Solutions.
1.
R
3t2
1
(2t3 +1) 2
R
− 12
dt = (2t3 + 1)
3t2 dt
1. U-sub appropriate?
1. Composite function? Yes!
−1
(2t3 + 1) 2
Let u = 2t3 + 1
1. Approximate function/derivative pair? Yes!
3
+ 1 → |{z}
3t2
2t
| {z }
function
deriv
R ³
´− 1
Let u = 2t3 + 1
2. Compute du
u = 2t3 + 1
du
= 6t2
dt
du = 6t2 dt
1
du = 3t2 dt
2
3. Analyze in terms of u and du
|
3
2t + 1
{z
−1
2
u
4. Integrate:
=
1
2
R
2
R
1
2
−2 1
3t
| {zdt} = u 2 du =
}
1
du
2
1
u− 2 du =
1
2
∙
1
u2
1
2
¸
1
2
R
1
u− 2 du
1
+ C = u2 + C
5. Restate in terms
of t
1
= (2t3 + 1) 2 + C
2.
3.
⎡
⎤
´
⎢
⎥
d ⎢
cos 4x2 + 3x + 2 ⎥
⎦
dx ⎣ |{z}
{z
}
outer|
d
dx
³
inner
⎡
⎤
⎢
sin cos (x)⎥
⎣ |{z}
⎦
| {z }
outer inner
h
³
´i
= − sin 4x2 + 3x + 2
|
{z
deriv of outer;
e v u la ted a t in n e r
|
{z
(8x + 3)
} |
{z
}
deriv of inner
}
By Chain Rule
=[cos (cos x)] (− sin x) = − cos (cos x) sin x
|
{z
}|
{z
}
deriv of outer; deriv of inner
eva lu a ted a t in n e r
For problems 4 and 5, suppose P (x) has the property that P 0 (x) =
4. Compute:
d
dx
1
|sin
{z x}
[P (sin (x))] =
·
d
dx
[P (4x3 + 7x2 )] =
=
cos(x)
sin(x)
= cot (x)
deriv of inner
deriv of outer
evaluated at inner
5. Compute:
cos
| {z x}
³
´
1
2
·
12x
+
14x
=
4x3 +
7x2} |
{z
}
|
{z
deriv of outer
evaluated at inner
3
1
x
dreiv of inner
12x2 +14x
4x3 +7x2
6.
R
4x
dx
(9x2 +3)2
R
−2
= (9x2 + 3)
4xdx
1. U-sub appropriate?
1. Composite function? Yes!
−2
(9x2 + 3)
Let u = 9x2 + 3
2. ³Approximate
funct/deriv pair? Yes!
´
2
9x + 3 → |{z}
4x
|
{z
}
function
deriv.
Let u = 9x2 + 3
2. Compute du
u = 9x2 + 3
du = 18xdx
1
du = xdx
18
4
du = 4xdx
18
2
du
= 4xdx
9
3. Analyze
in´terms of u and du.
−2
R ³ 2
R −2 2
2 R −2
9x + 3 4xdx
| {z }= u 9 du = 9 u du
|
{z
}
u−2
2
du
9
4. Integrate:
h
i
2 R −2
2 u−1
u
du
=
+ C = − 29 u−1 + C
9
9 −1
5. Restate in terms of x
−1
= − 29 (9x2 + 3) + C
7. f (x) = 3 sin (x) − 4 cos x; f 0 (x) = 3 [cos (x)] − 4 [− sin (x)] = 3 cos (x) + 4 sin (x)
8.
d
dx
i
d h
(cos (x))5 =5 [cos (x)]4 · (− sin x)= −5 cos4 (x) sin x
|
{z
} | {z }
|dx
{z
}
[cos5 (x)] =
9.
d
dx
h
2
i
2
sin (6x + 3x) =
h
³
´i
=2 sin 6x2 + 3x
|
|
{z
n[g(x)]n−1
d
dx
∙³
³
2
{z
[g(x)]n
´i
d h ³ 2
sin 6x + 3x
} |dx
{z
}
·
{z
d
[g(x)]
dx
By Chain Rule
h
|
R
3
´´2 ¸
sin 6x + 3x
|
³
´i
= 2 [sin (6x2 + 3x)] cos 6x2 + 3x
10.
g 0 (x)
n[g(x)]n−1
[g(x)]n
|
{z
deriv of outer
evaluated at inner
{z
}
[12x + 3]
}|
{z
}
deriv of inner
By Chain Rule
(1 + cos x) 2 sin x dx
}
1. U-sub appropriate?
4
}
1. Composite function? Yes!
3
(1 + cos (x)) 2
let u = 1 + cos (x)
2. Approx. funct/deriv Pair? Yes!
(1 + cos (x)) → sin (x)
|
{z
}
function
| {z }
deriv.
let u = 1 + cos (x)
2. Compute du
u = 1 + cos x
du
= − sin (x)
dx
du = − sin (x) dx
−du = sin (x) dx
3. Analyze in terms of u and du
3
R
R 3
R 3
(1 + cos (x)) 2 sin (x) dx= u 2 (−du) = − u 2 du
|
{z
}|
3
u2
{z
}
−du
4. Integrate
5
R 3
5
− u 2 du = − u52 + C = − 25 u 2 + C
2
5. Restate in terms of5 x
= − 25 (1 + cos (x)) 2 + c
11.
