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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