The Scenario
A pendulum swings with small oscillations. From physics, its angular acceleration is proportional to the negative of its displacement — this is Simple Harmonic Motion (SHM). A sensor measures the angular acceleration as a function of time:
$\alpha(t) = -\dfrac{\pi^2}{4}\sin\!\left(\dfrac{\pi}{2}t\right) \;\text{rad/s}^2$
The constant $\dfrac{\pi}{2}$ inside the sine is the pendulum's angular frequency $\omega$. The pendulum's motion has period $T = \dfrac{2\pi}{\omega} = 4$ seconds — one full swing every 4 s.
To recover the pendulum's angular velocity $v(t)$, we integrate the acceleration: $v(t) = \int \alpha(t)\,\text{d}t$. To recover the angular position $\theta(t)$, we integrate again: $\theta(t) = \int v(t)\,\text{d}t$. Each integration of $\sin\!\left(\dfrac{\pi}{2}t\right)$ or $\cos\!\left(\dfrac{\pi}{2}t\right)$ introduces a factor of $\dfrac{2}{\pi}$.
The Question
Without yet knowing the integration rule for $\sin(ax+b)$, attempt to find the pendulum's velocity $v(t)$ given that $v(0) = \dfrac{\pi}{2}$.
Specifically: What is $\displaystyle\int -\dfrac{\pi^2}{4}\sin\!\left(\dfrac{\pi}{2}t\right)\,\text{d}t$? And where do the factors of $\dfrac{2}{\pi}$ come from in the answer?
Hint: think back to last week's differentiation rule. If $\dfrac{\text{d}}{\text{d}x}\!\left[\cos(ax)\right] = -a\sin(ax)$, what must $\int\sin(ax)\,\text{d}x$ be?
💡 The Bridge (click to reveal)
Reversing $\dfrac{\text{d}}{\text{d}x}\!\left[\cos(ax+b)\right] = -a\sin(ax+b)$ gives the integration rule:
$$\int\sin(ax+b)\,\text{d}x = -\dfrac{1}{a}\cos(ax+b) + C$$
The factor $\dfrac{1}{a}$ "undoes" the chain rule's multiplier $a$. For the pendulum, $a = \dfrac{\pi}{2}$, so $\dfrac{1}{a} = \dfrac{2}{\pi}$. Each integration introduces a $\dfrac{2}{\pi}$ factor — that is exactly the $\dfrac{1}{a}$ in the general rule.
Today we learn the three special integral forms:
$$\int\sin(ax+b)\,\text{d}x = -\dfrac{1}{a}\cos(ax+b) + C, \quad \int\cos(ax+b)\,\text{d}x = \dfrac{1}{a}\sin(ax+b) + C, \quad \int\sec^2(ax+b)\,\text{d}x = \dfrac{1}{a}\tan(ax+b) + C$$
Each of the three forms below comes from reversing a differentiation rule you learned in Week 5 (P3 §3.4). The chain rule in differentiation introduces a multiplier $a$; reversing it in integration introduces the compensating factor $\dfrac{1}{a}$.
Since $\dfrac{\text{d}}{\text{d}x}\!\left[\tan(ax+b)\right] = a\sec^2(ax+b)$, reversing gives:
$$\boxed{\;\int\sec^2(ax+b)\,\text{d}x = \dfrac{1}{a}\tan(ax+b) + C\;}\qquad(a \neq 0)$$
Since $\dfrac{\text{d}}{\text{d}x}\!\left[\sin(ax+b)\right] = a\cos(ax+b)$, reversing gives:
$$\boxed{\;\int\cos(ax+b)\,\text{d}x = \dfrac{1}{a}\sin(ax+b) + C\;}\qquad(a \neq 0)$$
Since $\dfrac{\text{d}}{\text{d}x}\!\left[\cos(ax+b)\right] = -a\sin(ax+b)$, reversing (with the minus sign) gives:
$$\boxed{\;\int\sin(ax+b)\,\text{d}x = -\dfrac{1}{a}\cos(ax+b) + C\;}\qquad(a \neq 0)$$
Watch the minus sign — this is the most common source of errors. Always check that differentiating your answer returns the original integrand.
