MATH 12002 - CALCULUS I 5.6: More Integral Examples Professor - - PowerPoint PPT Presentation

MATH 12002 - CALCULUS I §5.6: More Integral Examples

Professor Donald L. White

Department of Mathematical Sciences Kent State University

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Integration Formulas

• un du =

1 n+1un+1 + C

1 u du = ln |u| + C

• sin u du = − cos u + C
• cos u du = sin u + C
• eu du = eu + C
• 1

√ 1 − u2 du = sin−1 u + C

• 1

1 + u2 du = tan−1 u + C

• 1

u √ u2 − 1 du = sec−1 u + C

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Examples

1 Evaluate

• 1

4 + 25x2 dx and

• x

4 + 25x2 dx. We evaluated the first integral in the previous lecture:

• 1

4 + 25x2 dx = 1 4

• 1

1 + ( 5

2x)2 dx = 1

10 tan−1 5 2x

• + C.

For the second, let u = 4 + 25x2, so du = 50x dx and

1 50 du = x dx:

• x

4 + 25x2 dx = 1 50 1 u du = 1 50 ln |u| + C = 1 50 ln(4 + 25x2) + C.

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Examples

2 Evaluate

• ex

√1 − ex dx and

• ex

√ 1 − e2x dx. For the first integral, let u = 1 − ex, so du = −ex dx and (−1) du = ex dx:

• ex

√1 − ex dx = −

• 1

√u du = −

• u−1/2 du

= −2u1/2 + C = −2 √ 1 − ex + C. For the second integral,

• bserve that e2x = (ex)2, and let u = ex, so du = ex dx:
• ex

√ 1 − e2x dx =

• ex
• 1 − (ex)2 dx =
• 1

√ 1 − u2 du = sin−1 u + C = sin−1(ex) + C.

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Examples

3 Evaluate

sin x cos x 1 + cos4 x dx and sin x cos3 x 1 + cos4 x dx. To simplify these integrals, first let u = cos x so du = − sin x dx and (−1) du = sin x dx. We then have sin x cos x 1 + cos4 x dx = −

• u

1 + u4 du and sin x cos3 x 1 + cos4 x dx = −

• u3

1 + u4 du. Each integral will now require a second substitution. [Continued →]

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Examples

[Example 3, continued] For the integral sin x cos x 1 + cos4 x dx = −

• u

1 + u4 du,

• bserve that u4 = (u2)2 and

d duu2 = 2u.

Now let t = u2 so that dt = 2u du and 1

2 dt = u du:

sin x cos x 1 + cos4 x dx = −

• u

1 + u4 du = −

• u

1 + (u2)2 du = −1 2

• 1

1 + t2 dt = −1

2 tan−1 t + C

= −1

2 tan−1(u2) + C

= −1

2 tan−1(cos2 x) + C.

[Continued →]

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Examples

[Example 3, continued] For the integral sin x cos3 x 1 + cos4 x dx = −

• u3

1 + u4 du, let t = 1 + u4 so that dt = 4u3 du and 1

4 dt = u3 du:

sin x cos3 x 1 + cos4 x dx = −

• u3

1 + u4 du = −1 4 1 t dt = −1

4 ln |t| + C

= −1

4 ln(1 + u4) + C

= −1

4 ln(1 + cos4 x) + C.

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