(a) Parts #1 on $I$ M1·A1·A1
With $u = e^{x}$, $dv = \sin x\, dx$: $du = e^{x}\, dx$, $v = -\cos x$. Parts formula:
$$ I \;=\; uv - \int v\, du \;=\; e^{x} \cdot (-\cos x) - \int (-\cos x)\, e^{x}\, dx \;=\; -e^{x} \cos x + \int e^{x} \cos x\, dx. $$
(b) Parts #2 on $\int e^{x} \cos x\, dx$ M1·A1·A1
With $u = e^{x}$, $dv = \cos x\, dx$: $du = e^{x}\, dx$, $v = \sin x$. Parts formula:
$$ \int e^{x} \cos x\, dx \;=\; e^{x} \sin x - \int \sin x \cdot e^{x}\, dx \;=\; e^{x} \sin x - I. $$
(c) Algebraic loop-back M1·A1·A1
Substitute (b) into (a):
$$ I \;=\; -e^{x} \cos x + \bigl(e^{x} \sin x - I\bigr) \;=\; e^{x}(\sin x - \cos x) - I. $$
Solve for $I$: $\; 2 I = e^{x}(\sin x - \cos x)$, hence
$$ I \;=\; \tfrac{1}{2}\, e^{x}(\sin x - \cos x) + C. $$
(d) Verify by differentiating M1·A1·A1
Let $F(x) = \tfrac{1}{2}\, e^{x}(\sin x - \cos x)$. Product rule:
$$ F'(x) = \tfrac{1}{2} \bigl[e^{x}(\sin x - \cos x) + e^{x}(\cos x + \sin x)\bigr] = \tfrac{1}{2}\, e^{x}\bigl[(\sin x - \cos x) + (\cos x + \sin x)\bigr] = \tfrac{1}{2}\, e^{x}(2 \sin x) = e^{x} \sin x. \;\checkmark $$
(e) Definite integral on $[0, \pi]$ M1·A1·A1
Use the anti-derivative from (c):
$$ \int_{0}^{\pi} e^{x} \sin x\, dx \;=\; \Bigl[\tfrac{1}{2}\, e^{x}(\sin x - \cos x)\Bigr]_{0}^{\pi}. $$
Upper limit: $\tfrac{1}{2}\, e^{\pi}(\sin \pi - \cos \pi) = \tfrac{1}{2}\, e^{\pi}(0 - (-1)) = \tfrac{1}{2}\, e^{\pi}$.
Lower limit: $\tfrac{1}{2}\, e^{0}(\sin 0 - \cos 0) = \tfrac{1}{2}(0 - 1) = -\tfrac{1}{2}$.
Subtract: $\tfrac{1}{2}\, e^{\pi} - (-\tfrac{1}{2}) = \tfrac{1}{2}(e^{\pi} + 1)$.
Decimal: $\tfrac{1}{2}(e^{\pi} + 1) \approx \tfrac{1}{2}(23.140 + 1) \approx 12.1$ (3 s.f.).
The "loop back" works because $e^{x}$ is its own derivative. Apply parts to $\int e^{x} f(x)\, dx$ with $u = e^{x}$ and $f(x)$ is $\sin x$ or $\cos x$; each parts step (i) keeps $u = e^{x}$ unchanged (since $\dfrac{d}{dx} e^{x} = e^{x}$) and (ii) cycles $\sin \to \cos \to -\sin$, so after two parts the inner integral is $\pm I$ — and you solve algebraically. The trick generalises to any integrand $e^{ax} \cos(bx)$ or $e^{ax} \sin(bx)$: two parts, loop back, solve. Classical Paper 3 framing because the structural insight ("$e^{x}$ is a fixed point of $\dfrac{d}{dx}$") is more important than the algebra. Always sanity-check by differentiation, and on a definite integral always commit to the bracket evaluation $[\cdot]_{a}^{b}$ on a single line — sign errors in (e) commonly cost an A1.
Complex-exponential view: one-line derivation. If you've met Euler's formula, $\sin x = \operatorname{Im}(e^{ix})$, so $e^{x} \sin x = \operatorname{Im}(e^{x} \cdot e^{ix}) = \operatorname{Im}(e^{(1+i)x})$. Then $\int e^{(1+i)x}\, dx = \dfrac{e^{(1+i)x}}{1 + i} + C = \dfrac{1 - i}{2}\, e^{x}(\cos x + i \sin x) + C$. Take the imaginary part: $\dfrac{1}{2}\, e^{x}(\sin x - \cos x) + C$. Same answer, zero parts, zero loop-back. This is HL-level territory and not always accepted on Paper 1B unless explicitly invited, but on a Paper 3 modelling/exploration question that allows complex methods, it cuts a 15-mark problem to two lines. Mention this idea in your IA if integration techniques fit your topic.