R sin x
√
cos x
R
1
dx = (cos (x))− 2 sin (x) dx
1. U-sub appropriate?
1. Composite Function? Yes!
1
(cos (x))− 2
Let u = cos (x)
2. Approx funct/deriv pair? Yes!
(cos (x)) → (sin (x))
|
{z
function
}
|
{z
}
deriv
Let u = cos (x)
2. Compute du
u = cos (x)
du
= − sin (x)
dx
du = − sin (x) dx
−du = sin (x) dx
3. Analyze in terms of u and du.
1
R
R
R
1
1
(cos (x))− 2 sin (x) dx= u− 2 (−du) = − u− 2 du
|
{z
1
u− 2
4. Integrate
R
− 12
− u
}|
du = −
{z
}
−du
∙
1
u− 2
1
2
¸
1
+ C = −2u 2 + C
5. Restate in terms
of x
1
2
= −2 (cos (x)) + C
5
12.
R
(1 + sin x) cos x dx
1. U-sub appropriate?
1. Composite function? If there is, I don’t see it!
2. Approx funct/deriv pair? Yes!
(1 + sin (x)) → cos d (x)
|
{z
}
function
|
{z
}
deriv
Let u = 1 + sin (x)
2. Compute du
u = 1 + sin (x)
du = cos (x) dx
3. RAnalyze in terms of u andR du
(1 + sin (x))cos (x) dx= u du
|
{z
}|
u
{z
}
du
4. RIntegrate:
2
u du = u2 + C
5. Re-express in terms of x
(1+sin(x))2
+C
2
13.
d
dx
√
[csc ( x)] =
d
dx
⎡
⎤
⎢
³ 1 ´⎥
⎢
⎥
⎢ csc x 2 ⎥
⎣ ↑
⎦
outer
³
=− csc x
|
1
2
´
{z
³
cot x
1
2
deriv of outer
evaluated at inner
↑
inner
´
1
=−
x
} |2 {z }
− 12
³
1
csc x 2
´
³
|
{z
[g(x)]n
1 + (sec (x))3
´− 1 h
− 12
= 12 (1 + sec3 (x))
15.
´1
2
}
g 0 (x)
n[g(x)]n−1
=
´
´− 1
h
i
1³
d
1 + sec (x)3 2 · dx
1 + (sec (x))3
{z
}
|
|2
{z
}
f 0 (x) =
³
1
deriv of inner
1
1
2
1
2x 2
14. f (x) = (1 + sec3 x) 2 ; f (x) = 1 + sec (x)3
=
³
cot x 2
2
|
3 (sec (x))2
{z
n[g(x)]n−1
i
d
[sec (x)]
}|dx {z
}
g 0 (x)
[3 sec2 (x)] [sec (x) tan (x)]
3
3 sec
√ (x) tan(x)
2
1+sec3 (x)
⎡
⎤
³
´⎥
⎢
d ⎢
cot x2 + 2x ⎥
⎦
dx ⎣ |{z}
{z
}
outer|
inner
h
³
´i
= − csc2 x2 + 2x
|
{z
deriv of outer
evaluated at inner
(2x + 2)
}|
{z
}
deriv of inner
6
16.
d
dx
⎡
⎤
⎢
sin (tan (3x))⎥
⎣ |{z}
⎦
|
{z
}
outer
|
{z
deriv of outer
evaluated at inner
h
|
i
sec2 (3x)
{z
·
}
deriv of outer
evaluated at inner
17.
R tan2 (√x)
√
x
⎤
d ⎢
⎥
⎣ tan (3x)⎦
} dx |{z} | {z }
outer
=[cos (tan (3x))]
inner
= [cos (tan (3x))]
⎡
|
{z
inner
deriv of inner
}
= 3 [cos (tan (3x))] [sec2 (3x)]
3
|{z}
deriv of inner
dx =
Remark 1 Since we don’t know how to integrate tan2 (x) , we’ll convert it to something
that we CAN integrate, using the identity tan2 (x) = sec2 (x) − 1.
