Trig Identities in Integration

Using double-angle formulae to integrate $\sin^2 x$, $\cos^2(2x)$, $\tan^2 x$ and beyond
9709 Paper 3 ยท ยง3.5 Integration ยท Week 6 Lesson 3
๐Ÿ’ก Lead In: A Power the Table Cannot Reach

Quick Recall

Quick check โ€” pick the correct statement in each question before we meet today's wall.

Recall 1 โ€” Which row of the integral table is correct?
$\displaystyle\int \sin x \, \text{d}x = \cos x + C, \quad \int \cos x \, \text{d}x = -\sin x + C$
$\displaystyle\int \sin x \, \text{d}x = -\cos x + C, \quad \int \cos x \, \text{d}x = \sin x + C$
$\displaystyle\int \sin x \, \text{d}x = -\sin x + C, \quad \int \cos x \, \text{d}x = \cos x + C$
$\displaystyle\int \sin x \, \text{d}x = -\cos x + C, \quad \int \cos x \, \text{d}x = -\sin x + C$

$\displaystyle\int \sin x \, \text{d}x = -\cos x + C$ and $\displaystyle\int \cos x \, \text{d}x = \sin x + C$ โ€” each reverses the corresponding differentiation rule, including the sign of the sine case.

Recall 2 โ€” Which form of the double-angle identity is most useful for integrating $\sin^2 x$ or $\cos^2 x$?
$\cos 2A = \cos^2 A - \sin^2 A$
$\cos 2A = 2\cos^2 A - 1$
Solving $\cos 2A = 2\cos^2 A - 1 = 1 - 2\sin^2 A$ for $\cos^2 A$ or $\sin^2 A$
$\sin 2A = 2\sin A \cos A$

Rearranging gives $\sin^2 A = \dfrac{1-\cos 2A}{2}$ and $\cos^2 A = \dfrac{1+\cos 2A}{2}$ โ€” linear in $\cos 2A$, so both integrate cleanly. This is exactly the identity needed for $\displaystyle\int \sin^2 x \, \text{d}x$.

The Wall

Now try to integrate this:

$\displaystyle \int \sin^2 x \, \text{d}x = \;?$

A student reaches for the โ€œraise the power and divideโ€ rule from polynomial integration and writes:

$\displaystyle \int \sin^2 x \, \text{d}x \stackrel{?}{=} -\dfrac{\cos^3 x}{3} + C$

But differentiating the right-hand side gives $\sin^2 x \cos x$, not $\sin^2 x$. The chain rule has smuggled in an extra factor of $\cos x$. The polynomial rule fails because $\sin x$ is not a free variable โ€” it is a function with its own derivative.

So $\sin^2 x$ is not in our integral table. We need a way to rewrite $\sin^2 x$ using only first powers of trig functions, which are in the table.

The Need

We need a strategy for converting powers of trig functions into first powers of multiple angles โ€” and the double-angle identities from Week 3 are exactly the conversion tool.

๐Ÿ‘ The Bridge (click to reveal)

From $\cos 2A = 1 - 2\sin^2 A$, rearrange to solve for $\sin^2 A$:

$$\sin^2 A = \dfrac{1 - \cos 2A}{2}$$

So $\displaystyle \int \sin^2 x \, \text{d}x = \int \dfrac{1 - \cos 2x}{2} \, \text{d}x = \dfrac{x}{2} - \dfrac{\sin 2x}{4} + C$.

Same trick for $\cos^2 A$: from $\cos 2A = 2\cos^2 A - 1$, we get $\cos^2 A = \dfrac{1 + \cos 2A}{2}$. Today we generalise this strategy and apply it to $\cos^2(2x)$, $\tan^2 x$, and even $(a\cos x + b\sin x)^2$.

๐Ÿ“– Key Concepts: Converting Powers to Multiple Angles

The integral table from Lesson 2 covers first powers of $\sin$, $\cos$, $\sec^2$ and linear arguments like $\sin(ax+b)$. Squared (or higher-power) trig functions are not in the table. The strategy: use a trig identity to replace the power with a multiple angle, then integrate term by term.

