Math 152-copyright Joe Kahlig, 21A Page 1

Math 152-copyright Joe Kahlig, 21A

Section 7.3:Additional Problems Solutions

Page 1

For all of these problems we try to match up one of these trig identities for the substitution.

cos2 = 1 - sin2

sec2 = 1 + tan2

tan2 = sec2 - 1

1. The expression under the square root matches the form of the identity, sec2 = 1+tan2 , except the constant is a 25. So multiply the identity by 25 to fix the constant.

25 sec2 = 25 + 25 tan2

we need x2 = 25 tan2 so let x = 5 tan and then dx = 5 sec2 d

x3

dx =

(5 tan )3

5 sec2 d =

54 tan3 sec2

d =

54 tan3 sec2 d

x2 + 25

25 tan2 + 25

25 sec2

5 sec

= 53 tan3 sec d = 125 tan2 tan sec d

Now we want to use the u-sub of u = sec with du = tan sec d so use the identity tan2 = sec2 - 1 to get.

= 125

(sec2 - 1) tan sec d = 125

(u2 - 1)du = 125

u3 -u

+C

3

= 125 1 sec3 - sec + C 3

x2 + 25

From the triangle we see that sec =

.

5

x2 + 25 x

125

x2 + 25

3

125 x2 + 25

1

Answer: =

-

+ C or

3

5

5

3

5

3

x2 + 25 - 25 x2 + 25 + C

2. The expression under the square root matches the form of the identity, sec2 = 1+tan2 , except the constant is a 25. So multiply the identity by 16 to fix the constant.

16 sec2 = 16 + 16 tan2

we need 25x2 = 16 tan2 so let 5x = 4 tan or x = 4 tan and then dx = 4 sec2 d

5

5

1

dx =

x2 25x2 + 16

4 tan

1 2

16 tan2 + 16

4 sec2 d = 5

5

4 sec2

4

5

2

d

tan2 16 sec2

5

1

=

sec2 4 tan2 ? 4 sec d =

sec 16 tan2 d =

5 cos 16 sin2

d =

5 cos 16 sin2 d

5

5

cos2

Math 152-copyright Joe Kahlig, 21A

Page 2

Now we want to use the u-sub of u = sin with du = cos d. This will give the integral

51

-5 1

-5 1

=

16 u2

du =

16

+C = u

16

sin

+C

5x

From the triangle we see that sin =

.

25x2 + 16

1

25x2 + 16

So

=

.

sin

5x

-5 25x2 + 16

- 25x2 + 16

Answer: ?

+C =

+C

16

5x

16x

15x2 + 16 5x 4

3. This integral does not have a square root but we can still use the identity, sec2 = 1 + tan2 . Let x = tan and then dx = sec2 d

1 (1 + x2)2 dx =

(1

+

1 tan2

)2

sec2

d

=

sec2 (sec2 )2 d =

1 sec2 d =

=

cos2 d =

1

1

1

(1 + cos 2) d = + sin 2 + C

2

2

2

11

11

11

= + sin 2 + C = + ? 2 sin cos + C = + sin cos + C

24

24

22

x From the triangle we see that sin =

x2 + 1 1 and cos = x2 + 1

x2 + 1 x

1

1

1x

1

1

11

Answer:

arctan(x) + 2

+C

2 x2 + 1 x2 + 1

=

2 arctan(x) + 2 x2 + 1 + C

4. The expression under the square root matches the form of the identity, cos2 = 1 - sin2 , except the constant is a 4. So multiply the identity by 4 to fix the constant.

4 cos2 = 4 - 4 sin2

we need 9x2 = 4 sin2 so let 3x = 2 sin or x = 2 sin and then dx = 2 cos d

3

3

x3 4 - 9x2 dx =

2

3

sin

4 - 4 sin2 ? 2 cos d =

3

3

2

4

sin3 cos 4 cos2 d

3

= 32 sin3 cos2 d = 32 sin2 cos2 sin d = 32 (1 - cos2 ) cos2 sin d

81

81

81

Now we want to use the u-sub of u = cos with du = - sin d. This will give the integral

-32 =

(1 - u2)u2 du = -32

(u2 - u4) du = -32

1 u3 - 1 u5 + C

81

81

81 3 5

Math 152-copyright Joe Kahlig, 21A

Page 3

-32 =

1 cos3 - 1 cos5

+C

81 3

5

4 - 9x2

From the triangle we see that cos = 2

-32 1

4 - 9x2

3

1

4 - 9x2

5

Answer:

-

+C

81 3

2

5

2

2

3x

4 - 9x2

5. The expression under the square root matches the form of the identity, tan2 = sec2 -1, except the constant is a 9. So multiply the identity by 9 to fix the constant.

