MATH 12002 - CALCULUS I 4.5: Integration by Substitution Indefinite - - PowerPoint PPT Presentation

MATH 12002 - CALCULUS I §4.5: Integration by Substitution — Indefinite Integrals

Professor Donald L. White

Department of Mathematical Sciences Kent State University

D.L. White (Kent State University) 1 / 7

Examples

1 Evaluate the integral

• (3x2 + 10x)
• x3 + 5x2 dx.
• (3x2 + 10x)
• x3 + 5x2 dx

= x3 + 5x2 (3x2 + 10x) dx Let u = x3 + 5x2 so du = (3x2 + 10x) dx = √u du =

• u1/2 du

=

2 3u3/2 + C

=

2 3(x3 + 5x2)3/2 + C.

D.L. White (Kent State University) 2 / 7

Examples

2 Evaluate the integral

• x4

(x5 + 7)10 dx.

• x4

(x5 + 7)10 dx =

• x4(x5 + 7)−10 dx

Let u = x5 + 7 so du = 5x4 dx and 1

5du = x4 dx

=

• u−10 · 1

5 du

=

1 5

• u−10 du

=

1 5 · 1 −9u−9 + C

= − 1

45(x5 + 7)−9 + C.

D.L. White (Kent State University) 3 / 7

Examples

3 Evaluate the integral

sin √x √x dx. sin √x √x dx =

• (sin √x) 1

√x dx Let u = √x = x1/2 so du = 1

2x−1/2 dx = 1 2 · 1 √x dx

and 2 du =

1 √x dx

=

• (sin u) · 2 du

= 2

• sin u du

= 2(− cos u) + C = −2 cos √x + C.

D.L. White (Kent State University) 4 / 7

Examples

4 Evaluate the integral

• cos7 x sin x dx.
• cos7 x sin x dx

=

• (cos x)7 sin x dx

Let u = cos x so du = − sin x dx and (−1) du = sin x dx =

• u7(−1) du

= −

• u7 du

= −1

8u8 + C

= −1

8 cos8 x + C.

D.L. White (Kent State University) 5 / 7

Examples

5 Evaluate the integral

• x

√x + 7 dx. In this case, the inside of the composite function is x + 7, but what is outside is not exactly the derivative of x + 7. A sum under a radical can complicate an integral because, for example, √ x + 7 = √x + √ 7. We can handle a sum in a numerator because, for example, x5 + x x3 dx = x5 x3 + x x3 dx =

• x2 + x−2 dx,

which can be evaluated easily, while a sum in a denominator is a problem: x3 x5 + x = x3 x5 + x3 x . [Continued→]

D.L. White (Kent State University) 6 / 7

Examples

[Example 5,

• x

√x+7 dx, continued]

Both of these issues can be avoided by letting u = x + 7, so that du = dx. In the integral, √x + 7 becomes √u and dx becomes du. Since u = x + 7, we can write x in terms of u as x = u − 7. We have

• x

√x + 7 dx = u − 7 √u du =

• u

√u − 7 √u du =

• u

u1/2 − 7 u1/2 du =

• u1/2 − 7u−1/2 du

=

2 3u3/2 − 7 · 2u1/2 + C

=

2 3(x + 7)3/2 − 14(x + 7)1/2 + C.

D.L. White (Kent State University) 7 / 7