3.3 部分積分法

置換積分法でうまく解けなかったときに用いる．

1.

(a) $\left\{\begin{array}{lcl} u = x & & dv = e^{x}dx\\ &\searrow& \\ du = dx &\leftarrow& v = e^{x} \end{array}\right.$ より

$$\int xe^{x}\; dx = xe^{x} - \int e^{x}dx$$

$$= xe^{x} - e^{x} + c$$

(b) $\left\{\begin{array}{ll} u = x & dv = sin{x}dx\\ du = dx & v = -\cos{x} \end{array}\right.$ より

$$\int xsin{x}\; dx = -x\cos{x} - \int (-cos{x})dx$$

$$= -x \cos{x} + \sin{x} + c$$

(c) $\left\{\begin{array}{ll} u = x & dv = e^{2x}dx\\ du = dx & v = \frac{1}{2}e^{2x} \end{array}\right.$ より

$$\int xe^{2x}\; dx = \frac{1}{2}xe^{2x} -\frac{1}{2} \int e^{2x}dx$$

$$= \frac{1}{2}xe^{2x} - \frac{1}{4}e^{2x} + c$$

(d) $\left\{\begin{array}{ll} u = x^2 & dv = e^{x}dx\\ du = 2x\;dx & v = e^{x} \end{array}\right.$ より

$$\int x^2 e^{x}\; dx = x^2 e^{x} - 2\int xe^{x}dx \left(\begin{array}{ll} u = x & dv = e^{x}dx\\ du = dx & v = e^{x} \end{array}\right]$$

$$= x^2 e^{x} - 2[xe^{x} - e^{x}] + c$$

$$= x^2 e^{x} - 2xe^{x} + 2e^{x} + c = (x^2 - 2x + 2)e^{x} + c$$

(e) $\left\{\begin{array}{ll} u = x^2 & dv = \sin{x}dx\\ du = 2x\;dx & v = -\cos{x} \end{array}\right.$ より

$$\int x^2 \sin{x}\; dx = -x^2 \cos{x} + 2 \int x\cos{x}dx \left(\begin{array}{ll} u = x & dv = \cos{x}\;dx\\ du = dx & v = \sin{x} \end{array}\right]$$

$$= -x^2 \cos{x} + 2[x\sin{x} + \cos{x}] + c$$

(f)

まず， $\int x^{5}e^{x^3}\; dx = \int x^3 \cdot x^2 e^{x^3}\; dx$と書き直し，$t = x^3$とおくと，

$\left\{\begin{array}{ll} u = x^3 & dv = x^2 e^{x^3}dx\\ du = 3x^2\;dx & v = \frac{1}{3}e^{3} \end{array}\right.$ より

$$\int x^3 \cdot x^2 e^{x^3}\; dx = \frac{1}{3}x^3 e^{x^3} - \int x^2 e^{x^3}\; dx$$

$$= \frac{1}{3}x^3 e^{x^3} - \frac{1}{3}e^{x^3} + c$$

$$= \frac{1}{3}e^{x^3}(x^3 - 1) + c$$

(g) $\left\{\begin{array}{ll} u = x & dv = \cos{x}dx\\ du = dx & v = \sin{x} \end{array}\right.$ より

$$\int x \cos{x}\; dx = x \sin{x} - \int \sin{x}\;dx$$

$$= x\sin{x} + \cos{x} + c$$