\sin^3x\,dx

  1. Lower the power with the triple-angle formula

    From sin⁡3x=3sin⁡x−4sin⁡3x\sin 3x = 3\sin x - 4\sin^3x⁠, solve for sin⁡3x\sin^3x⁠:

    sin⁡3x=3sin⁡x−sin⁡3x4.\sin^3x = \frac{3\sin x - \sin 3x}{4}.
    This turns a cube into a combination of first powers of sin⁡x\sin x and sin⁡3x\sin 3x⁠, both integrable.
    sin⁡3x=3sin⁡x−4sin⁡3x\sin 3x = 3\sin x - 4\sin^3x (Class XI). Basics: trigonometric identities

  2. Integrate

    ∫sin⁡3x dx=34∫sin⁡x dx−14∫sin⁡3x dx=−34cos⁡x+112cos⁡3x+C.\begin{aligned} \int\sin^3x\,dx &= \frac34\int\sin x\,dx - \frac14\int\sin 3x\,dx \\ &= -\frac34\cos x + \frac{1}{12}\cos 3x + C. \end{aligned}
    ∫sin⁡x dx=−cos⁡x\int\sin x\,dx = -\cos x and ∫sin⁡3x dx=−13cos⁡3x\int\sin 3x\,dx = -\frac13\cos 3x⁠, so −14⋅(−13)=+112-\frac14\cdot(-\frac13) = +\frac{1}{12}⁠.
    Standard formula (iii), Section 7.2; Example 5(i). Lesson: integration as anti-differentiation

  3. Alternative: substitution t=cos⁡xt = \cos x

    Save one sin⁡x\sin x⁠: sin⁡3x=(1−cos⁡2x)sin⁡x\sin^3x = (1 - \cos^2x)\sin x⁠. Put t=cos⁡xt = \cos x⁠, dt=−sin⁡x dxdt = -\sin x\,dx⁠:

    ∫sin⁡3x dx=−∫(1−t2) dt=−t+t33+C=−cos⁡x+13cos⁡3x+C.\begin{aligned} \int\sin^3x\,dx &= -\int(1 - t^2)\,dt \\ &= -t + \frac{t^3}{3} + C \\ &= -\cos x + \frac13\cos^3x + C. \end{aligned}
    The same "odd power" method as Example 6(i): one sin⁡x\sin x goes with dxdx to make −dt-dt⁠, and the even power sin⁡2x\sin^2x becomes 1−t21 - t^2⁠.
    Substitution, Section 7.3.1. Basics: substitution

  4. The two answers are the same function

    The book remarks that both answers are equivalent. Use cos⁡3x=4cos⁡3x−3cos⁡x\cos 3x = 4\cos^3x - 3\cos x⁠:

    −34cos⁡x+112cos⁡3x=−34cos⁡x+13cos⁡3x−14cos⁡x=−cos⁡x+13cos⁡3x.\begin{aligned} &-\frac34\cos x + \frac{1}{12}\cos 3x \\ &= -\frac34\cos x + \frac13\cos^3x - \frac14\cos x \\ &= -\cos x + \frac13\cos^3x. \end{aligned}
    Two correct anti derivatives can differ at most by a constant (Section 7.2, Remark). Here the difference is exactly 0: the two expressions are identical, so even the same CC can be used.
    Triple-angle formula for cosine (Class XI). Basics: trigonometric identities

  5. Check: differentiate the answer

    ddx(−cos⁡x+13cos⁡3x)=sin⁡x−cos⁡2xsin⁡x=sin⁡x (1−cos⁡2x)=sin⁡3x. ✓\begin{aligned} &\frac{d}{dx}\left(-\cos x + \frac13\cos^3x\right) \\ &= \sin x - \cos^2x\sin x \\ &= \sin x\,(1 - \cos^2x) = \sin^3x. \ \checkmark \end{aligned}
    ddxcos⁡3x=3cos⁡2x⋅(−sin⁡x)\frac{d}{dx}\cos^3x = 3\cos^2x\cdot(-\sin x)⁠. Since Part (iii), Step 4 shows the first answer is the same function, this checks both.
Result

(i) ∫cos⁡2x dx=x2+14sin⁡2x+C\displaystyle\int\cos^2x\,dx = \frac x2 + \frac14\sin 2x + C⁠.

