微积分学习笔记-积分法则：替换积分法

2019-11-25  本文已影响0人  LonnieQ

不定积分法则

  1. 常倍数法则: \int kf(x) dx = k\int f(x) dx(k不依赖于x)
  2. 负法则:\int -f(x) dx = -\int f(x)dx
  3. 和与差法则: \int [f(x)\pm g(x)dx = \int f(x) dx \pm \int g(x) dx

例1 改写积分常数

\int 5 sec(x)tan (x) dx = 5\int sec(x)tan(x)dx = 5 sec(x) + C

例2 求积分 \int (x^2 - 2x + 5) dx

\int (x^2 - 2x + 5) dx = \int x^2dx - \int 2xdx + \int 5dx = \frac{1}{3}x^3 - x^2 + 5x + c

例3 求sin^2(x)和cos^2(x)的积分

\int sin^2(x) dx = \int \frac{1 - cos(2x)}{2} dx = \frac{1}{2} \int (1 -cos(2x)) dx = \frac{1}{2}x - \frac{sin(2x)}{4} + C
\int sin^2(x) dx = \int \frac{1 + cos(2x)}{2} dx = \frac{1}{2} \int (1 +cos(2x)) dx = \frac{1}{2}x + \frac{sin(2x)}{4} + C

微分形式的幂法则

如果u是任一可微函数，则 \int u^n du = \frac{u^{n+1}}{n+1} + C (n \not = -1, n为有理数)

例4. 使用幂法则

\int \sqrt {1 + y^2}. 2y dy = \int u^{\frac{1}{2}} du = \frac{2u^{\frac{3}{2}}}{3} + C = \frac{2}{3} (1+y^2)^{\frac{3}{2}} + C

例5. 用一个常数调整积分

\int \sqrt {4t - 1} dt =\frac{1}{4} \int u^{\frac{1}{2}} du (令u = 4t - 1)= \frac{1}{4} * \frac{2u^{\frac{3}{2}}}{3} + C = \frac{u^{\frac{3}{2}}}{6} + C = \frac{(4t-1)^{\frac{3}{2}}}{6} + C

替换积分法

当f和g'都是连续函数时，为求积分\int f(g(x))g'(x)dx，采用以下步骤。

  1. 作替换u = g(x), 则du = g'(x)dx, 得到积分\int f(u)du.
  2. 对u积分
  3. 结果用g(x)替代u

例6 解\int cos(7\theta+5)d\theta

令u = 7\theta+5, 得
\int cos(7\theta+5)d\theta =\frac{1}{7} \int cos(u) du = \frac{1}{7} sin(u) + C = \frac{1}{7}sin(7\theta+5) + C

例7 用替换法解\int x^2 sin(x^3) dx

令u=x^3, 得
\int x^2 sin(x^3) dx = \frac{1}{3} \int sin(u) du = \frac{1}{3}(-cos(u))+C = -\frac{1}{3} cos(x^3)+C

例8. 解\int \frac{1}{cos^2(2x)}dx

令u = 2x, 得:
\int \frac{1}{cos^2(2x)}dx= \int sec^2 (2x) dx
= \frac{1}{2}\int sec^2(u) du = \frac{1}{2} tan(u) + C
= \frac{1}{2} tan(2x) + C

例9. 求积分\int \frac{2z}{\sqrt[3] {z^2 + 1}}

令u = z^2 + 1
\int \frac{2z}{\sqrt[3] z^2 + 1} = \int \frac{du}{u^{\frac{1}{3}}}
= \int u^{-\frac{1}{3}}du = \frac{2}{3} u ^{\frac{2}{3}} + C
=\frac{2}{3}(z^2+1)^{\frac{2}{3}} + C

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