R tan2 (√x)
√
x
R
dx =
³
1
´
R³
³
´
1
´
1
sec2 x 2 − 1 x− 2 dx
1
R
1
= sec2 x 2 · x− 2 dx − x− 2 dx
R
³
1
´
1
= sec2 x 2 · x− 2 dx −
1
x2
1
2
+C
1. U-sub appropriate?
1. Composite
³ 1 ´ Function? Yes!
2
sec x 2
1
Let u = x 2
2. Approx funct/deriv pair? Yes!
1
− 12
x2 → x
|{z}
|{z}
function
deriv
1
Let u = x 2
2. Compute du
1
u = x2
1
⇒ du
= 12 x− 2
dx
1
⇒ du = 12 x− 2 dx
1
⇒ 2du = x− 2 dx
3. ZAnalyze
in terms of u and du
³ 1´
R
1
1
2
− 12
2
2
2
sec x 2 x
| {zdx} −2x + C = sec u 2u du − 2x + C
|
18.
R
{z
sec2 u
R
}
2du
1
= 2 sec2 u du − 2x 2 + C
4. Integrate
1
= 2 tan (u) − 2x 2 + C
5. Rewrite³in terms
of x
´
1
1
= 2 tan x 2 − 2x 2 + C
tan (3x) sec2 (3x) dx
(a) The FIRST WAY:
7
1. Is U-sub appropriate?
1. Composite function? Yes!
tan (3x)
Let u = tan (3x)
2. Approx funct/deriv pair? Yes!
tan (3x) → sec2 (3x)
|
{z
function
}
|
{z
derivative
}
Let u = tan (3x)
2. Compute du
u = tan (3x)
⇒ du
= sec2 (3x) · 3
dx
⇒ du = 3 sec2 (3x) dx
⇒ 13 du = sec2 (3x) dx
3. ZAnalyze in terms of u and du
R
tan (3x)sec2 (3x) dx= u 13 du = 13 u du
|
{z
u
|
}
{z
}
1
du
3
4. Integrate
2
1R
u du = 13 u2 + C =
3
5. Rewrite
in terms of x
tan2 (3x)
=
+C
6
u2
6
+C
(b) The SECOND WAY
R
R
tan (3x) sec2 (3x) dx = sec (3x) sec (3x) tan (3x)
1. Is U-sub appropriate?
1. Composite function??? I there is, I don’t see it!
Let u =???
2. Approx. function/deriv pair? Yes!
sec (3x) ↔ sec (3x) tan (3x)
Let u = sec (3x)
2. Compute du
u = sec (3x)
⇒ du
sec (3x) tan (3x) · 3
dx
⇒ du = 3 sec (3x) tan (3x)
⇒ 13 du = sec (3x) tan (3x)
3. ZAnalyze in terms of u and du
R
sec (3x)sec (3x) tan (3x) dx= 13 u du
|
{z
u
|
}
{z
1
du
3
R
2
}
4. Integrate 13 u du = 13 u2 + C =
5. Re-express in terms of x.
2
= sec 6(3x) + C
19.
R
R
u2
6
tan7 (x) sec2 (x) dx = (tan x)7 sec2 (x) dx
1. U-sub appropriate?
8
+C
1. Composite function? Yes!
(tan x)7
Let u = tan x
2. Approx funct/deriv pair? Yes!
tan (x) → sec2 (x)
| {z }
| {z }
funct
deriv
Let u = tan (x)
2. Compute du
u = tan (x)
= sec2 (x)
⇒ du
dx
⇒ du = sec2 (x) dx
3. ZAnalyze in terms of u and du
R
(tan (x))7 sec2 (x) dx= u7 du
|
|
}
{z
u7
{z
}
du
4. RIntegrate
8
u7 du = u8 + C
5. Re-express in terms of x
(tan(x))8
+C
8
20.
Z
|
sec3 (x) tan (x) dx =
Z
(sec (x))2 sec (x) tan (x) dx
{z
Get this trick!
1. U-sub appropriate?
1. Composite function? Yes!
(sec x)2
Let u = sec (x)
2. Approx funct/deriv pair? Yes!
sec (x) → sec (x) tan (x)
| {z }
funct.
|
{z
}
deriv.
Let u = sec (x)
2. Compute du
u = sec (x)
= sec (x) tan (x)
⇒ du
dx
⇒ du = sec (x) tan (x) dx
3. RAnalyze in terms of u and du R
(sec (x))2 sec (x) tan (x) dx = u2 du
|
{z
u2
}|
{z
du
}
4. RIntegrate
3
u2 du = u3 + C
5. Re-express in terms of u and du
3
+ C = 13 sec3 (x) + C
= (sec(x))
3
9
}