Sections 1 to 6: trig integrals (sin/cos/sec/tan powers), trig substitution (all three cases), partial fractions (linear, repeated, irreducible quadratic), rationalising substitutions, and integration strategyCALC II第 1 至 6 节：三角积分（sin/cos/sec/tan 的幂次）、三角换元法（三种情形）、部分分数分解（线性因子、重复因子、不可约二次因子）、有理化换元，以及积分策略CALC II
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PART I · CORE TECHNIQUES第I部分 · 核心技巧Computational fluency · 28 marks计算能力 · 28分
Trigonometric Integrals and Substitutions三角积分与三角换元
Show all working. State the substitution or identity used at each step. Carry the constant of integration $C$ throughout indefinite integrals. All back-substitutions must be completed using a reference triangle.展示全部过程。在每一步注明所用的换元或恒等式。在不定积分中全程保留积分常数 $C$。所有反代换必须借助参考三角形完成。
Q1MEDIUMCOREtrig integrals: powers of sine and cosine三角积分：正弦与余弦的幂次[8 marks]
Evaluate each integral, stating clearly which parity strategy you use.计算下列各积分，并清楚说明所采用的奇偶策略。
(a) $\displaystyle\int \sin^4 x\,dx$ [3]
(b) $\displaystyle\int \sin^3 x \cos^4 x\,dx$ [3]
(c) $\displaystyle\int \tan^4 x\,dx$ (Hint: write $\tan^4 x = \tan^2 x(\sec^2 x - 1)$.提示：将 $\tan^4 x$ 写成 $\tan^2 x(\sec^2 x - 1)$。) [2]
Evaluate each integral. Identify the relevant substitution ($u = \tan x$ or $u = \sec x$) and convert all remaining factors before integrating.计算下列各积分。确定相关换元（$u = \tan x$ 或 $u = \sec x$），并在积分前将所有剩余因子转换完毕。
(a) $\displaystyle\int \tan^3 x \sec^4 x\,dx$ [3]
(b) $\displaystyle\int \sec^6 x\,dx$ [3]
Q3MEDIUMCOREtrig substitution: sine and tangent cases三角换元：正弦与正切情形[8 marks]
Evaluate each integral using the appropriate trigonometric substitution. State the substitution, change the limits or carry out a full back-substitution, and simplify completely.使用适当的三角换元法计算下列各积分。注明换元式，转换积分限或完整执行反代换，并化简至最简形式。
Q4MEDIUMCOREpartial fractions: distinct linear factors部分分数：不同线性因子[6 marks]
Decompose each rational function into partial fractions and evaluate the integral. Verify your coefficients by recombining over a common denominator.将各有理函数分解为部分分数，并计算积分。通过通分还原验证系数。
(a) $\displaystyle\int \frac{3x + 1}{x^2 - x - 6}\,dx$ [3]
PART II · DEFINITIONS AND PROOF第II部分 · 定义与证明Rigorous arguments · 26 marks严格论证 · 26分
Derivations, Reductions, and Structural Arguments推导、降幂与结构论证
These items are graded on the logic of the argument, not just the final line. State every identity you invoke. In derivations, work from the left-hand side to the right-hand side without assuming the result. Verify partial-fraction coefficients algebraically.本部分按论证逻辑评分，而非仅看最终结果。写出每一个引用的恒等式。在推导中，从左边出发推至右边，不得假设结论。用代数方法验证部分分数的系数。
Q5HARDPROOFtrig power reduction: deriving the half-angle identity route三角降幂：推导半角恒等式路径[8 marks]
This question asks you to derive and use a reduction result for $\int \sin^n x\,dx$ when $n$ is a positive even integer.本题要求推导并使用当 $n$ 为正偶数时 $\int \sin^n x\,dx$ 的降幂结果。
(a)Starting only from $\sin^2 x = \tfrac{1}{2}(1 - \cos 2x)$, show that
$$ \sin^4 x = \tfrac{3}{8} - \tfrac{1}{2}\cos 2x + \tfrac{1}{8}\cos 4x. $$
You must expand $(\sin^2 x)^2$ and apply the half-angle identity a second time; no other identity may be assumed.仅从 $\sin^2 x = \tfrac{1}{2}(1 - \cos 2x)$ 出发，证明
$$ \sin^4 x = \tfrac{3}{8} - \tfrac{1}{2}\cos 2x + \tfrac{1}{8}\cos 4x. $$
须展开 $(\sin^2 x)^2$ 并第二次应用半角恒等式；不得假设其他恒等式。[4]
(b)Hence derive the formula
$$ \int_0^{\pi/2} \sin^4 x\,dx = \frac{3\pi}{16}. $$
Show the antiderivative explicitly before evaluating the limits.由此推导公式
$$ \int_0^{\pi/2} \sin^4 x\,dx = \frac{3\pi}{16}. $$
在代入积分限之前，明确写出原函数。[2]
(c)Using the identity from (a) and an analogous expansion of $\cos^4 x = \tfrac{3}{8} + \tfrac{1}{2}\cos 2x + \tfrac{1}{8}\cos 4x$, deduce that
$$ \int_0^{\pi/2} \sin^4 x\,dx = \int_0^{\pi/2} \cos^4 x\,dx. $$
Justify why this equality holds without evaluating the right-hand side from scratch.利用 (a) 中的恒等式以及 $\cos^4 x = \tfrac{3}{8} + \tfrac{1}{2}\cos 2x + \tfrac{1}{8}\cos 4x$ 的类似展开，推断
$$ \int_0^{\pi/2} \sin^4 x\,dx = \int_0^{\pi/2} \cos^4 x\,dx. $$
说明该等式成立的理由，无须重新计算右边的积分。[2]