Key relation 1 โ€” integrate $\sin^2 x$:

From $\cos 2A = 1 - 2\sin^2 A$, solve for $\sin^2 A$:

$$\sin^2 A = \dfrac{1 - \cos 2A}{2}$$

Substitute $A = x$ (or $A = kx$) and integrate:

$$\int \sin^2 x \, \text{d}x = \int \dfrac{1 - \cos 2x}{2} \, \text{d}x = \dfrac{x}{2} - \dfrac{\sin 2x}{4} + C$$

Key relation 2 โ€” integrate $\cos^2 x$:

From $\cos 2A = 2\cos^2 A - 1$, solve for $\cos^2 A$:

$$\cos^2 A = \dfrac{1 + \cos 2A}{2}$$

Substitute $A = x$ (or $A = kx$) and integrate:

$$\int \cos^2 x \, \text{d}x = \int \dfrac{1 + \cos 2x}{2} \, \text{d}x = \dfrac{x}{2} + \dfrac{\sin 2x}{4} + C$$

Notice the only difference from $\sin^2 x$ is the sign in front of the cosine term. The two identities are the same equation, just rearranged differently.

Key relation 3 โ€” the general strategy:

Whenever you see a power of a trig function inside an integral, convert it to first powers of multiple angles using an identity, then integrate term by term.

Three identities you will use most:

$\sin^2(kx) = \dfrac{1 - \cos(2kx)}{2}$, โ€ƒ $\cos^2(kx) = \dfrac{1 + \cos(2kx)}{2}$, โ€ƒ $\tan^2 x = \sec^2 x - 1$.

For higher even powers (e.g. $\cos^4 x$), apply the identity repeatedly or use a product-to-sum formula. For odd powers, a one-off factorisation often suffices (e.g. $\cos^3 x = \cos^2 x \cdot \cos x$ and then a substitution $u = \sin x$ โ€” though substitution is out of scope here, the factorisation $\cos^3 x = \tfrac{1}{4}(\cos 3x + 3\cos x)$ from the triple-angle identity handles it without substitution).

Common pattern: the angle always doubles. $\sin^2(kx) \to \cos(2kx)$, $\cos^2(kx) \to \cos(2kx)$, $\cos^4(kx) \to \cos(2kx)$ and $\cos(4kx)$. Each doubling corresponds to halving the period, which is what makes higher-power trig functions oscillate faster.

Group Activity 1: Pattern Detective โ€” What Is the Same Here?

Work in a group of three. Each level shows three trig-power integrals. Your job is not to calculate them but to name the identity they share โ€” and then to say what the angle becomes after the conversion. Level 3 only opens when level 2 is cleared, and only a clean first-attempt run at every level earns all three stars.

Level: 1 / 3 First-attempt clear: 0 / 3

Group rule: every member must be able to justify the shared identity in one sentence before the group presses an option. If members disagree, they say why โ€” that disagreement is the point of the game.

Individual Activity 2: Sort the Squares โ€” Which Identity Family?

Work on your own. Drag each integral โ€” or tap it and then tap a box โ€” into the identity family it needs. A card placed in the wrong box flashes red and returns to the pile, so read the function and its power before you move. Six integrals, three families.

Sorted: 0 / 6 First-time correct: 0 Wrong drops: 0
Integrals to sort

Hint: look at the trig function and its power only. $\sin^{2}$ and $\sin^{4}$ both subtract a cosine after the conversion; $\cos^{2}$ and $\cos^{4}$ both add one; $\tan^{2}$ becomes $\sec^{2}A - 1$. After the conversion the angle always doubles.

Group ยท Guided Worked Example 1: $\int \cos^2 x \, \text{d}x$ and a Definite Integral Exercise 5B Q7

(i) Express $\cos^2 x$ in terms of $\cos 2x$.

(ii) Hence show that $\displaystyle \int_{0}^{\frac{1}{3}\pi} \cos^2 x \, \text{d}x = \dfrac{1}{6}\pi + \dfrac{1}{8}\sqrt{3}$.

(iii) By using an appropriate trigonometrical identity, deduce the exact value of $\displaystyle \int_{0}^{\frac{1}{3}\pi} \sin^2 x \, \text{d}x$.

Cambridge International AS & A Level Mathematics 9709 Paper 2 Q6 June 2007

๐Ÿ‘ Reveal Solution
Part (i): Express $\cos^2 x$ in terms of $\cos 2x$

From $\cos 2x = 2\cos^2 x - 1$, rearrange:

$$2\cos^2 x = 1 + \cos 2x \;\Longrightarrow\; \cos^2 x = \dfrac{1 + \cos 2x}{2}$$

Part (ii): Show $\displaystyle \int_{0}^{\pi/3} \cos^2 x \, \text{d}x = \dfrac{\pi}{6} + \dfrac{\sqrt{3}}{8}$

