Calc 2, Test #1A - Solutions
Spring 1988
Pat Rossi
R 3t2
1.
3
(2t
(a)
1
+1) 2
Name
dt =
R
− 12
(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
6t2
|2t {z+ 1} → |{z}
deriv
function
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
¢− 1 2
R
R
R¡ 3
1
1
dt} = u− 2 12 du = 12 u− 2 du
2t + 1 2 |3t{z
|
{z
} 1
1
u− 2
2
du
4. Integrate:
h 1i
R −1
1
1
1 u2
2
= 2 u du = 2 1 + C = u 2 + C
2
5. Restate in terms
of t
1
= (2t3 + 1) 2 + C
⎡
⎤
¢
¢¤
¡
£
¡
2
d ⎣
⎦ = − sin 4x2 + 3x + 2 (8x + 3)
2. dx
+
3x
+
2
4x
cos
|{z}|
|
{z
}
{z
} | {z }
outer
3.
d
dx
⎡
inner
|
⎤
deriv of outer;
deriv of inner
evulated at inner
}
{z
By Chain Rule
⎣ sin cos (x)⎦ = [cos (cos x)] (− sin x) = − cos (cos x) sin x
|{z}| {z }
| {z } | {z }
outer
inner
deriv of outer; deriv of inner
evaluated at inner
For problems 4 and 5, suppose P (x) has the property that P 0 (x) =
4. Compute:
d
dx
[P (sin (x))] =
1
{zx}
|sin
deriv of outer
evaluated at inner
·
cos
| {zx}
deriv of inner
=
cos(x)
sin(x)
1
x
= cot (x)
5. Compute:
d
dx
[P (4x3 + 7x2 )] =
4x3
|
¡
¢
1
· 12x2 + 14x =
2
+ 7x } |
{z
}
{z
deriv of outer
evaluated at inner
6.
R
4x
dx
(9x2 +3)2
(a)
=
R
−2
(9x2 + 3)
4xdx
12x2 +14x
4x3 +7x2
dreiv of inner
1. U-sub appropriate?
1. Composite function? Yes!
−2
(9x2 + 3)
Let u = 9x2 + 3
2. Approximate
funct/deriv pair? Yes!
¡ 2
¢
9x + 3 → |{z}
4x
| {z }
deriv.
function
2
Let u = 9x + 3
2. Compute du
u = 9x2 + 3
du = 18xdx
1
du = 4xdx
18
4
du = 4xdx
18
2
du = 4xdx
9
3. RAnalyze
terms of uRand du.
R
¡ 2 in¢−2
= u−2 29 du = 29 u−2 du
9x + 3 4xdx
{z
}
|
|
{z
} 2
u−2
9
du
4. Integrate:
h i
R −2
2
2 u−1
+ C = − 29 u−1 + C
du
=
u
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
[cos5 (x)] =
¤
d £
(cos (x))5 = 5 [cos (x)]4 · (− sin x) = −5 cos4 (x) sin x
|dx {z
} | {z } | {z }
9.
d
dx
¤
£ 2
sin (6x2 + 3x) =
d
dx
g 0 (x)
n[g(x)]n−1
[g(x)]n
h¡ ¡
¢¢2 i
sin 6x2 + 3x
{z
}
|
[g(x)]n
£ ¡
¢¤ d £ ¡ 2
¢¤
sin 6x + 3x
= 2 sin 6x2 + 3x ·
|
{z
} |dx
{z
}
n[g(x)]n−1
|
{z
d
[g(x)]
dx
By Chain Rule
}
£ ¡
¢¤
= 2 [sin (6x2 + 3x)] cos 6x2 + 3x [12x + 3]
|
{z
} | {z }
|
deriv of inner
deriv of outer
evaluated at inner
{z
By Chain Rule
2
}
10.
R
3
(1 + cos x) 2 sin x dx
(a)
1. U-sub appropriate?
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 }
|
{z
}
function
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
R
R 3
R 3
3
(1 + cos (x)) 2 sin (x) dx = u 2 (−du) = − u 2 du
|
{z
}| {z }
3
−du
u2
4. Integrate
5
R 3
5
− u 2 du = − u52 + C = − 25 u 2 + C
2
11.
R
5. Restate in terms of5 x
= − 25 (1 + cos (x)) 2 + c
√sin x dx
cos x
(a)
=
R
1
(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 }
| {z }
function
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.
R
R
R
1
1
1
(cos (x))− 2 sin (x) dx = u− 2 (−du) = − u− 2 du
| {z }| {z }
1
u− 2
−du
4. Integrate
h −1 i
R
1
1
− u− 2 du = − u 1 2 + C = −2u 2 + C
2
3
12.
