AP Calculus Chapter 05 Antidifferentiation 不定积分

Chapter 5 Antidifferentiation不定积分【Vocabulary · 词汇梳理】Indefinite Integral [ɪnˈdefɪnət 'ɪntɪɡrəl] 不定积分antiderivative [,æntidɪ'rɪvətɪv] 不定积分[引]integration ['ɪntə'greʃən] 积分, 积分法integrand ['ɪntə,grænd] 被积函数U-substitution [ju: ,sʌbstə'tjʊʃən] U换元法corollary [ˈkɔːrəleri] 推论natural logarithm [ˈnætʃrəl ˈlɔːɡərɪðəm] 自然对数(以e为底的)absolute value mark 绝对值符号eliminate [ɪˈlɪmɪneɪt] 除去, 消去【导图】A. ANTIDERIV ATIVES 不定积分The antiderivative or indefinite integral of a function f(x)is a function F(x)whose derivative is f(x). Since the derivative of a constant equals zero, the antiderivative of f(x)is not unique; that is, if F(x)is an integral of f(x), then so is F(x)+C, where C is any constant. The arbitrary constant C is called the constant of integration. The indefinite integral of f(x)is written as ∫f(x)dx; thus∫f(x)dx=F(x)+C if dF(x)dx=f(x)The function f(x)is called the integrand. The Mean Value Theorem can be used to show that, if two functions have the same derivative on an interval, then they differ at most by a constant; thatis, if dF(x)dx =dG(x)dx, thenF(x)−G(x)=C (C is a constant)B. BASIC FORMULAS 基本公式Familiarity with the following fundamental integration formulas is essential.All the references in the following set of examples are to the preceding basic formulas. In all of these, whenever u is a function of x , we define du to be u ′(x )dx ; when u is a function of t , we define du to be u ′(t )dt ; and so on.Example 1∫5xdx =5∫xdx by (1)=5(x 22)+C by (3)Example 2 ∫(x 4+√x 23−2x 23√x3)dx =∫(x 4+x23−2x −2−13x −13)dx =∫x 4dx +∫x 23dx −2∫x −2dx −13∫x −13dx by (1) and (2)=x 55+x 5353−2x −1−1−13x 2323+C by (3) =x 5+3x 53+2−1x 23+CExample 3∫(3−4x +2x 3)dx =∫3dx −4∫xdx +2∫x 3dx=3x −42x 2+24x 4+C=3x −2x 2+12x 4+CExample 4∫2(1−3x )2dx is integrated most efficiently by using formula (3) with u =1−3x and du =u ′(x )dx =−3dx .2∫(1−3x )2dx =−23∫(1−3x )2(−3dx )=−29u 3+C by (3)=−2(1−3x )3+CExample 5∫(2x3−1)5∙x2dx=16∫(2x3−1)5∙(6x2dx)=16∫u5duwhere u=2x3−1and du=u′(x)dx, this, by formula (3), equals1 6∙u66+C=136(2x3−1)6+CExample 6∫√1−x3dx=∫(1−x)1/3dx=−∫(1−x)13(−1dx)=−∫u13du where u=1−x and du=−1dx, this, bu formula (3), equals−u4343+C=−34(1−x)43+CExample 7x√3−4x2=∫(3−4x2)−12∙(xdx)=−18∫(3−4x2)−12(−8xdx) =−18∫u−12du=−18u1212+C by (3)=−14√3−4x+CExample 8∫4x2(x3−1)3dx=4∫(x3−1)−3∙x2dx=43∫(x3−1)−3(3x2dx) =4(x3−1)−2+C=−21(3)2+CExample 9(1+√x)4√x =∫(1+x12)4∙1x12dxLet u=1+x12, and note that du=12x−12dx,2∫(1+x 12)4∙(12x12dx)=2∫u4du=25u5+C=25(1+x12)5+CExample 