Substitute the result of part (i):

$$\int_{0}^{\pi/3} \cos^2 x \, \text{d}x = \int_{0}^{\pi/3} \dfrac{1 + \cos 2x}{2} \, \text{d}x = \dfrac{1}{2}\!\int_{0}^{\pi/3}\! 1 \, \text{d}x + \dfrac{1}{2}\!\int_{0}^{\pi/3}\! \cos 2x \, \text{d}x$$

$$= \left[\dfrac{x}{2}\right]_{0}^{\pi/3} + \dfrac{1}{2}\cdot\dfrac{1}{2}\Big[\sin 2x\Big]_{0}^{\pi/3} = \dfrac{\pi}{6} + \dfrac{1}{4}\!\left(\sin \dfrac{2\pi}{3} - \sin 0\right)$$

Since $\sin(2\pi/3) = \sqrt{3}/2$ and $\sin 0 = 0$:

$$= \dfrac{\pi}{6} + \dfrac{1}{4}\cdot\dfrac{\sqrt{3}}{2} = \dfrac{\pi}{6} + \dfrac{\sqrt{3}}{8}$$

$\boxed{\displaystyle \int_{0}^{\pi/3} \cos^2 x \, \text{d}x = \dfrac{\pi}{6} + \dfrac{\sqrt{3}}{8}}$

Part (iii): Deduce $\displaystyle \int_{0}^{\pi/3} \sin^2 x \, \text{d}x$

Use $\sin^2 x + \cos^2 x \equiv 1$, so $\sin^2 x = 1 - \cos^2 x$. Then:

$$\int_{0}^{\pi/3} \sin^2 x \, \text{d}x = \int_{0}^{\pi/3} (1 - \cos^2 x) \, \text{d}x = \int_{0}^{\pi/3} 1 \, \text{d}x - \int_{0}^{\pi/3} \cos^2 x \, \text{d}x$$

$$= \dfrac{\pi}{3} - \left(\dfrac{\pi}{6} + \dfrac{\sqrt{3}}{8}\right) = \dfrac{\pi}{6} - \dfrac{\sqrt{3}}{8}$$

$\boxed{\displaystyle \int_{0}^{\pi/3} \sin^2 x \, \text{d}x = \dfrac{\pi}{6} - \dfrac{\sqrt{3}}{8}}$

Sanity check: the two results add to $\pi/3$, which equals $\int_0^{\pi/3} 1 \, \text{d}x$ โ€” consistent with $\sin^2 x + \cos^2 x = 1$.

Individual ยท Guided Worked Example 2: A Squared Linear Combination Exercise 5B Q8

(i) Prove the identity $(\cos x + 3\sin x)^2 \equiv 5 - 4\cos 2x + 3\sin 2x$.

(ii) Using the identity, or otherwise, find the exact value of $\displaystyle \int_{0}^{\frac{1}{4}\pi} (\cos x + 3\sin x)^2 \, \text{d}x$.

๐Ÿ‘ Reveal Solution
Part (i): Prove $(\cos x + 3\sin x)^2 \equiv 5 - 4\cos 2x + 3\sin 2x$

Expand the square:

$$(\cos x + 3\sin x)^2 = \cos^2 x + 6\sin x \cos x + 9\sin^2 x$$

Apply three identities: $\cos^2 x = \tfrac{1}{2}(1 + \cos 2x)$, $\sin^2 x = \tfrac{1}{2}(1 - \cos 2x)$, and $2\sin x \cos x = \sin 2x$:

$$= \dfrac{1 + \cos 2x}{2} + 3(2\sin x \cos x) + 9 \cdot \dfrac{1 - \cos 2x}{2}$$

$$= \dfrac{1 + \cos 2x}{2} + 3\sin 2x + \dfrac{9 - 9\cos 2x}{2}$$

$$= \dfrac{(1 + \cos 2x) + (9 - 9\cos 2x)}{2} + 3\sin 2x = \dfrac{10 - 8\cos 2x}{2} + 3\sin 2x$$

$$= 5 - 4\cos 2x + 3\sin 2x$$

$\boxed{(\cos x + 3\sin x)^2 \equiv 5 - 4\cos 2x + 3\sin 2x}$ โ€‚ $\checkmark$

Part (ii): Find $\displaystyle \int_{0}^{\pi/4} (\cos x + 3\sin x)^2 \, \text{d}x$

Use the identity from part (i):

$$\int_{0}^{\pi/4} (\cos x + 3\sin x)^2 \, \text{d}x = \int_{0}^{\pi/4} (5 - 4\cos 2x + 3\sin 2x) \, \text{d}x$$

