Math 16B: Homework 3
Due: July 16
1. Find the equation of the tangent line at the given point for each of the following curves:
(a) y = tan(x) sec(x) at x = π/4:
√
At x = π/4, we get y = tan(π/4) sec(π/4) = 2. By the product rule, we have
dy
= (tan(x))(sec(x) tan(x)) + (sec2 (x))(sec(x)) = sec(x) tan2 (x) + sec3 (x).
dx
2
At x = π/4, the
of √
the tangent
√ line√is therefore
√ m = sec(π/4) tan (π/4) +
√ slope
2
3
3
sec (π/4) = ( 2)(1) + ( 2) = 2 + 2 2 = 3 2. Thus, the equation of the
tangent line is
√
√
y − 2 = (3 2)(x − π/4)
(
)
√
√ 3π
3 2x − y =
2
−1
4
(b) y = cosec(x) + sin2 (x) at x = π/6:
( )2
At x = π/6, we get y = cosec(π/6) + sin2 (π/6) = 2 + 21 = 49 . Furthermore, we
have
dy
= −cosec(x) cot(x) + 2 sin(x) cos(x)
dx
√
( ) ( √3 )
At x = π/6, the slope of the tangent line is therefore m = −(2)( 3)+2 21
=
2
√
−3 3
.
2
Thus, the equation of the tangent line is
( √ )
9
−3 3
y−
=
(x − π/6)
4
2
√
√
9+π 3
3 3
x+y =
2
4
2. (a) Recall that sin(π/2 − x) = cos(x). Thus, cos(π/2 − x) = sin(π/2 − (π/2 − x)) =
sin(x) and so
tan(π/2 − x) =
cos(x)
sin(π/2 − x)
=
= cot(x).
cos(π/2 − x)
sin(x)
1
(b) Recall that cos(x) = sin(π/2 − x) and sin(A + B) = sin(A) cos(B) + cos(A) sin(B)
so
cos(x + y) =
=
=
=
sin(π/2 − (x + y))
sin((π/2 − x) + (−y))
sin(π/2 − x) cos(−y) + cos(π/2 − x) sin(−y)
cos(x) cos(y) − sin(x) sin(y).
(c) Use sin(A + B) = sin(A) cos(B) + cos(A) sin(B) with A = B = x to get
sin(2x) = sin(x + x)
= sin(x) cos(x) + cos(x) sin(x)
= 2 sin(x) cos(x).
3. (a)
∫
(u2 + u + 1)eu du:
We integrate this by parts. Let f (u) = u2 + u + 1 and g(u) = eu . Then, f ′ (u) =
2u + 1 and G(u) = eu . Then, integrating by parts gives
∫
∫
2
u
2
u
(u + u + 1)e du = (u + u + 1)e − (2u + 1)eu du
(1)
We apply integration by parts again. Let f (u) = 2u + 1 and g(u) = eu . Then,
f ′ (u) = 2 and G(u) = eu and so
∫
∫
u
u
(2u + 1)e du = (2u + 1)e − (2)eu du
= (2u + 1)eu − 2eu
Plug this into (1) to get
∫
(u2 + u + 1)eu du = (u2 + u + 1)eu − (2u + 1)eu + 2eu + C.
(b)
∫π
cos(x)
0 1+sin(x)
dx :
We use the substitution u = 1 + sin(x). Then, du
= cos(x) ⇒ du = cos(x)dx.
dx
Also, x = 0 corresponds to u = 1 + sin(0) = 1 and x = π gives u = 1 + sin(π) = 1.
Thus,
∫ u=1
∫ π
1
cos(x)
dx =
du
u=1 u
0 1 + sin(x)
Since the upper and lower limits are the same, we can infer even without integrating that the result is 0.