By the same reverse-differentiation principle, since $\dfrac{\text{d}}{\text{d}x}\!\left[\tan^{-1}\!\left(\dfrac{x}{a}\right)\right] = \dfrac{a}{x^2+a^2}$, we obtain $\displaystyle\int\dfrac{1}{x^2+a^2}\,\text{d}x = \dfrac{1}{a}\tan^{-1}\!\left(\dfrac{x}{a}\right) + C$. The same $\dfrac{1}{a}$ factor appears. This form is the inverse-tangent analogue of the secant-squared rule (since $\dfrac{\text{d}}{\text{d}x}\!\left[\tan(ax+b)\right] = a\sec^2(ax+b)$ and $\dfrac{\text{d}}{\text{d}x}\!\left[\tan^{-1}(x/a)\right] = \dfrac{a}{x^2+a^2}$ are inverse operations). It is referenced here for completeness; today's worked examples and practice focus on the three trig forms above.
Work in a group. Each card shows an integral of a special form; drag it — or tap it and then tap a box — into the family of its antiderivative. A $\sec^2(ax+b)$ integral belongs to the tan family, a $\cos(ax+b)$ integral to the sin family, and a $\sin(ax+b)$ integral to the negative-cos family. A card dropped in the wrong box flashes red and returns to the pile, so name the coefficient $a$ and decide the sign before the card moves.
Group rule: nobody clicks until all members agree on the family. Read the integrand aloud, name the coefficient $a$, then name the sign — all three, every time.
On your own. Each level shows three integrals. Your job is to name the rule they share, not to work them out one by one. Level 3 only opens when level 2 is cleared, and only a run with no wrong pick at a level earns that level's star.
Detective habit: before you choose, ask yourself "what is the same in all three?" — the shape of the integrand, the position of $a$, and the sign of the answer.
Integrate the following with respect to $x$.
(i) $\sin x - 2\cos x$ (ii) $3\cos x + 2\sin x$ (iii) $5\sin x + 4\cos x$
(iv) $4\sec^2 x$ (v) $\sin(2x+1)$ (vi) $\cos(5x-\pi)$
(vii) $6\sec^2 2x$ (viii) $3\sec^2 3x - \sin 2x$ (ix) $4\sec^2 x - \cos 2x$
👁 Reveal Solution
Here $a = 1$ for both terms. $\int\sin x\,\text{d}x = -\cos x + C$, $\int\cos x\,\text{d}x = \sin x + C$.
$\boxed{-\cos x - 2\sin x + C}$
$a = 1$. $\int 3\cos x\,\text{d}x = 3\sin x$, $\int 2\sin x\,\text{d}x = -2\cos x$.
$\boxed{3\sin x - 2\cos x + C}$
$a = 1$. $\int 5\sin x\,\text{d}x = -5\cos x$, $\int 4\cos x\,\text{d}x = 4\sin x$.
$\boxed{-5\cos x + 4\sin x + C}$
$a = 1$. $\int 4\sec^2 x\,\text{d}x = 4\tan x$.
$\boxed{4\tan x + C}$
$a = 2$, $b = 1$. $\int\sin(ax+b)\,\text{d}x = -\dfrac{1}{a}\cos(ax+b) + C$.
$\boxed{-\dfrac{1}{2}\cos(2x+1) + C}$
$a = 5$, $b = -\pi$. $\int\cos(ax+b)\,\text{d}x = \dfrac{1}{a}\sin(ax+b) + C$.
$\boxed{\dfrac{1}{5}\sin(5x-\pi) + C}$
$a = 2$, $b = 0$. $\int 6\sec^2(2x)\,\text{d}x = 6 \cdot \dfrac{1}{2}\tan(2x) = 3\tan(2x)$.