Integrate term by term:

$$= \Big[5x - 2\sin 2x - \tfrac{3}{2}\cos 2x\Big]_{0}^{\pi/4}$$

Evaluate at the upper limit $x = \pi/4$ (so $2x = \pi/2$, $\sin(\pi/2) = 1$, $\cos(\pi/2) = 0$):

$$5\cdot\dfrac{\pi}{4} - 2(1) - \tfrac{3}{2}(0) = \dfrac{5\pi}{4} - 2$$

Evaluate at the lower limit $x = 0$ (so $2x = 0$, $\sin 0 = 0$, $\cos 0 = 1$):

$$5(0) - 2(0) - \tfrac{3}{2}(1) = -\dfrac{3}{2}$$

Subtract:

$$\left(\dfrac{5\pi}{4} - 2\right) - \left(-\dfrac{3}{2}\right) = \dfrac{5\pi}{4} - 2 + \dfrac{3}{2} = \dfrac{5\pi}{4} - \dfrac{1}{2}$$

$\boxed{\displaystyle \int_{0}^{\pi/4} (\cos x + 3\sin x)^2 \, \text{d}x = \dfrac{5\pi}{4} - \dfrac{1}{2}}$

โœ๏ธ Concept Quiz
Score: 0 / 4
1. Which identity converts $\sin^2 x$ into a form you can integrate directly?
$\sin^2 x = \dfrac{1 + \cos 2x}{2}$
$\sin^2 x = \dfrac{1 - \cos 2x}{2}$
$\sin^2 x = \dfrac{1 + \sin 2x}{2}$
$\sin^2 x = \sec^2 x - 1$
From $\cos 2x = 1 - 2\sin^2 x$, rearrange: $2\sin^2 x = 1 - \cos 2x$, so $\sin^2 x = \tfrac{1}{2}(1 - \cos 2x)$. The minus sign is the distinguishing feature โ€” $\sin^2$ and $\cos^2$ differ only in the sign in front of the cosine term. The fourth option is the identity for $\tan^2 x$, not $\sin^2 x$.
2. After rewriting $\cos^2(2x) = \tfrac{1}{2}(1 + \cos 4x)$, what is $\displaystyle \int \cos^2(2x) \, \text{d}x$?
$\dfrac{x}{2} + \dfrac{\sin 4x}{4} + C$
$\dfrac{x}{2} + \dfrac{\sin 4x}{8} + C$
$x + \dfrac{\sin 4x}{2} + C$
$2x + \sin 4x + C$
$\int \cos^2(2x) \, \text{d}x = \tfrac{1}{2}\int (1 + \cos 4x) \, \text{d}x = \tfrac{x}{2} + \tfrac{1}{2}\int \cos 4x \, \text{d}x$. The chain rule gives $\int \cos(4x) \, \text{d}x = \tfrac{1}{4}\sin(4x)$, so the second term is $\tfrac{1}{2} \cdot \tfrac{1}{4}\sin 4x = \tfrac{\sin 4x}{8}$. The most common error is forgetting the chain-rule factor of $1/4$ โ€” that gives the trap answer $\sin 4x / 4$.
3. To convert $\tan^2 x$ into an integrable form, use:
$\tan^2 x = \sec^2 x - 1$
$\tan^2 x = 1 - \sec^2 x$
$\tan^2 x = \dfrac{1 - \cos 2x}{2}$
$\tan^2 x = \dfrac{1 + \cos 2x}{2}$
Divide $\sin^2 x + \cos^2 x = 1$ by $\cos^2 x$ to get $\tan^2 x + 1 = \sec^2 x$, hence $\tan^2 x = \sec^2 x - 1$. Then $\int \tan^2 x \, \text{d}x = \int (\sec^2 x - 1) \, \text{d}x = \tan x - x + C$. The double-angle identities in the third and fourth options apply to $\sin^2 x$ and $\cos^2 x$, not $\tan^2 x$.
4. To integrate $(a\cos x + b\sin x)^2$, the first step is:
Apply integration by substitution $u = a\cos x + b\sin x$.
Expand the square, then rewrite $\cos^2 x$, $\sin^2 x$ and $\sin x \cos x$ using double-angle identities.
Apply integration by parts with $u = (a\cos x + b\sin x)^2$.
Use the R-form $a\cos x + b\sin x = R\cos(x - \alpha)$ and then square $R$.
Expand $(a\cos x + b\sin x)^2 = a^2\cos^2 x + 2ab\sin x\cos x + b^2\sin^2 x$, then apply $\cos^2 x = \tfrac{1}{2}(1+\cos 2x)$, $\sin^2 x = \tfrac{1}{2}(1-\cos 2x)$, and $2\sin x\cos x = \sin 2x$. The result is a linear combination of $1$, $\cos 2x$ and $\sin 2x$, which integrates term by term. (The R-form approach in the fourth option does work but is more steps โ€” you would still need to square and expand, so it is not the first step.)
๐ŸŽ‰
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๐Ÿ“ More Practice

Attempt each question on paper first, then click Reveal Solution to check your work. Questions are ordered from easiest to hardest.

