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             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             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

 

  1. 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.

  2. Compute du= g (x) dx.

  3. Rewrite the integral in terms of variable u.

  4. Find the resulting integral in terms of u.

  5. Replace u by g(x) to obtain and antiderivative in terms of s.

  6. 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         du = 2x dx  

 

Before substituting, determine the new upper and lower limits of integration.

Lower limits when x = 0, u = 0^2 +1 = 1.

Upper limits 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    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            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    2x = u^2 +1    x = (u^2 +1) / 2

u du = dx   Differentiate both sides.

 

Before substituting, determine the new upper and lower limits of integration.

Lower limits when x = 1, u =(2(1) -1)^(1/2) = (2-1)^(1/2) = 1.

Upper limits 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)