$\boxed{3\tan(2x) + C}$
Split: $\int 3\sec^2(3x)\,\text{d}x = 3 \cdot \dfrac{1}{3}\tan(3x) = \tan(3x)$; $\int\sin(2x)\,\text{d}x = -\dfrac{1}{2}\cos(2x)$.
$\boxed{\tan(3x) + \dfrac{1}{2}\cos(2x) + C}$
$\int 4\sec^2(x)\,\text{d}x = 4\tan(x)$; $\int\cos(2x)\,\text{d}x = \dfrac{1}{2}\sin(2x)$.
$\boxed{4\tan(x) - \dfrac{1}{2}\sin(2x) + C}$
Find the exact value of the following.
(i) $\displaystyle\int_{0}^{\frac{\pi}{3}} \sin x \, \text{d}x$ (ii) $\displaystyle\int_{0}^{\frac{\pi}{4}} \sec^{2} x \, \text{d}x$ (iii) $\displaystyle\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \cos x \, \text{d}x$
(iv) $\displaystyle\int_{0}^{\frac{2\pi}{3}} \sin 2x \, \text{d}x$ (v) $\displaystyle\int_{0}^{\frac{5\pi}{6}} \cos 3x \, \text{d}x$ (vi) $\displaystyle\int_{\frac{\pi}{8}}^{\frac{\pi}{6}} \sec^{2} 2x \, \text{d}x$
(vii) $\displaystyle\int_{0}^{\pi} \cos\!\left(2x + \frac{\pi}{2}\right) \, \text{d}x$ (viii) $\displaystyle\int_{0}^{\frac{\pi}{4}} \left(\sec^{2}x + \cos 4x\right) \, \text{d}x$ (ix) $\displaystyle\int_{0}^{\frac{\pi}{6}} \left(\cos x + \sin 2x\right) \, \text{d}x$
👁 Reveal Solution
$a = 1$. $[-\cos x]_0^{\pi/3} = -\cos(\pi/3) - (-\cos 0) = -\tfrac{1}{2} + 1 = \tfrac{1}{2}$.
$\boxed{\dfrac{1}{2}}$
$a = 1$. $[\tan x]_0^{\pi/4} = \tan(\pi/4) - \tan(0) = 1 - 0 = 1$.
$\boxed{1}$
$a = 1$. $[\sin x]_{\pi/6}^{\pi/3} = \sin(\pi/3) - \sin(\pi/6) = \tfrac{\sqrt{3}}{2} - \tfrac{1}{2} = \tfrac{\sqrt{3}-1}{2}$.
$\boxed{\dfrac{\sqrt{3}-1}{2}}$
$a = 2$. $\int\sin(2x)\,\text{d}x = -\tfrac{1}{2}\cos(2x)$. $\left[-\tfrac{1}{2}\cos(2x)\right]_0^{2\pi/3} = -\tfrac{1}{2}\cos\!\left(\tfrac{4\pi}{3}\right) - \left(-\tfrac{1}{2}\cos 0\right)$.
$\cos\!\left(\tfrac{4\pi}{3}\right) = -\tfrac{1}{2}$, $\cos 0 = 1$. So $= -\tfrac{1}{2}\!\left(-\tfrac{1}{2}\right) + \tfrac{1}{2}(1) = \tfrac{1}{4} + \tfrac{1}{2} = \tfrac{3}{4}$.
$\boxed{\dfrac{3}{4}}$
$a = 3$. $\int\cos(3x)\,\text{d}x = \tfrac{1}{3}\sin(3x)$. $\left[\tfrac{1}{3}\sin(3x)\right]_0^{5\pi/6} = \tfrac{1}{3}\sin\!\left(\tfrac{5\pi}{2}\right) - \tfrac{1}{3}\sin(0)$.
$\sin\!\left(\tfrac{5\pi}{2}\right) = \sin\!\left(\tfrac{\pi}{2}\right) = 1$, $\sin 0 = 0$. So $= \tfrac{1}{3}(1) - 0 = \tfrac{1}{3}$.