2
(c)
∫
x sec2 (x) dx:
We use integration by parts. Let f (x) = x and g(x) = sec2 (x). Then, f ′ (x) = 1
and G(x) = tan(x) so
∫
∫
2
x sec (x) dx = x tan(x) − tan(x) dx
sin(x)
Recall that tan(x) = cos(x)
. To integrate tan(x), we use the substitution u =
du
cos(x) ⇒ dx = − sin(x) ⇒ −du = sin(x) dx. Thus,
∫
∫
sin(x)
−1
dx =
du = − ln |u|
cos(x)
u
∫
so
x sec2 (x) dx = x tan(x) + ln | cos(x)| + C
(d)
∫ 16
1
x3/2 ln(x) dx:
We use integration by parts. Let f (x) = ln(x) and g(x) = x3/2 ⇒ f ′ (x) =
G(x) = 52 x5/2 . Thus,
∫
[
16
3/2
x
ln(x) dx =
1
=
=
=
=
=
(e)
(
and
)
( )(
2 5/2
1
(ln(x))
−
x
dx
x
5
1
1
∫
] 2 16 3/2
2[
5/2
5/2
x dx
(16) ln(16) − (1) ln(1) −
5
5 1
[
]16
2
2 2 5/2
[1024 ln(16)] −
x
5
5 5
1
]
2048 ln(16)
4 [
−
(16)5/2 − (1)5/2
5
25
(512)(4) ln(16)
4
−
[1024 − 1]
5
25
4
(2560 ln(16) − 1023)
25
2 5/2
x
5
)]16
1
x
∫
16
∫
x(ln(x))2 dx:
We integrate
Let f (x) = (ln(x))2 and g(x) = x so that f ′ (x) =
( 1 ) this by parts.
x2
2 ln(x) x and G(x) = 2 . Hence,
( )( 2)
∫
∫
x2
1
x
2
2
x(ln(x)) dx =
(ln(x)) − 2 ln(x)
dx
2
x
2
∫
(x ln(x))2
=
− x ln(x) dx
(2)
2
To evaluate this integral, we again integrate by parts. Use f (x) = ln(x) and
3
g(x) = x ⇒ f ′ (x) =
1
x
∫
and G(x) =
x2
2
so that
∫ 2( )
x2
x
1
x ln(x) dx =
ln(x) −
dx
2
2 x
∫
1
x2
=
ln(x) −
xdx
2
2
x2
x2
ln(x) −
=
2
4
Plug into (2) to get
∫
(x ln(x))2 x2
x2
x(ln(x))2 dx =
−
ln(x) +
+C
2
2
4
(f)
∫ 25
√
e
t
dt:
√
1
1
We use the substitution u = t ⇒ du
= 2√
= 2u
⇒ 2u du = dt. Furthermore,
dt
t
t = 9 gives u = 3 and t = 25 gives u = 5 so
∫ 25 √
∫ 5
t
e dt =
eu (2u) du
9
9
3
To evaluate this, we use integration by parts. Let f (u) = u and g(u) = eu ⇒
f ′ (u) = 1 and G(u) = eu . Thus,
∫ 5
∫ 5
u
u 5
2
ue du = 2 [ue ]3 − 2
eu du
3
[ 5
]3
3
= 2 5e − 3e − 2[eu ]53
= 2(5e5 − 3e3 ) − 2(e5 − e3 )
= 2(4e5 − 2e3 )
= 4e3 (2e2 − 1)
(g)
∫
cos(π/x)
x2
dx:
Use the substitution u =
∫
π
x
⇒
du
dx
=
−π
x2
cos(π/x)
dx =
x2
=
=
=
(h)
∫e
√
e
ln(x + x2 ) dx:
4
⇒
−1
π
du = x12 dx. Thus,
( )
∫
−1
cos(u)
du
π
∫
−1
cos(u) du
π
−1
sin(u) + C
π
−1
sin(π/x) + C
π
Note that we can write ln(x + x2 ) = ln(x(1 + x)) = ln(x) + ln(1 + x). Thus, we
simply need to integrate the two terms independently and add the results.