R
5. Restate in terms
of x
1
2
= −2 (cos (x)) + C
(1 + sin x) cos x dx
(a)
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 }
{z
}
function
deriv
Let u = 1 + sin (x)
2. Compute du
u = 1 + sin (x)
du = cos (x) dx
3. RAnalyze in terms of u and Rdu
(1 + sin (x))cos (x) dx = u du
|
{z
}| {z }
u
du
4. RIntegrate:
2
u du = u2 + C
5. Re-express in terms of x
(1+sin(x))2
+C
2
⎡
⎤
³ 1 ´⎥
⎢
√
d
d ⎢
13. dx [csc ( x)] = dx ⎣ csc x 2 ⎥
⎦
↑
outer
↑
inner
³ 1´ 1 1
³ 1´
2
x− 2
= − csc x cot x 2
{z
} |2 {z }
|
deriv of outer
evaluated at inner
=−
³ 1´
³ 1´
csc x 2 cot x 2
1
2x 2
deriv of inner
¡
¢1
1
14. f (x) = (1 + sec3 x) 2 ; f (x) = 1 + sec (x)3 2
|
{z
}
[g(x)]n
f 0 (x) =
¢− 1
1¡
1 + sec (x)3 2 ·
|2
{z
}
n[g(x)]n−1
=
1
2
d
dx
£
¤
1 + (sec (x))3
|
{z
}
¡
¢− 1 £
¤d
[sec (x)]
1 + (sec (x))3 2 3 (sec (x))2
|
{z
}|dx {z }
n[g(x)]n−1
− 12
= 12 (1 + sec3 (x))
=
g 0 (x)
g 0 (x)
[3 sec2 (x)] [sec (x) tan (x)]
3
3 sec
√ (x) tan(x)
2
1+sec3 (x)
4
⎡
⎤
¢
¢¤
¡
¡
£
d ⎣
15. dx
cot x2 + 2x ⎦ = − csc2 x2 + 2x (2x + 2)
|{z}
| {z }
|
{z
} | {z }
outer
inner
16.
⎡
⎤
⎡
⎤
⎣ sin (tan (3x))⎦ = [cos (tan (3x))] d ⎣ tan (3x)⎦
|{z}| {z }
{z
} dx |{z}|{z}
|
outer
outer inner
inner
deriv of outer
{z
}
evaluated at inner |
deriv of inner
£
¤
= [cos (tan (3x))] sec2 (3x) · |{z}
3
= 3 [cos (tan (3x))] [sec2 (3x)]
| {z }
d
dx
deriv of outer
evaluated at inner
17.
deriv of inner
deriv of outer
evaluated at inner
R
√
tan2 ( x)
√
dx
x
deriv of inner
=
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.
´ 1
R ³ 2 ³ 1´
2
=
sec x − 1 x− 2 dx
³ 1´
R
R
1
1
= sec2 x 2 · x− 2 dx − x− 2 dx
³ 1´
1
R
1
= sec2 x 2 · x− 2 dx − x12 + C
R
√
tan2 ( x)
√
dx
x
2
(a)
1. U-sub appropriate?
1. Composite
³ 1 ´ Function? Yes!
sec2 x 2
1
Let u = x 2
2. Approx funct/deriv pair? Yes!
1
1
x− 2
x 2 → |{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. Z
Analyze in terms of u and du
³ 1´
R
1
1
− 12
2
sec2 x 2 x
| {zdx} − 2x 2 + C = sec u 2u du − 2x 2 + C
|
{z
} 2du
sec2 u
R
1
= 2 sec2 u du − 2x 2 + C
4. Integrate
1
= 2 tan (u) − 2x 2 + C
5
18.
R
5. Rewrite³in terms
of x
´
1
1
2
= 2 tan x − 2x 2 + C
tan (3x) sec2 (3x) dx
(a) The FIRST WAY:
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 }
| {z }
function
derivative
Let u = tan (3x)
2. Compute du
u = tan (3x)
= sec2 (3x) · 3
⇒ du
dx
⇒ du = 3 sec2 (3x) dx
⇒ 13 du = sec2 (3x) dx
3. Z
Analyze in terms of u and du
tan (3x)sec2 (3x) dx = u 13 du =
|
{z
}
| {z }
1
du
u
3
4. Integrate
R
2
1
u du = 13 u2 + C =
3
5. Rewrite
in terms of x
tan2 (3x)
=
+C
6
u2
6
1
3
R
u du
+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??? If 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. Z
Analyze in terms of u and du
R
sec (3x)sec (3x) tan (3x) dx = 13 u du
|
{z
}
| {z }
1
du
u
3
6
R
2
4. Integrate 13 u du = 13 u2 + C =
5. Re-express in terms of x.
2
= sec 6(3x) + C
19.
R
tan7 (x) sec2 (x) dx =
(a)
R
u2
6
+C
(tan x)7 sec2 (x) dx
1. U-sub appropriate?
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. Z
Analyze in terms of u and du
R
(tan (x))7 sec2 (x) dx = u7 du
| {z }
|
{z
}
du
u7
4. RIntegrate
8
u7 du = u8 + C
5. Re-express in terms of x
(tan(x))8
+C
8
20.
Z
|
3
sec (x) tan (x) dx =
(a)
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 }
{z
}
funct.
deriv.
Let u = sec (x)
2. Compute du
u = sec (x)
= sec (x) tan (x)
⇒ du
dx
⇒ du = sec (x) tan (x) dx
7
3. RAnalyze in terms of u and du R
(sec (x))2 sec (x) tan (x)dx = u2 du
| {z }|
{z
}
u2
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
8