Exercise 5B Q9 Standard

(i) Show that $\displaystyle \int_{0}^{\frac{1}{4}\pi} \cos 2x \, \text{d}x = \dfrac{1}{2}$.

(ii) By using an appropriate trigonometrical identity, find the exact value of $\displaystyle \int_{\frac{1}{6}\pi}^{\frac{1}{3}\pi} 3\tan^{2}x \, \text{d}x$.

๐Ÿ’ก Hint

For part (i): use $\int \cos(ax) \, \text{d}x = \dfrac{1}{a}\sin(ax) + C$.

For part (ii): use $\tan^2 x = \sec^2 x - 1$, so $3\tan^2 x = 3\sec^2 x - 3$. Recall $\tan(\pi/3) = \sqrt{3}$ and $\tan(\pi/6) = 1/\sqrt{3}$.

๐Ÿ‘ Reveal Solution
Part (i): Show $\displaystyle \int_{0}^{\pi/4} \cos 2x \, \text{d}x = \dfrac{1}{2}$

$$\int_{0}^{\pi/4} \cos 2x \, \text{d}x = \left[\dfrac{1}{2}\sin 2x\right]_{0}^{\pi/4} = \dfrac{1}{2}\sin\!\left(\dfrac{\pi}{2}\right) - \dfrac{1}{2}\sin 0 = \dfrac{1}{2}(1) - 0 = \dfrac{1}{2}$$

$\boxed{\displaystyle \int_{0}^{\pi/4} \cos 2x \, \text{d}x = \dfrac{1}{2}}$ โ€‚ $\checkmark$

Part (ii): Find $\displaystyle \int_{\pi/6}^{\pi/3} 3\tan^2 x \, \text{d}x$

Use $\tan^2 x = \sec^2 x - 1$:

$$3\tan^2 x = 3(\sec^2 x - 1) = 3\sec^2 x - 3$$

$$\int_{\pi/6}^{\pi/3} 3\tan^2 x \, \text{d}x = \Big[3\tan x - 3x\Big]_{\pi/6}^{\pi/3}$$

Evaluate at $x = \pi/3$ ($\tan(\pi/3) = \sqrt{3}$): โ€‚ $3\sqrt{3} - 3\cdot\dfrac{\pi}{3} = 3\sqrt{3} - \pi$.

Evaluate at $x = \pi/6$ ($\tan(\pi/6) = 1/\sqrt{3}$): โ€‚ $3\cdot\dfrac{1}{\sqrt{3}} - 3\cdot\dfrac{\pi}{6} = \dfrac{3}{\sqrt{3}} - \dfrac{\pi}{2} = \sqrt{3} - \dfrac{\pi}{2}$.

Subtract:

$$\Big(3\sqrt{3} - \pi\Big) - \Big(\sqrt{3} - \dfrac{\pi}{2}\Big) = 2\sqrt{3} - \pi + \dfrac{\pi}{2} = 2\sqrt{3} - \dfrac{\pi}{2}$$

$\boxed{\displaystyle \int_{\pi/6}^{\pi/3} 3\tan^2 x \, \text{d}x = 2\sqrt{3} - \dfrac{\pi}{2}}$

Exercise 5B Q10 Medium

(i) Find $\displaystyle \int 4\mathrm{e}^{x}(3 + \mathrm{e}^{2x}) \, \text{d}x$.

(ii) Show that $\displaystyle \int_{-\frac{1}{4}\pi}^{\frac{1}{4}\pi} (3 + 2\tan^2\theta) \, \text{d}\theta = \tfrac{1}{2}(8 + \pi)$.

๐Ÿ’ก Hint

For part (i): expand $4\mathrm{e}^x(3 + \mathrm{e}^{2x}) = 12\mathrm{e}^x + 4\mathrm{e}^{3x}$, then integrate each term using $\int \mathrm{e}^{ax} \, \text{d}x = \dfrac{1}{a}\mathrm{e}^{ax} + C$.