For the first term, we use the result derived in class:
∫ e
ln(x) dx = [x ln(x) − x]e√e
√
e
[√
√
√ ]
= [e ln(e) − e] − e ln( e) − e
[√
]
e √
− e
= [e − e] −
2
√
e
=
2
To integrate ln(x + 1), we use integration by parts: take f (x) = ln(x + 1) and
1
g(x) = 1 ⇒ f ′ (x) = x+1
and G(x) = x + 1 (recall that we need G(x) to be an
anti-derivative of g(x); hence G(x) should be of the form x + C where the choice
of C is up to us. Usually C = 0 is used but here C = 1 is preferable). Then,
)
∫ e(
∫ e
1
e√
ln(x + 1) dx = [(x + 1) ln(x + 1)] e − √
(x + 1) dx
√
x+1
e
e
∫ e
e√
= [(x + 1) ln(x + 1)] e − √ 1 dx
= [(x + 1) ln(x +
1)]e√
e
= [(x + 1) ln(x + 1) −
−
e
e√
[x]
x]e√
e
e
[√
√
√ ]
= [(e + 1) ln(e + 1) − e] − ( e + 1) ln( e + 1) − e
Add the two results to get
√
∫ e
[√
√
√ ]
e
2
ln(x
+
x
)
dx
=
+
[(e
+
1)
ln(e
+
1)
−
e]
−
(
e
+
1)
ln(
e
+
1)
−
e
√
2
e
√
√
√
3 e
=
− e + (e + 1) ln(e + 1) − ( e + 1) ln( e + 1)
2
4. Determine whether each integral is convergent or divergent. Evaluate those that are
convergent.
∫∞
(a) 1 ln(x)
dx:
x
We use the substitution u = ln(x) ⇒ du
= x1 ⇒ du = x1 dx. Then,
dx
∫ ∞
∫ b
ln(x)
ln(x)
dx = lim
dx
b→∞ 1
x
x
1
∫ u=ln(b)
= lim
u du
b→∞
[
u=ln(1)
]
2 ln(b)
u
b→∞
2 0
(ln(b))2
= lim
b→∞
2
=
5
lim
Since this limit is ∞, the integral is divergent.
∫∞
(b) 1 ln(x)
dx:
x2
We apply integration by parts. Let f (x) = ln(x) and g(x) = x12 ⇒ f ′ (x) =
G(x) = −1
. Then,
x
∫ ∞
∫ b
ln(x)
ln(x)
dx
=
lim
dx
b→∞ 1
x2
x2
1
([
)
]b ∫ b ( ) ( )
1
−1
−1
= lim
ln(x) −
dx
b→∞
x
x
x
1
1
)
([
]b ∫ b
1
− ln(x)
dx
= lim
+
2
b→∞
x
1 x
1
([
]b [ ]b )
− ln(x)
−1
= lim
+
b→∞
x
x 1
1
([
] [
])
− ln(b) ln(1)
−1 1
= lim
+
+
+
b→∞
b
1
b
1
(
)
− ln(b) 1
= lim
− +1
b→∞
b
b
1
x
and
= 0 and limb→∞ 1b = 0, it follows that the above limit is 1.
Since limb→∞ ln(b)
b
Thus, the integral is convergent and has value 1.
∫∞
(c) 4 xx+1
2 +2x dx:
= 2x+2 = 2(x+1) ⇒ 21 du = (x+1) dx.
We use the substitution u = x2 +2x ⇒ du
dx
Then,
∫ ∞
∫ b
x+1
x+1
dx = lim
dx
2
2
b→∞
x + 2x
4
4 x + 2x
∫ u=b2 +2b
1
= lim
du
(limits changed)
b→∞ u=24
u
=
=
2
lim [ln(u)]b24+2b
b→∞
lim ln(b2 + 2b) − ln(24)
b→∞
Since limb→∞ ln(b2 + 2b) = ∞, this integral is divergent.
∫0
(d) −∞ x2 ex dx:
We use integration by parts. Let f (x) = x2 and g(x) = ex ⇒ f ′ (x) = 2x and
G(x) = ex . Thus,
∫ 0
∫ 0
2 x
x2 ex dx
x e dx = lim
a→−∞
a
−∞
(
)
∫ 0
[ 2 x ]0
x
= lim
x e a−
(2x)e dx
(3)
a→−∞
6
a
We use integration by parts again for the integral left over: take f (x) = x and
g(x) = ex ⇒ f ′ (x) = 1 and G(x) = ex . Then, (3) becomes
(
)
∫ 0
∫ 0
[ 2 x ]0
2 x
x 0
x
x e a − 2 [xe ]a + 2
x e dx = lim
e dx
a→−∞
−∞
a
([
)
]
2 x 0
x 0
x 0
= lim
x e a − 2 [xe ]a + 2 [e ]a
a→−∞
[
]0
= lim (x2 − 2x + 2)ex a
a→−∞
=
lim 2 − (a2 − 2a + 2)ea
a→−∞
Since lima→−∞ (a2 − 2a + 2)ea = 0, the integral is convergent with value 2.
7