$\boxed{\dfrac{1}{3}}$
$a = 2$. $\int\sec^2(2x)\,\text{d}x = \tfrac{1}{2}\tan(2x)$. $\left[\tfrac{1}{2}\tan(2x)\right]_{\pi/8}^{\pi/6} = \tfrac{1}{2}\tan\!\left(\tfrac{\pi}{3}\right) - \tfrac{1}{2}\tan\!\left(\tfrac{\pi}{4}\right)$.
$\tan\!\left(\tfrac{\pi}{3}\right) = \sqrt{3}$, $\tan\!\left(\tfrac{\pi}{4}\right) = 1$. So $= \tfrac{\sqrt{3}}{2} - \tfrac{1}{2} = \tfrac{\sqrt{3}-1}{2}$.
$\boxed{\dfrac{\sqrt{3}-1}{2}}$
$a = 2$, $b = \tfrac{\pi}{2}$. $\int\cos(2x+\tfrac{\pi}{2})\,\text{d}x = \tfrac{1}{2}\sin(2x+\tfrac{\pi}{2})$. $\left[\tfrac{1}{2}\sin(2x+\tfrac{\pi}{2})\right]_0^{\pi} = \tfrac{1}{2}\sin\!\left(2\pi+\tfrac{\pi}{2}\right) - \tfrac{1}{2}\sin\!\left(\tfrac{\pi}{2}\right)$.
$\sin\!\left(2\pi+\tfrac{\pi}{2}\right) = \sin\!\left(\tfrac{\pi}{2}\right) = 1$. So $= \tfrac{1}{2}(1) - \tfrac{1}{2}(1) = 0$.
$\boxed{0}$ (Alt: $\cos(2x+\tfrac{\pi}{2}) = -\sin(2x)$; over a full period of $\sin(2x)$, the integral is zero.)
$\int\sec^2(x)\,\text{d}x = \tan(x)$; $\int\cos(4x)\,\text{d}x = \tfrac{1}{4}\sin(4x)$.
$\left[\tan x + \tfrac{1}{4}\sin(4x)\right]_0^{\pi/4} = \left(\tan\!\left(\tfrac{\pi}{4}\right) + \tfrac{1}{4}\sin(\pi)\right) - (\tan 0 + \tfrac{1}{4}\sin 0) = (1 + 0) - (0 + 0) = 1$.
$\boxed{1}$
$\int\cos x\,\text{d}x = \sin x$; $\int\sin(2x)\,\text{d}x = -\tfrac{1}{2}\cos(2x)$.
$\left[\sin x - \tfrac{1}{2}\cos(2x)\right]_0^{\pi/6} = \left(\sin\!\left(\tfrac{\pi}{6}\right) - \tfrac{1}{2}\cos\!\left(\tfrac{\pi}{3}\right)\right) - \left(\sin 0 - \tfrac{1}{2}\cos 0\right)$.
$= \left(\tfrac{1}{2} - \tfrac{1}{2}\cdot\tfrac{1}{2}\right) - \left(0 - \tfrac{1}{2}\right) = \left(\tfrac{1}{2} - \tfrac{1}{4}\right) + \tfrac{1}{2} = \tfrac{1}{4} + \tfrac{1}{2} = \tfrac{3}{4}$.
$\boxed{\dfrac{3}{4}}$
Attempt each question on paper first, then click Reveal Solution to check your work. Questions are ordered from easiest to hardest.
(i) Show that $\sin x \cos x = \dfrac{1}{2}\sin 2x$.
(ii) Hence find the exact value of $\displaystyle\int_{0}^{\frac{\pi}{3}} \sin x \cos x \,\text{d}x$
💡 Hint
For part (i): use the double-angle identity $\sin(2x) = 2\sin x\cos x$, then divide both sides by 2.
For part (ii): after rewriting $\sin x\cos x$ as $\tfrac{1}{2}\sin(2x)$, integrate using the $\sin(ax+b)$ rule with $a = 2$, $b = 0$. Don't forget the $\dfrac{1}{a} = \dfrac{1}{2}$ factor — combined with the $\tfrac{1}{2}$ outside, this gives $\tfrac{1}{4}$ overall.
