题目
(2) (int )_(0)^dfrac (pi {2)}sin varphi (cos )^3varphi dvarphi ;
题目解答
答案
解析
步骤 1：代换
令 $u = \cos \varphi$，则 $du = -\sin \varphi d\varphi$。当 $\varphi = 0$ 时，$u = \cos 0 = 1$；当 $\varphi = \dfrac{\pi}{2}$ 时，$u = \cos \dfrac{\pi}{2} = 0$。因此，原积分可以写为：
$$
\int_{0}^{\frac{\pi}{2}} \sin \varphi \cos^3 \varphi d\varphi = -\int_{1}^{0} u^3 du
$$
步骤 2：计算积分
计算积分 $-\int_{1}^{0} u^3 du$：
$$
-\int_{1}^{0} u^3 du = -\left[ \frac{u^4}{4} \right]_{1}^{0} = -\left( \frac{0^4}{4} - \frac{1^4}{4} \right) = -\left( 0 - \frac{1}{4} \right) = \frac{1}{4}
$$
令 $u = \cos \varphi$，则 $du = -\sin \varphi d\varphi$。当 $\varphi = 0$ 时，$u = \cos 0 = 1$；当 $\varphi = \dfrac{\pi}{2}$ 时，$u = \cos \dfrac{\pi}{2} = 0$。因此，原积分可以写为：
$$
\int_{0}^{\frac{\pi}{2}} \sin \varphi \cos^3 \varphi d\varphi = -\int_{1}^{0} u^3 du
$$
步骤 2：计算积分
计算积分 $-\int_{1}^{0} u^3 du$：
$$
-\int_{1}^{0} u^3 du = -\left[ \frac{u^4}{4} \right]_{1}^{0} = -\left( \frac{0^4}{4} - \frac{1^4}{4} \right) = -\left( 0 - \frac{1}{4} \right) = \frac{1}{4}
$$