📚 Edexcel A-Level Mathematics Question 248 Walkthrough | Edexcel A-Level 数学第248题详解
que of integration by substitution. The problem asks: evaluate the definite integral I = ∫0π/2 ach stage in detail, from setting up the substitution to the final simplification, and discuss alternative verification methods.
这篇分步详解将攻克 Edexcel A-Level 数学历年真题中的第248题，重点考察代换积分法。题目要求：用代换 t = tan(x/2) 计算定积分 I = ∫0π/2 (sin x)/(1+cos x) dx。我们将逐一剖析，从设定代换到最终化简，并探讨验证方法。
1. Reading the Question Carefully | 仔细审题
is is a clue that straightforward methods might be unavailable or that the examiner wants to test the Weierstrass substitution. Always note such directives, as marks are allocated for the substitution process.
积分限从 0 到 π/2。被积函数为 (sin x)/(1+cos x)。题目明确要求用代换 t = tan(x/2)，这暗示直接方法可能不行，或考官想考查万能代换。务必留意此类指令，因为过程分往往分配给代换步骤。
2. Setting Up the Substitution | 设定代换关系
t²) dt. This relationship for dx will be used to replace the differential in the integral.
令 t = tan(x/2)，则 x = 2 arctan t。两边对 t 求导得 dx/dt = 2/(1 + t²)。整理得 dx = 2/(1 + t²) dt。这个 dx 关系式将用于替换积分中的微分。
3. Expressing sin x and cos x in Terms of t | 用 t 表示 sin x 和 cos x
These identities are fundamental and must be memorised or derived quickly. They transform the trigonometric integrand into a rational function of t.
利用万能代换的半角公式，sin x = 2t/(1 + t²)，cos x = (1 - t²)/(1 + t²)。这些基本恒等式需熟记或能快速推导。它们将三角被积函数化为关于 t 的有理函数。
4. Changing the Limits of Integration | 变换积分上下限
onvert to limits in t from 0 to 1. A common mistake is forgetting to update the limits, which would make the final answer incorrect. Always write the new limits clearly.
当 x = 0 时，t = tan(0/2) = 0。当 x = π/2 时，t = tan(π/4) = 1。x 的积分限变成 t 的积分限 0 到 1。常见错误是忘记更新积分限，导致最后答案错误。务必清晰写出新限。
5. Constructing the Substituted Integral | 构建代换后的积分
Substituting sin x, cos x, dx, and the limits into I gives: I = ∫01 [ (2t/(1+t²)) / (1 + (1-t²)/(1+t²)) ] · (2/(1+t²)) dt. The integrand looks messy but will simplify dramatically once the denominator is combined.
将 sin x、cos x、dx 及积分限代入 I 得：I = ∫01 [ (2t/(1+t²)) / (1 + (1-t²)/(1+t²)) ] · (2/(1+t²)) dt。被积函数看似杂乱，但合并分母后将大幅化简。
6. Simplifying the Denominator 1 + cos x | 化简分母 1 + cos x
. This simplification is very tidy. Now the fraction (sin x)/(1+cos x) becomes (2t/(1+t²)) ÷ (2/(1+t²)), which equals t.
先计算 1 + cos x = 1 + (1 - t²)/(1 + t²) = (1 + t² + 1 - t²)/(1 + t²) = 2/(1 + t²)。该化简非常干净。于是分式 (sin x)/(1+cos x) 变为 (2t/(1+t²)) ÷ (2/(1+t²))，等于 t。
7. Rewriting the Integral in a Simple Form | 将积分重写为简洁形式
es to t multiplied by the dx factor: t · (2/(1+t²)). Therefore, the integral becomes I = ∫01 2t/(1 + t²) dt. This is a standard form that can be integrated directly by recognising the derivative of the denominator.
约分后，被积函数化简为 t 乘以 dx 因子：t · (2/(1+t²))。因此积分变为 I = ∫01 2t/(1 + t²) dt。这是标准形式，可通过识别分母的导数直接积分。
8. Evaluating the Integral ∫ 2t/(1 + t²) dt | 计算积分 ∫ 2t/(1 + t²) dt
derivative of the denominator divided by the denominator. The integral is of the form ∫ f'(t)/f(t) dt, which integrates to ln|f(t)| + C. So, ∫ 2t/(1+t²) dt = ln(1 + t²) + C.
注意到 d/dt (1 + t²) = 2t。故被积函数恰为分母的导数除以分母。积分形如 ∫ f'(t)/f(t) dt，结果为 ln|f(t)| + C。因此 ∫ 2t/(1+t²) dt = ln(1 + t²) + C。
9. Applying the Limits and Obtaining the Result | 代入上下限并得出结果
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