👁 Reveal Solution
The double-angle identity $\sin(2x) = 2\sin x\cos x$ is a standard result (from §3.2 Trigonometry). Dividing both sides by 2:
$\sin x\cos x = \dfrac{1}{2}\sin(2x)$. $\square$
Using part (i): $\int_0^{\pi/3}\sin x\cos x\,\text{d}x = \int_0^{\pi/3}\tfrac{1}{2}\sin(2x)\,\text{d}x = \tfrac{1}{2}\int_0^{\pi/3}\sin(2x)\,\text{d}x$.
$\int\sin(2x)\,\text{d}x = -\tfrac{1}{2}\cos(2x)$, so the overall coefficient is $\tfrac{1}{2}\cdot\!\left(-\tfrac{1}{2}\right) = -\tfrac{1}{4}$.
$= \left[-\tfrac{1}{4}\cos(2x)\right]_0^{\pi/3} = -\tfrac{1}{4}\cos\!\left(\tfrac{2\pi}{3}\right) - \left(-\tfrac{1}{4}\cos 0\right)$.
$\cos\!\left(\tfrac{2\pi}{3}\right) = -\tfrac{1}{2}$, $\cos 0 = 1$. So $= -\tfrac{1}{4}\!\left(-\tfrac{1}{2}\right) + \tfrac{1}{4}(1) = \tfrac{1}{8} + \tfrac{2}{8} = \tfrac{3}{8}$.
$\boxed{\dfrac{3}{8}}$
Use a suitable trigonometrical identity to help you find these.
(i) (a) $\displaystyle\int \cos^2 x \, \text{d}x$ (b) $\displaystyle\int_0^{\frac{\pi}{2}} \cos^2 x \, \text{d}x$
(ii) (a) $\displaystyle\int \sin^2 x \, \text{d}x$ (b) $\displaystyle\int_0^{\frac{\pi}{3}} \sin^2 x \, \text{d}x$
💡 Hint
Use the half-angle identities: $\cos^2 x = \dfrac{1+\cos 2x}{2}$ and $\sin^2 x = \dfrac{1-\cos 2x}{2}$.
After substitution, split the integral into a constant term (which integrates to $\tfrac{1}{2}x$) and a $\cos(2x)$ term (which integrates to $\tfrac{1}{4}\sin(2x)$ using the $\dfrac{1}{a} = \dfrac{1}{2}$ factor).
👁 Reveal Solution
Use $\cos^2 x = \dfrac{1+\cos(2x)}{2} = \tfrac{1}{2} + \tfrac{1}{2}\cos(2x)$.
$\int\!\left(\tfrac{1}{2} + \tfrac{1}{2}\cos(2x)\right)\text{d}x = \tfrac{1}{2}x + \tfrac{1}{2}\cdot\tfrac{1}{2}\sin(2x) + C$.
$\boxed{\dfrac{1}{2}x + \dfrac{1}{4}\sin(2x) + C}$
$\left[\tfrac{1}{2}x + \tfrac{1}{4}\sin(2x)\right]_0^{\pi/2} = \left(\tfrac{1}{2}\cdot\tfrac{\pi}{2} + \tfrac{1}{4}\sin(\pi)\right) - (0 + 0) = \tfrac{\pi}{4} + 0 = \tfrac{\pi}{4}$.
$\boxed{\dfrac{\pi}{4}}$
Use $\sin^2 x = \dfrac{1-\cos(2x)}{2} = \tfrac{1}{2} - \tfrac{1}{2}\cos(2x)$.
$\int\!\left(\tfrac{1}{2} - \tfrac{1}{2}\cos(2x)\right)\text{d}x = \tfrac{1}{2}x - \tfrac{1}{2}\cdot\tfrac{1}{2}\sin(2x) + C$.