For part (ii): use $\tan^2\theta = \sec^2\theta - 1$, so $3 + 2\tan^2\theta = 3 + 2(\sec^2\theta - 1) = 1 + 2\sec^2\theta$. Note $\tan$ is an odd function, so the integral over a symmetric interval $[-\pi/4, \pi/4]$ simplifies โ€” the $\tan\theta$ term cancels.

๐Ÿ‘ Reveal Solution
Part (i): Find $\displaystyle \int 4\mathrm{e}^{x}(3 + \mathrm{e}^{2x}) \, \text{d}x$

Expand the integrand:

$$4\mathrm{e}^{x}(3 + \mathrm{e}^{2x}) = 12\mathrm{e}^{x} + 4\mathrm{e}^{3x}$$

Integrate term by term:

$$\int 12\mathrm{e}^{x} \, \text{d}x = 12\mathrm{e}^{x}, \qquad \int 4\mathrm{e}^{3x} \, \text{d}x = 4 \cdot \dfrac{1}{3}\mathrm{e}^{3x} = \dfrac{4}{3}\mathrm{e}^{3x}$$

$\boxed{\displaystyle \int 4\mathrm{e}^{x}(3 + \mathrm{e}^{2x}) \, \text{d}x = 12\mathrm{e}^{x} + \dfrac{4}{3}\mathrm{e}^{3x} + C}$

Part (ii): Show $\displaystyle \int_{-\pi/4}^{\pi/4} (3 + 2\tan^2\theta) \, \text{d}\theta = \tfrac{1}{2}(8 + \pi)$

Use $\tan^2\theta = \sec^2\theta - 1$:

$$3 + 2\tan^2\theta = 3 + 2(\sec^2\theta - 1) = 1 + 2\sec^2\theta$$

Integrate:

$$\int_{-\pi/4}^{\pi/4} (1 + 2\sec^2\theta) \, \text{d}\theta = \Big[\theta + 2\tan\theta\Big]_{-\pi/4}^{\pi/4}$$

At $\theta = \pi/4$: โ€‚ $\dfrac{\pi}{4} + 2\tan\!\left(\dfrac{\pi}{4}\right) = \dfrac{\pi}{4} + 2(1) = \dfrac{\pi}{4} + 2$.

At $\theta = -\pi/4$: โ€‚ $-\dfrac{\pi}{4} + 2\tan\!\left(-\dfrac{\pi}{4}\right) = -\dfrac{\pi}{4} + 2(-1) = -\dfrac{\pi}{4} - 2$.

Subtract:

$$\left(\dfrac{\pi}{4} + 2\right) - \left(-\dfrac{\pi}{4} - 2\right) = \dfrac{\pi}{4} + 2 + \dfrac{\pi}{4} + 2 = \dfrac{\pi}{2} + 4 = \dfrac{\pi + 8}{2} = \dfrac{1}{2}(8 + \pi)$$

$\boxed{\displaystyle \int_{-\pi/4}^{\pi/4} (3 + 2\tan^2\theta) \, \text{d}\theta = \dfrac{1}{2}(8 + \pi)}$ โ€‚ $\checkmark$

Exercise 5B Q11 Medium

(i) Prove the identity $\cos 4\theta + 4\cos 2\theta \equiv 8\cos^{4}\theta - 3$.

(ii) Hence (a) solve the equation $\cos 4\theta + 4\cos 2\theta = 1$ for $-\dfrac{1}{2}\pi \leqslant \theta \leqslant \dfrac{1}{2}\pi$.

(b) Find the exact value of $\displaystyle \int_{0}^{\frac{1}{4}\pi} \cos^{4}\theta \, \text{d}\theta$.

Cambridge International AS & A Level Mathematics 9709 Paper 31 Q9 June 2011

๐Ÿ’ก Hint

For part (i): start with $\cos 4\theta = 2\cos^2 2\theta - 1$ and then apply $\cos 2\theta = 2\cos^2\theta - 1$ again to remove the $\cos^2 2\theta$ term.

For part (ii)(a): set $8\cos^4\theta - 3 = 1$ and solve $\cos^4\theta = \tfrac{1}{2}$. Equivalently, let $u = \cos 2\theta$ and solve the quadratic $u^2 + 2u - 1 = 0$.

For part (ii)(b): rearrange the identity from part (i) to express $\cos^4\theta$ in terms of $\cos 4\theta$, $\cos 2\theta$ and a constant. Then integrate term by term.

