早稲田大学 創造理工学研究科 経営システム工学専攻 2014年8月実施 微积分

标签：

• Waseda-University
• Definite-Integral
• Indefinite-Integral
• Integration-by-Parts
• Integration

Author

思齐塾, 祭音Myyura

Description

次の不定積分(indefinite integral) を計算せよ.

$$I = \int e^x \cos x \, dx$$

题目描述

计算不定积分

$$I=\int e^x\cos x\,dx.$$

Kai

Let $I = \int e^x \cos x \, dx$ . We can use integration by parts twice. Let $u = \cos x$ and $dv = e^x \, dx$ . Then $du = -\sin x \, dx$ and $v = e^x$ . So, $I = e^x \cos x - \int e^x (-\sin x) \, dx = e^x \cos x + \int e^x \sin x \, dx$ . Now, we use integration by parts again for $\int e^x \sin x \, dx$ . Let $u = \sin x$ and $dv = e^x \, dx$ . Then $du = \cos x \, dx$ and $v = e^x$ . So, $\int e^x \sin x \, dx = e^x \sin x - \int e^x \cos x \, dx = e^x \sin x - I$ . Substituting this back into the expression for $I$ , we get: $I = e^x \cos x + e^x \sin x - I$ . Therefore, $2I = e^x (\cos x + \sin x)$ . So, $I = \frac{1}{2} e^x (\cos x + \sin x) + C$ , where C is the constant of integration.