$\boxed{\dfrac{1}{2}x - \dfrac{1}{4}\sin(2x) + C}$
$\left[\tfrac{1}{2}x - \tfrac{1}{4}\sin(2x)\right]_0^{\pi/3} = \left(\tfrac{1}{2}\cdot\tfrac{\pi}{3} - \tfrac{1}{4}\sin\!\left(\tfrac{2\pi}{3}\right)\right) - (0 - 0) = \tfrac{\pi}{6} - \tfrac{1}{4}\cdot\tfrac{\sqrt{3}}{2}$.
$\boxed{\dfrac{\pi}{6} - \dfrac{\sqrt{3}}{8}}$
(i) Express $\sqrt{3}\cos x + \sin x$ in the form $R\cos(x - \alpha)$, where $R > 0$ and $0 < \alpha < \dfrac{1}{2}\pi$, giving the exact values of $R$ and $\alpha$.
(ii) Hence show that $\displaystyle\int_{\frac{1}{6}\pi}^{\frac{1}{2}\pi} \dfrac{1}{\left(\sqrt{3}\cos x + \sin x\right)^2} \text{d}x = \dfrac{1}{4}\sqrt{3}$
💡 Hint
For part (i): recall Lesson 1 (W6-L1) on R-form. Match $\sqrt{3}\cos x + \sin x$ with $R\cos(x-\alpha) = R\cos x\cos\alpha + R\sin x\sin\alpha$. Read off $R\cos\alpha = \sqrt{3}$ and $R\sin\alpha = 1$.
For part (ii): after substituting $R\cos(x-\alpha)$ into the integrand, the denominator becomes $R^2\cos^2(x-\alpha)$. The integral becomes $\int\dfrac{1}{R^2\cos^2(x-\alpha)}\,\text{d}x = \dfrac{1}{R^2}\int\sec^2(x-\alpha)\,\text{d}x$ — which uses this lesson's $\sec^2$ rule.
👁 Reveal Solution
Match with $R\cos(x-\alpha) = R\cos x\cos\alpha + R\sin x\sin\alpha$:
$R\cos\alpha = \sqrt{3}$, $R\sin\alpha = 1$.
$R = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3+1} = 2$. $\tan\alpha = \dfrac{1}{\sqrt{3}} \Rightarrow \alpha = \dfrac{\pi}{6}$.
$\boxed{\sqrt{3}\cos x + \sin x = 2\cos\!\left(x - \dfrac{\pi}{6}\right)}$
From part (i): $\sqrt{3}\cos x + \sin x = 2\cos(x - \tfrac{\pi}{6})$, so $\left(\sqrt{3}\cos x + \sin x\right)^2 = 4\cos^2(x - \tfrac{\pi}{6})$.
$\int_{\pi/6}^{\pi/2}\dfrac{1}{4\cos^2(x-\pi/6)}\,\text{d}x = \dfrac{1}{4}\int_{\pi/6}^{\pi/2}\sec^2\!\left(x - \tfrac{\pi}{6}\right)\,\text{d}x$.
Using $\int\sec^2(ax+b)\,\text{d}x = \dfrac{1}{a}\tan(ax+b) + C$ with $a = 1$, $b = -\tfrac{\pi}{6}$: $\int\sec^2\!\left(x - \tfrac{\pi}{6}\right)\text{d}x = \tan\!\left(x - \tfrac{\pi}{6}\right) + C$.
$= \dfrac{1}{4}\left[\tan\!\left(x - \tfrac{\pi}{6}\right)\right]_{\pi/6}^{\pi/2} = \dfrac{1}{4}\!\left(\tan\!\left(\tfrac{\pi}{2} - \tfrac{\pi}{6}\right) - \tan(0)\right) = \dfrac{1}{4}\!\left(\tan\!\left(\tfrac{\pi}{3}\right) - 0\right) = \dfrac{1}{4}\cdot\sqrt{3}$.
$\boxed{\dfrac{\sqrt{3}}{4}}$ (as required). $\square$
(i) Prove that $\tan^2\theta + \cot^2\theta \equiv \dfrac{2}{\sin 2\theta}$.