๐Ÿ‘ Reveal Solution
Part (i): Prove $\cos 4\theta + 4\cos 2\theta \equiv 8\cos^{4}\theta - 3$

Start with the double-angle identity applied twice. First, $\cos 4\theta = 2\cos^2 2\theta - 1$. Then apply $\cos 2\theta = 2\cos^2\theta - 1$ to the $\cos^2 2\theta$ term:

$$\cos^2 2\theta = (2\cos^2\theta - 1)^2 = 4\cos^4\theta - 4\cos^2\theta + 1$$

Substitute:

$$\cos 4\theta = 2(4\cos^4\theta - 4\cos^2\theta + 1) - 1 = 8\cos^4\theta - 8\cos^2\theta + 1$$

Add $4\cos 2\theta = 4(2\cos^2\theta - 1) = 8\cos^2\theta - 4$:

$$\cos 4\theta + 4\cos 2\theta = (8\cos^4\theta - 8\cos^2\theta + 1) + (8\cos^2\theta - 4) = 8\cos^4\theta - 3$$

$\boxed{\cos 4\theta + 4\cos 2\theta \equiv 8\cos^{4}\theta - 3}$ โ€‚ $\checkmark$

Part (ii)(a): Solve $\cos 4\theta + 4\cos 2\theta = 1$ for $-\pi/2 \leq \theta \leq \pi/2$

Using the identity from part (i): $8\cos^4\theta - 3 = 1$, so $\cos^4\theta = \tfrac{1}{2}$. Equivalently (and more cleanly), substitute $u = \cos 2\theta$:

$$\cos 4\theta + 4\cos 2\theta = 1 \;\Longrightarrow\; 2\cos^2 2\theta - 1 + 4\cos 2\theta = 1 \;\Longrightarrow\; 2u^2 + 4u - 2 = 0 \;\Longrightarrow\; u^2 + 2u - 1 = 0$$

$$u = \dfrac{-2 \pm \sqrt{4 + 4}}{2} = -1 \pm \sqrt{2}$$

Since $|u| = |\cos 2\theta| \leq 1$ and $-1 - \sqrt{2} \approx -2.414 < -1$, only $u = -1 + \sqrt{2} \approx 0.414$ is valid. So:

$$\cos 2\theta = \sqrt{2} - 1 \;\Longrightarrow\; 2\theta = \pm \arccos(\sqrt{2} - 1) \;\Longrightarrow\; \theta = \pm \dfrac{1}{2}\arccos(\sqrt{2} - 1)$$

Numerical check: $\arccos(\sqrt{2}-1) \approx 1.144\,\text{rad}$, so $\theta \approx \pm 0.572\,\text{rad} \approx \pm 32.8^{\circ}$, both within $[-\pi/2, \pi/2]$.

$\boxed{\theta = \pm \dfrac{1}{2}\arccos(\sqrt{2} - 1)}$

Part (ii)(b): Find $\displaystyle \int_{0}^{\pi/4} \cos^{4}\theta \, \text{d}\theta$

From part (i): $8\cos^4\theta = \cos 4\theta + 4\cos 2\theta + 3$, so:

$$\cos^4\theta = \dfrac{1}{8}\big(\cos 4\theta + 4\cos 2\theta + 3\big)$$

Integrate term by term:

$$\int_{0}^{\pi/4} \cos^4\theta \, \text{d}\theta = \dfrac{1}{8}\!\int_{0}^{\pi/4}\! \big(\cos 4\theta + 4\cos 2\theta + 3\big) \, \text{d}\theta = \dfrac{1}{8}\!\left[\dfrac{1}{4}\sin 4\theta + 2\sin 2\theta + 3\theta\right]_{0}^{\pi/4}$$

At $\theta = \pi/4$: โ€‚ $\dfrac{1}{4}\sin\pi + 2\sin\!\left(\dfrac{\pi}{2}\right) + 3\cdot\dfrac{\pi}{4} = 0 + 2(1) + \dfrac{3\pi}{4} = 2 + \dfrac{3\pi}{4}$.

At $\theta = 0$: โ€‚ $\dfrac{1}{4}\sin 0 + 2\sin 0 + 0 = 0$.

Subtract:

$$\int_{0}^{\pi/4} \cos^4\theta \, \text{d}\theta = \dfrac{1}{8}\!\left(2 + \dfrac{3\pi}{4}\right) = \dfrac{1}{8} \cdot \dfrac{8 + 3\pi}{4} = \dfrac{8 + 3\pi}{32}$$

$\boxed{\displaystyle \int_{0}^{\pi/4} \cos^{4}\theta \, \text{d}\theta = \dfrac{8 + 3\pi}{32}}$

Exercise 5B Q12 Challenging

(i) By first expanding $\cos(2x + x)$, show that $\cos 3x \equiv 4\cos^{3}x - 3\cos x$.