(ii) ∫sin⁡2xcos⁡3x dx\displaystyle\int\sin 2x\cos 3x\,dx =−110cos⁡5x+12cos⁡x+C= -\frac{1}{10}\cos 5x + \frac12\cos x + C⁠.

(iii) ∫sin⁡3x dx\displaystyle\int\sin^3x\,dx =−34cos⁡x+112cos⁡3x+C= -\frac34\cos x + \frac{1}{12}\cos 3x + C⁠, which is the same function as −cos⁡x+13cos⁡3x+C-\cos x + \frac13\cos^3x + C⁠.

All are valid for every real xx⁠, and agree with the textbook's solution.

Check it with numbers

By Section 7.8, ∫abf(x) dx=F(b)−F(a)\int_a^b f(x)\,dx = F(b) - F(a)⁠. The middle column is a computer's sum of very many thin strips, found without our answers:

PartIntervalSum of thin stripsF(b)−F(a)F(b) - F(a)
(i)[0,π2][0, \frac{\pi}{2}]0.78539820.7853982π4≈0.7853982\frac{\pi}{4} \approx 0.7853982
(ii)[0,π2][0, \frac{\pi}{2}]−0.4-0.40−(−110+12)=−0.40 - (-\frac{1}{10} + \frac12) = -0.4
(iii)[0,π][0, \pi]1.33333331.333333332−16=43\frac32 - \frac16 = \frac43

For (iii) the alternative form gives (1−13)−(−1+13)=43(1 - \frac13) - (-1 + \frac13) = \frac43 too. At x=1x = 1 the two forms of (iii) both equal −0.4877261…-0.4877261\ldots⁠.

Common mistakes
  1. ∫cos⁡2x dx=cos⁡3x3\int\cos^2x\,dx = \frac{\cos^3x}{3}⁠. The power rule is for powers of xx⁠, not of cos⁡x\cos x⁠; differentiating cos⁡3x3\frac{\cos^3x}{3} gives −cos⁡2xsin⁡x-\cos^2x\sin x⁠.
  2. Losing the sign of sin⁡(−x)\sin(-x) in part (ii).
  3. Forgetting the 1k\frac1k when integrating sin⁡kx\sin kx or cos⁡kx\cos kx⁠.
  4. Thinking two different-looking answers must disagree. Check whether they differ by a constant (here, by nothing at all).
What to take away
  1. sin⁡2\sin^2⁠, cos⁡2\cos^2⁠: use the double-angle formulas. sin⁡3\sin^3⁠, cos⁡3\cos^3⁠: use the triple-angle formulas, or save one factor and substitute.
  2. Products sin⁡Acos⁡B\sin A\cos B⁠, cos⁡Acos⁡B\cos A\cos B⁠, sin⁡Asin⁡B\sin A\sin B⁠: use product-to-sum formulas.
  3. Different methods can give answers that look different; they differ by a constant at most.

Every fact used, and where it comes from

FactUsed inSource
cos⁡2x=2cos⁡2x−1\cos 2x = 2\cos^2x - 1Part (i), Step 1Class XI, Trigonometric Functions
sin⁡Acos⁡B\sin A\cos B =12[sin⁡(A+B)+sin⁡(A−B)]= \frac12[\sin(A + B) + \sin(A - B)]Part (ii), Steps 1, 2Class XI compound angles; Section 7.3.2
sin⁡3x=3sin⁡x−4sin⁡3x\sin 3x = 3\sin x - 4\sin^3x⁠, cos⁡3x=4cos⁡3x−3cos⁡x\cos 3x = 4\cos^3x - 3\cos xPart (iii), Steps 1, 4Class XI, Trigonometric Functions (Basics: trigonometric identities)
∫sin⁡kx dx=−1kcos⁡kx+C\int\sin kx\,dx = -\frac1k\cos kx + C⁠, ∫cos⁡kx dx=1ksin⁡kx+C\int\cos kx\,dx = \frac1k\sin kx + CAll partsExample 5(i), Example 1(i)
Substitution t=cos⁡xt = \cos xPart (iii), Step 3Section 7.3.1 (Lesson: integration by substitution)
Anti derivatives differ by a constantPart (iii), Step 4Remark, Section 7.2; Remark after Example 7
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