Chapter 4 notes from the Text Book Section 5

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

∫_a^b 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)