Chapter 4 notes from the Text Book Section 4.5  Integration by Substitution Theorem 4.12 - Antidifferentiation of a Composite Function   Let g be a function whose range is an interval I, and let f be a function that is continuous on I.   If f is differentiable on its domain and F is an antiderivative of f on I, then   ∫ f(g(x)) g (x) dx = F(g(x)) + C   if u= g(x), then du = g (x) dx     and   ∫ f(u) du = F(u) + C Example 1 recognizing the f(g(x)) g (x) pattern   Find ∫ f(x^2 +1)^2 (2x)dx   Solution  Letting g(x) = x^2 +1, you obtain             g(x) = 2x  and   f(g(x)) = f(x^2 +1) = (x^2 +1)^2     From this, you can recognize that the integrand follows the ∫ f(g(x)) g (x) pattern.  Using the Power Rule for integration and Theorem 4.12, you can write             ∫ f(x^2 +1)^2 (2x)dx = (1/3) (x^2+1)^3 + C   (note that the du or g(x) became the constant C, then you got the antiderivative of u or f(g(x)) )   Try using the Chain Rule to check the derivative of (1/3) (x^2+1)^3 + C is the integrand of the original integral.   You will see that you get (x^2 +1)^2 (2x) so it checks Example 2 recognizing the f(g(x)) g (x) pattern   Find ∫ 5 cos 5x dx   Solution  Letting g(x) = 5, you obtain             g(x) = 5  and   f(g(x)) = f(5x) = cos 5x   From this, you can recognize that the integrand follows the ∫ f(g(x)) g (x) pattern.  Using the Cosine Rule for integration and Theorem 4.12, you can write             ∫ f(cos 5x) (5)dx = sin 5x + C   You can check this by differentiation sin (5x) + C to obtain the original integrand. Example 3  Multiplying and Dividing by a Constant   Find ∫ x(x^2 + 1)^2 dx   Solution - This is similar to the integral given in Example 1, except that the integrand is missing a factor of 2.  Recognizing that 2x is the derivative of x^2 + 1, you can let g(x) = x^2 +1 and supply the 2x as follows.   ∫ x *(x^2 + 1)^2 dx = ∫ 1*(x^2 + 1)^2 ((1/2) (2x)) dx               Multiply & Divide by 2                           = (1/2) ∫ (x^2 + 1)^2  (2x) dx                         Constant Multiple Rule                                            = (1/2) [(x^2 + 1)^3/3] + C                              Got Antiderivative and Integrate.                                     = (1/6) [(x^2 + 1)^3 +C                               Simplified Example 4  Change of Variables   Find  ∫ sqrt(2x-1) dx   Solution  First, let u be the inner function, u = 2x -1.  Then calculate the differential du to be du = 2 dx . Now, using sqrt(2x-1) = sqrt u and dx = du/2 , substitute to obtain the following.   ∫ sqrt(2x-1) dx = ∫ sqrt(u) (du/2)                         integral in terms of u                         = (1/2) ∫ u^(1/2) du                                   = (1/2) ∫ (u^(3/2))/(3/2) + C               Antiderivative in terms of u                         = (1/3) (u^(3/2)) + C                         =(1/3) (2x -1)^(3/2) + C          Antiderivative in terms of x Example 5  Change of Variables   Find  ∫ x sqrt(2x-1) dx           à  sqrt = the square root of the enclosed.          Solution  As in the previous example, let  u = 2x -1 and obtain dx = du/2.  Because the integrand contains a factor of x. you must also solve for x in terms of u, as follows: u = 2x -1      à       x = (u+1)/2       solve for x in terms of  u.   Now, using substitution, you obtain the following.   ∫ x sqrt(2x-1) dx = ∫ ((u+1)/2 )* u^(1/2)*( du/2)                                                 = (1/4) ∫ (u^(3/2)) + (u^(1/2)) * du               Antiderivative in terms of u                         = (1/4 (u^(5/2)/(5/2)) +(u^(3/2)/(3/2)) + C                         =(1/10) (2x -1)^(5/2) + (1/6) (2x-1)^(3/2)+ C          Antiderivative in terms of x Example 6  Change of Variables   Find  ∫ sin^2 3x cos 3x dx   Solution  Because sin^2 3x = (sin 3x)^2, you can let  u = sin 3x.   and     du = (cos 3x)(3) dx   Now, because cos 3x dx is part of the given integral, you can write du/3 = cos 3x dx.   Substituting u and du/3 in the given integral yields the following.   ∫ sin^2 3x cos 3x dx = ∫ (u^2)(du/3)                         = (1/3) ∫ (u^2) du            ( get fraction out front)                         = (1/3) ∫ (u^3)/3 + C            (get antideriv and du becomes C)                         =(1/9) (sin^3 3x )+ C            (Simplify then replace substititute) Guidelines for Making a Change of Variables   Choose a substitution u = g(x).  Usually, it is best to choose the inner part of a composite function, such as a quantity raised to a power. Compute du= g (x) dx. Rewrite the integral in terms of variable u. Find the resulting integral in terms of u. Replace u by g(x) to obtain and antiderivative in terms of s. Check your answer by differentiating. Example 7  Substitution and the General Power Rule    The u will be in this color,  the du will be in this color.   a)      ∫ 3(3x -1)^4 dx = ∫ (3x -1)^4 * (3) dx  =  ((3x-1)^5) /5 + C b)      ∫ (2x +1)(x^2 +x) dx = ∫ (x^2 +x) * (2x +1) dx =   ((x^2 + x)^2) /2 + C c)      ∫ (3x^2)(x^3-2)^(1/2) dx = ∫ (x^3 - 2)^(1/2) * (3x^2) dx = ((x^3 - 2 )^(3/2) )/ (3/2) + C = (2/3) ((x^3 - 2 )^(3/2) ) + C d)      ∫ ((-4x) / ((1 -2x^2)^2)) dx = ∫ ((1 -2x^2)^ (-2)) * (-4x) /dx = ((1 -2x^2)^ (-1))/(-1) + C e)      ∫ cos^2 * sin(x)  dx = ∫ (cos x)^2 * (- sin (x))dx = (cos x^3) /3 + C Theorem 4.1  Change of Variables for Definite Integrals If the function u = g(x) has a continuous derivative on the closed interval [a,b] and f is continuous on the range of g, then ∫ab  f(g(x)) * g (x) dx  =  ∫g(a)g(b)  f(u) du Example 8  Change of Variables   Evaluate  ∫01 x(x^2 + 1) ^3  dx   Solution  To evaluate this integral, let u = x^2 +1. Then you obtain         u = x^2 +1    à     du = 2x dx     Before substituting, determine the new upper and lower limits of integration. Lower limits à when x = 0, u = 0^2 +1 = 1. Upper limits à when x = 1, u = 1^2 +1 = 2.   Now, you can substitute to obtain   ∫01 x(x^2 + 1) ^3  dx= (1/2) ∫12 (x^2 + 1) ^3 (2x) dx    à needed to make du = 2x therefore *(1/2)   = (1/2) ∫12 (u) ^3 (du) =   ( 1 /2)*[u^4/4] 12 = (1/2) * (4- (1/4)) = (15/8) Example 9  Change of Variables   Evaluate A =  ∫15 x/ sqrt(x^2 - 1)   dx          à  sqrt = the square root of the enclosed.   Solution  To evaluate this integral, let u = sqrt(2x-1) Then , you obtain  u^2 = 2x -1    since u^2 = 2x -1 à   2x = u^2 +1 à   x = (u^2 +1) / 2 u du = dx   Differentiate both sides.   Before substituting, determine the new upper and lower limits of integration. Lower limits à when x = 1, u =(2(1) -1)^(1/2) = (2-1)^(1/2) = 1. Upper limits à when x = 5, u = (2(5)-1)^(1/2) = (10 -  1)^(1/2) = 3.   Now, substitute to obtain    ∫15 x/ sqrt(x^2 - 1)   dx =     ∫13 (1/u)*((u^2 +1)/2) u du = (1/2) ∫13  (u^2 +1) du  = (1/2) [(u^3 + u)] 13  = (1/2) (9 + 3 - (1/3) -1) = (16/3)