(a) 对 $I$ 第 1 轮分部 M1·A1·A1
取 $u = e^{x}$、$dv = \sin x\, dx$：$du = e^{x}\, dx$、$v = -\cos x$。分部公式：
$$ I \;=\; uv - \int v\, du \;=\; e^{x} \cdot (-\cos x) - \int (-\cos x)\, e^{x}\, dx \;=\; -e^{x} \cos x + \int e^{x} \cos x\, dx. $$
(b) 对 $\int e^{x} \cos x\, dx$ 第 2 轮分部 M1·A1·A1
取 $u = e^{x}$、$dv = \cos x\, dx$：$du = e^{x}\, dx$、$v = \sin x$。分部公式：
$$ \int e^{x} \cos x\, dx \;=\; e^{x} \sin x - \int \sin x \cdot e^{x}\, dx \;=\; e^{x} \sin x - I. $$
(c) 代数循环回 M1·A1·A1
把 (b) 代入 (a)：
$$ I \;=\; -e^{x} \cos x + \bigl(e^{x} \sin x - I\bigr) \;=\; e^{x}(\sin x - \cos x) - I. $$
解 $I$：$\; 2 I = e^{x}(\sin x - \cos x)$，故
$$ I \;=\; \tfrac{1}{2}\, e^{x}(\sin x - \cos x) + C. $$
(d) 求导验证 M1·A1·A1
设 $F(x) = \tfrac{1}{2}\, e^{x}(\sin x - \cos x)$。乘积法则：
$$ F'(x) = \tfrac{1}{2} \bigl[e^{x}(\sin x - \cos x) + e^{x}(\cos x + \sin x)\bigr] = \tfrac{1}{2}\, e^{x}\bigl[(\sin x - \cos x) + (\cos x + \sin x)\bigr] = \tfrac{1}{2}\, e^{x}(2 \sin x) = e^{x} \sin x. \;\checkmark $$
(e) $[0, \pi]$ 上的定积分 M1·A1·A1
用 (c) 中的反求导：
$$ \int_{0}^{\pi} e^{x} \sin x\, dx \;=\; \Bigl[\tfrac{1}{2}\, e^{x}(\sin x - \cos x)\Bigr]_{0}^{\pi}. $$
代上限：$\tfrac{1}{2}\, e^{\pi}(\sin \pi - \cos \pi) = \tfrac{1}{2}\, e^{\pi}(0 - (-1)) = \tfrac{1}{2}\, e^{\pi}$。
代下限：$\tfrac{1}{2}\, e^{0}(\sin 0 - \cos 0) = \tfrac{1}{2}(0 - 1) = -\tfrac{1}{2}$。
相减：$\tfrac{1}{2}\, e^{\pi} - (-\tfrac{1}{2}) = \tfrac{1}{2}(e^{\pi} + 1)$。
小数：$\tfrac{1}{2}(e^{\pi} + 1) \approx \tfrac{1}{2}(23.140 + 1) \approx 12.1$（3 位有效数字）。
"循环回"奏效的根因是 $e^{x}$ 求导不变。对 $\int e^{x} f(x)\, dx$（$f$ 为 $\sin x$ 或 $\cos x$）作分部，取 $u = e^{x}$：(i) 每轮分部后 $u = e^{x}$ 不变（因 $\dfrac{d}{dx} e^{x} = e^{x}$）；(ii) $\sin \to \cos \to -\sin$ 循环。两轮分部后内层积分变为 $\pm I$，代数解出即可。该技巧推广至任意 $e^{ax} \cos(bx)$ 或 $e^{ax} \sin(bx)$：两轮分部、循环回、解 $I$。Paper 3 经典：结构洞察（"$e^{x}$ 是 $\dfrac{d}{dx}$ 的不动点"）比代数操作更重要。务必求导自验；定积分须把 $[\cdot]_{a}^{b}$ 写在一行——(e) 中的符号失误是 A1 重灾区。
复指数视角：一行直得。若已学欧拉公式，$\sin x = \operatorname{Im}(e^{ix})$，故 $e^{x} \sin x = \operatorname{Im}(e^{x} \cdot e^{ix}) = \operatorname{Im}(e^{(1+i)x})$。则 $\int e^{(1+i)x}\, dx = \dfrac{e^{(1+i)x}}{1 + i} + C = \dfrac{1 - i}{2}\, e^{x}(\cos x + i \sin x) + C$。取虚部：$\dfrac{1}{2}\, e^{x}(\sin x - \cos x) + C$。同样答案，零分部、零循环回。HL 水平的方法，第一卷 B 节非显式邀请一般不接受；但若 Paper 3 建模/探究题允许复方法，15 分大题可压缩到两行。若你的 IA 主题契合，不妨提及此思路。