(ii) Hence (a) find the exact value of $\tan^2\frac{1}{8}\pi + \cot^2\frac{1}{8}\pi$
(b) evaluate $\displaystyle\int_{0}^{\frac{1}{2}\pi} \dfrac{6}{\tan^2\theta + \cot^2\theta} \text{d}\theta$
💡 Hint
The standard identity relating $\tan\theta + \cot\theta$ and $\sin(2\theta)$ is established by writing both in terms of $\sin\theta$ and $\cos\theta$:
$\tan\theta + \cot\theta = \dfrac{\sin\theta}{\cos\theta} + \dfrac{\cos\theta}{\sin\theta} = \dfrac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta} = \dfrac{1}{\sin\theta\cos\theta}$.
Then use $\sin(2\theta) = 2\sin\theta\cos\theta$ to write $\sin\theta\cos\theta = \tfrac{1}{2}\sin(2\theta)$. Substitute back to get the identity.
Note: The printed identity with $\tan^2 + \cot^2$ appears to be a transcription of the standard $\tan + \cot$ identity — the proof below establishes the standard form, which is what part (ii) actually requires.
👁 Reveal Solution
We establish the standard identity $\tan\theta + \cot\theta \equiv \dfrac{2}{\sin(2\theta)}$:
$\tan\theta + \cot\theta = \dfrac{\sin\theta}{\cos\theta} + \dfrac{\cos\theta}{\sin\theta} = \dfrac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta} = \dfrac{1}{\sin\theta\cos\theta}$.
Using $\sin(2\theta) = 2\sin\theta\cos\theta$, so $\sin\theta\cos\theta = \tfrac{1}{2}\sin(2\theta)$:
$\tan\theta + \cot\theta = \dfrac{1}{\tfrac{1}{2}\sin(2\theta)} = \dfrac{2}{\sin(2\theta)}$. $\square$
Note: The source prints the identity with $\tan^2 + \cot^2$, but that form does not hold numerically except at $\theta = \pi/4$ (e.g. at $\theta = \pi/6$: LHS $= \tfrac{1}{3} + 3 = \tfrac{10}{3}$, RHS $= \tfrac{2}{\sqrt{3}/2} = \tfrac{4}{\sqrt{3}} \approx 2.31$). The standard form (without squares) is what part (ii) requires, and we use it below.
Using the standard identity with $\theta = \tfrac{\pi}{8}$:
$\tan\!\left(\tfrac{\pi}{8}\right) + \cot\!\left(\tfrac{\pi}{8}\right) = \dfrac{2}{\sin\!\left(\tfrac{\pi}{4}\right)} = \dfrac{2}{\tfrac{\sqrt{2}}{2}} = \dfrac{4}{\sqrt{2}} = 2\sqrt{2}$.
(Interpreting the source as the standard form.)
$\boxed{2\sqrt{2}}$
Using the standard identity $\tan\theta + \cot\theta = \dfrac{2}{\sin(2\theta)}$, we have $\dfrac{1}{\tan\theta + \cot\theta} = \dfrac{\sin(2\theta)}{2}$.
$\int_0^{\pi/2}\dfrac{6}{\tan\theta + \cot\theta}\,\text{d}\theta = \int_0^{\pi/2}6\cdot\dfrac{\sin(2\theta)}{2}\,\text{d}\theta = 3\int_0^{\pi/2}\sin(2\theta)\,\text{d}\theta$.
Using $\int\sin(2\theta)\,\text{d}\theta = -\tfrac{1}{2}\cos(2\theta)$ (with $a = 2$):
$= 3\left[-\tfrac{1}{2}\cos(2\theta)\right]_0^{\pi/2} = 3\left(-\tfrac{1}{2}\cos(\pi) + \tfrac{1}{2}\cos(0)\right) = 3\left(-\tfrac{1}{2}(-1) + \tfrac{1}{2}(1)\right) = 3\!\left(\tfrac{1}{2} + \tfrac{1}{2}\right) = 3(1)$.
$\boxed{3}$
Exit Ticket
Before you leave, tick the objectives you are confident you can achieve. Be honest — this guides next lesson's review.