10∫(2−y)2∙√ydy=∫(4−4y+y2)∙y 12dy=∫(4y12−4y32+y52)dy=4∙23y32−4∙25∙y52+27∙y72+C by (2)=83y 32−85y 52+27y 72+CExample 11∫x 3−x −42x 2dx =12∫(x −1x −4x 2)dx =12(x 22−ln |x |+4x )+CExample 12√1−2x +3x 23 Let u =1−2x +3x 2,du =(6x −2)dx =2(3x −1)dx3x −1√1−2x +3x 23=12∫u −13du =12∙32u 23+C =34(1−2x +3x 2)23+CExample 13∫2x 2−4x +3()2dx =∫2(x −1)2+1()2dx =∫(2+1()2)dx=∫2dx +∫1(x −1)2dx =2x −1x −1+CThis example illustrates the following principle:If the degree of the numerator of a rational function is not less than that of the denominator, divide until a remainder of lower degree is obtained.Example 14∫duu −3=ln |u −3|+C by (4)Example 15∫zdz 1−4z 2=−18ln |1−4z 2|+C by (4)Example 16∫cos x 5+2sin x dx =12∫2cos x 5+2sin x dx =12∫d(5+2sin x)5+2sin x =12ln (5+2sin x )+C By (4) with u =5+2sin x . The absolute-value sign is not necessary here since (5+2sin x )>0 for all x .Example 17∫e x 1−2e x dx =−12∫−2e x 1−2e x dx =−12ln |1−2e x |+C by (4)Example 18∫x 1−x dx =∫(−1+11−x)dx (by long division )=−x −ln |1−x |+CExample 19∫sin(1−2y)dy=−1∫sin(1−2y)(−2dy)=−12[−cos(1−2y)]+C by (6)=12cos(1−2y)+CExample 20∫sin2x2cosx2dx=2∫sin2x2cosx2d(x2)=2∫sin2x2d(sinx2)=23sin3x2+C by (3)Example 21∫sin x1+3cos xdx=−13∫−3sin x dx1+3cos x=13ln|1+3cos x|+CExample 22∫e tan y sec2y dy=∫e tan y d(tan y)=e tan y+C by (15) witℎ u=tan yExample 23∫e x tan e x dx=ln|sec e x|+C by (7) witℎ u=e x Example 24∫cos xsin2xdx=∫csc x cot x dx=−csc x+C by (12)Example 25∫tan x sec2x dx=tan2x2+C by (3) witℎ u=tan x and du=u′(x)dx=sec2x dxExample 26 (a)√9−x2=13√(1−(x3)=3∙1313dx√(1−(x3)=arcsinx3+C by (17) witℎ u=x3(b)xdx √9−x2=−12∫(9−x2)−12(−2x)dx=−12(9−x2)1212+C by(3)=−√9−x2+Cwith u=9−x2,n=−12 (c)∫xdx9−x2=−12∫(−2xdx)9−x2=−12ln|9−x2|+C by (4) witℎ u=9−x2(d)∫xdx(9−x2)2=−12∫(9−x2)−2(−2xdx)=12(9−x2)+C by (3) witℎ u=9−x2(e)∫dx9+x2=19∫dx1+(x3)2=3∙19∫13dx1+(x3)2=13tan−1x3+C by (18) witℎ u=x3Example 27dx√x(1+2√x)=x−12dx1+2√x=ln(1+2√x)+C by (4) witℎ u=1+2√x and du=dx√xExample 28cos√x √x =2∫cos(x12)(12x−12dx)=2sin√x+C by (5) witℎ u=√xExample 29∫sin x cos x dx=12sin2x+C by (3) witℎ u=sin xor=−12cos2x+C by (3) witℎ u=cos xor=−14cos2x+C by (6),wℎere we use tℎe trigonometric identity sin2x=2sin x cos xExample 30∫sin3x dx=∫sin2x sin x dx=∫(1−cos2x)sin x dx=∫sin x dx−∫cos2x sin x dx Let u=cos x,tℎen du=−sin x dx∫sin3x dx=∫sin x dx−∫cos2x sin x dx=−cos x+∫u2du=−cos x+cos3x3+CExample 31∫sin2x dx=∫(12−cos2x2)dx=x2−sin2x4+Cusing the trigonometric identity sin2θ=1−cos2θ2.Tips: 求sin n x cos m x的积分时，若n, m至少有一个为奇数时，则以Ex29、Ex30处理，先考虑换元；若n, m全为偶数时，则以Ex31处理,先用倍角公式降幂。