Q6HARDPROOFpartial fractions: setting up and proving the form with a repeated and an irreducible-quadratic factor部分分数：建立并证明含重复因子与不可约二次因子的分解形式[10 marks]
(a)Write down the general partial-fraction form for $R(x)$, explaining in one sentence why each term has the form it does.写出 $R(x)$ 的一般部分分数形式，并用一句话解释每一项形式的原因。[2]
(b)Determine all four coefficients. You must show the system of equations arising from clearing denominators and justify each step.求出全部四个系数。须写出去分母后所得的方程组，并为每一步提供依据。[5]
(c)Verify your decomposition by recombining the partial fractions over a common denominator and confirming the numerator matches $4x^3 + 3x^2 + 8x + 3$.将部分分数通分还原，验证分子与 $4x^3 + 3x^2 + 8x + 3$ 一致，从而确认分解正确。[3]
Q7HARDPROOFtrig substitution: secant case and back-substitution via reference triangle三角换元：正割情形与参考三角形反代换[8 marks]
(a)State the appropriate substitution $x = a\sec\theta$, giving the value of $a$, and rewrite the entire integral (including $dx$ and the radical) in terms of $\theta$.给出适当的换元 $x = a\sec\theta$，注明 $a$ 的值，并将整个积分（含 $dx$ 和根号项）全部用 $\theta$ 表示。[3]
(b)Evaluate the resulting trigonometric integral.计算所得的三角积分。[2]
(c)Draw a clearly labelled reference triangle encoding the substitution from (a). Read off $\sin\theta$ in terms of $x$ and hence write the final answer in terms of $x$.画出清晰标注的参考三角形，体现 (a) 中的换元关系。从三角形读出以 $x$ 表示的 $\sin\theta$，进而将最终答案写成 $x$ 的函数。[3]
PART III · APPLICATIONS AND SYNTHESIS第III部分 · 应用与综合Extended problems · 28 marks综合题 · 28分
Definite Integrals, Areas, and Combined Techniques定积分、面积与综合方法
Set up each problem cleanly. Where a geometric interpretation is available, use it as a sanity check. Carry exact values throughout and simplify at the end. Diagrams earn method credit.清晰建立每道题的框架。在有几何意义的情况下，用几何解释作为合理性检验。全程保留精确值，最后化简。画图可获得方法分。
Q8HARDAPPLIEDtrig substitution with completing the square: definite integral配方后的三角换元：定积分[10 marks]
(a)Complete the square in the expression $x^2 - 2x + 5$ and hence identify the appropriate trigonometric substitution. State the new variable and the new limits of integration.对 $x^2 - 2x + 5$ 配方，进而确定适当的三角换元。写出新变量及新的积分限。[3]
(b)Rewrite the integrand (including $x^2$ in the numerator) entirely in terms of the new variable, and evaluate the resulting trigonometric integral.将被积式（含分子中的 $x^2$）全部用新变量表示，并计算所得的三角积分。[5]
(c)Convert back to a numerical answer, writing it in fully simplified exact form.反代换回原变量，将答案写成完全化简的精确形式。[2]
Q9HARDAPPLIEDpartial fractions with irreducible quadratic: area between curves含不可约二次因子的部分分数：曲线间面积[10 marks]
Find the area of the region bounded by the curves $y = \dfrac{6x^2 + 4}{(x^2+1)(x^2+4)}$ and $y = 0$ between $x = 0$ and $x = 2$.求曲线 $y = \dfrac{6x^2 + 4}{(x^2+1)(x^2+4)}$ 与 $y = 0$ 在 $x = 0$ 至 $x = 2$ 之间所围区域的面积。
(a)Write the partial-fraction decomposition of $\dfrac{6x^2 + 4}{(x^2+1)(x^2+4)}$, treating each quadratic as an irreducible factor. Determine all coefficients.将 $\dfrac{6x^2 + 4}{(x^2+1)(x^2+4)}$ 分解为部分分数，将每个二次式视为不可约因子。求出全部系数。[4]
(b)Evaluate $\displaystyle\int_0^2 \frac{6x^2 + 4}{(x^2+1)(x^2+4)}\,dx$ using the decomposition from (a). Express the antiderivative in terms of $\arctan$.利用 (a) 中的分解计算 $\displaystyle\int_0^2 \frac{6x^2 + 4}{(x^2+1)(x^2+4)}\,dx$。将原函数用 $\arctan$ 表示。[4]
(c)State the exact area and verify that it is positive.写出精确面积，并验证其为正值。[2]
Q10HARDAPPLIEDstrategy and combined techniques: rationalising substitution then partial fractions策略与综合方法：有理化换元后接部分分数[8 marks]
Consider the integral $\displaystyle\int \frac{1}{1 + \sqrt{x+1}}\,dx$.考虑积分 $\displaystyle\int \frac{1}{1 + \sqrt{x+1}}\,dx$。
(a)Use the rationalising substitution $u = \sqrt{x+1}$ (so $x = u^2 - 1$, $dx = 2u\,du$) to convert the integral to a rational function of $u$. Show the algebra fully.使用有理化换元 $u = \sqrt{x+1}$（即 $x = u^2 - 1$，$dx = 2u\,du$），将积分化为 $u$ 的有理函数。完整展示代数过程。[3]
(b)Perform polynomial long division on the rational integrand from (a) to separate it into a polynomial plus a proper fraction, then decompose the proper fraction.对 (a) 中的有理被积函数进行多项式长除法，将其分离为多项式加真分式，再对真分式作部分分数分解。[3]
(c)Integrate and back-substitute to give the final answer in terms of $x$, including $+C$.积分后反代换，给出以 $x$ 表示的最终答案，含 $+C$。[2]