(ii) Hence show that $\displaystyle \int_{0}^{\frac{1}{6}\pi} \left(2\cos^{3}x - \cos x \right) \text{d}x = \dfrac{5}{12}$.

๐Ÿ’ก Hint

For part (i): expand $\cos(2x + x)$ using the compound angle formula $\cos(A + B) = \cos A\cos B - \sin A\sin B$ with $A = 2x$ and $B = x$. Then use $\cos 2x = 2\cos^2 x - 1$ and $\sin 2x = 2\sin x \cos x$ to remove every appearance of $2x$.

For part (ii): from part (i), $4\cos^3 x = \cos 3x + 3\cos x$, so $2\cos^3 x = \tfrac{1}{2}(\cos 3x + 3\cos x)$. Substitute this into $2\cos^3 x - \cos x$ and simplify โ€” you should get a linear combination of $\cos 3x$ and $\cos x$, which integrates directly.

๐Ÿ‘ Reveal Solution
Part (i): Show $\cos 3x \equiv 4\cos^{3}x - 3\cos x$

Expand $\cos(2x + x)$ using the compound-angle formula $\cos(A + B) = \cos A\cos B - \sin A\sin B$:

$$\cos(2x + x) = \cos 2x \cos x - \sin 2x \sin x$$

Apply $\cos 2x = 2\cos^2 x - 1$ and $\sin 2x = 2\sin x \cos x$:

$$= (2\cos^2 x - 1)\cos x - (2\sin x \cos x)\sin x = 2\cos^3 x - \cos x - 2\sin^2 x \cos x$$

Replace $\sin^2 x = 1 - \cos^2 x$:

$$= 2\cos^3 x - \cos x - 2(1 - \cos^2 x)\cos x = 2\cos^3 x - \cos x - 2\cos x + 2\cos^3 x$$

$$= 4\cos^3 x - 3\cos x$$

Since $\cos(2x + x) = \cos 3x$:

$\boxed{\cos 3x \equiv 4\cos^{3}x - 3\cos x}$ โ€‚ $\checkmark$

Part (ii): Show $\displaystyle \int_{0}^{\pi/6} (2\cos^3 x - \cos x) \, \text{d}x = \dfrac{5}{12}$

From part (i): $\cos 3x = 4\cos^3 x - 3\cos x$, so $4\cos^3 x = \cos 3x + 3\cos x$, hence:

$$2\cos^3 x = \dfrac{1}{2}(\cos 3x + 3\cos x)$$

Substitute into the integrand $2\cos^3 x - \cos x$:

$$2\cos^3 x - \cos x = \dfrac{1}{2}(\cos 3x + 3\cos x) - \cos x = \dfrac{1}{2}\cos 3x + \dfrac{3}{2}\cos x - \cos x = \dfrac{1}{2}\cos 3x + \dfrac{1}{2}\cos x$$

Integrate term by term:

$$\int_{0}^{\pi/6} \!\left(\dfrac{1}{2}\cos 3x + \dfrac{1}{2}\cos x\right) \text{d}x = \left[\dfrac{1}{2}\cdot\dfrac{1}{3}\sin 3x + \dfrac{1}{2}\sin x\right]_{0}^{\pi/6} = \left[\dfrac{1}{6}\sin 3x + \dfrac{1}{2}\sin x\right]_{0}^{\pi/6}$$

At $x = \pi/6$: โ€‚ $\dfrac{1}{6}\sin\!\left(\dfrac{\pi}{2}\right) + \dfrac{1}{2}\sin\!\left(\dfrac{\pi}{6}\right) = \dfrac{1}{6}(1) + \dfrac{1}{2}\!\left(\dfrac{1}{2}\right) = \dfrac{1}{6} + \dfrac{1}{4}$.

At $x = 0$: โ€‚ $\dfrac{1}{6}\sin 0 + \dfrac{1}{2}\sin 0 = 0$.

Subtract:

$$\int_{0}^{\pi/6} (2\cos^3 x - \cos x) \, \text{d}x = \dfrac{1}{6} + \dfrac{1}{4} = \dfrac{2}{12} + \dfrac{3}{12} = \dfrac{5}{12}$$

$\boxed{\displaystyle \int_{0}^{\pi/6} \left(2\cos^{3}x - \cos x \right) \text{d}x = \dfrac{5}{12}}$ โ€‚ $\checkmark$

Exit Ticket

Before you leave, tick the objectives you are confident you can achieve. Be honest โ€” this guides next lesson's review.