9 tan2 = 9 sec2 - 9

we need 4x2 = 9 sec2 so let 2x = 3 sec or x = 3 sec and then dx = 3 sec tan d

2

2

4x2 - 9 dx = x

9 sec2 - 9 3 sec 2

3 sec tan d =

2

9 tan2 tan d

= 3 tan2 d = 3 sec2 - 1 d = 3 (tan - ) + C

2x

4x2 - 9

From the triangle we see that tan =

3

4x2 - 9

4x2 - 9

3

Answer: 3

- arccos

3

2x

3 + C = 4x2 - 9 - 3 arccos 3 + C

2x

6. Complete the square so we can rewrite the integral in a form that will work with trig sub.

6x - x2 = -x2 + 6x = -1(x2 - 6x) = -1 x2 - 6x+32 - 32 = -1 x2 - 6x + 9 - 9 = -1 (x - 3)2 - 9 = -(x - 3)2 + 9 = 9 - (x - 3)2

4

4

(6x - x2)3/2 dx = (9 - (x - 3)2)3/2 dx

The expression under the square root matches the form of the identity, cos2 = 1 - sin2 , except the constant is a 9. So multiply the identity by 9 to fix the constant.

9 cos2 = 9 - 9 sin2 we need (x - 3)2 = 9 sin2 so let x - 3 = 3 sin or x = 3 + 3 sin and then dx = 3 cos d

4

4

12 cos

(9 - (x - 3)2)3/2 dx = (9 - 9 sin2 )3/2 3 cos d = (9 cos2 )3/2 d

Math 152-copyright Joe Kahlig, 21A

Page 4

=

12 cos 27 cos3 d =

4 9 cos2 d =

4 sec2 d = 4 tan + C

9

9

x-3 From the triangle we see that tan =

6x - x2

3 x-3

9- (x - 3)2 or 6x - x2

4 x-3

Answer:

+C

9 6x - x2

7. Complete the square so we can rewrite the integral in a form that will work with trig sub.

9x2 + 18x + 13 = 9(x2 + 2x) + 13 = 9 x2 + 2x+12 - 12 + 13 = 9 x2 + 2x + 1 - 1 + 13 = 9 (x + 1)2 - 1 + 13 = 9(x + 1)2 - 9 + 13 = 9(x + 1)2 + 4

1

dx =

9x2 + 18x + 13

1 dx

9(x + 1)2 + 4

The expression under the square root matches the form of the identity, sec2 = 1+tan2 , except

the constant is a 4. So multiply the identity by 4 to fix the constant.

4 sec2 = 4 tan2 + 4 we need 9(x + 1)2 = 4 tan2 so let 3(x + 1) = 2 tan or x = 1 (2 tan - 3) and then

3 dx = 2 sec2 d

3

1

dx = 1

? 2 sec2 d = 2

sec2

d

9(x + 1)2 + 4

4 tan2 + 4 3

3 4 sec2

2 sec2

1

1

=

d = sec d = ln |sec + tan | + C

3 2 sec

3

3

From the triangle we see that sec = 3(x + 1)

and tan = 2

4 + 9(x + 1)2 2

4+ 9(x + 1)2 or 9x2 + 18x + 13

2

1

4 + 9(x + 1)2 3(x + 1)

Answer: ln

+

+C

3

2

2

3(x + 1)

Math 152-copyright Joe Kahlig, 21A

Page 5

8. Complete the square so we can rewrite the integral in a form that will work with trig sub.

4x - x2 - 3 = -x2 + 4x - 3 = -1(x2 - 4x) - 3 = -1 x2 - 4x+22 - 22 - 3 = -1 x2 - 4x + 4 - 4 - 3 = -1 (x - 2)2 - 4 - 3 = -(x - 2)2 + 4 - 3 = 1 - (x - 2)2

x2

x2

(4x - x2 - 3)3/2 dx = (1 - (x - 2)2)3/2 dx

Now procede with the trig substitution. The expression 1 - (x - 2)2 matches the form of the identity cos2 = 1 - sin2 .

We need (x - 2)2 = sin2 so let x - 2 = sin or x = 2 + sin and then dx = cos d

x2

(2 + sin )2

4 + 4 sin + sin2

(1 - (x - 2)2)3/2 dx = (1 - sin2 )3/2 cos d =

(cos2 )3/2

cos d

4 + 4 sin + sin2

4 + 4 sin + sin2

=

cos3

cos d =

cos2

d

=

4 4 sin sin2 cos2 + cos2 + cos2 d =

4 sec2 + 4 sin 1 + tan2 d cos cos

= 4 sec2 + 4 tan sec + sec2 - 1 d = 5 sec2 + 4 tan sec - 1 d

= 5 tan + 4 sec - + C

1 From the triangle we see that sec =

4x - x2 - 3 x+2 and tan = 4x - x2 - 3

1 x-2

1- (x - 2)2 or 4x - x2 - 3

5(x + 2)

4

Answer:

+

- arcsin(x - 2) + C

4x - x2 - 3 4x - x2 - 3

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