Quiz 3B for MATH 105 SECTION 205
February 11, 2015
Given Name
Family Name
Student Number
1. (a) (0.5 points) Is it true that any continuous function has an antiderivative?(just put ‘Yes’ or ‘No’)
Yes
(a)
Answer.
Let f (x) be a continuous function on an
Z interval [a, b], for any x0 ∈ (a, b), by the Fundamental
x
f (t) dt is an antiderivative of f (x).
Theorem of Calculus, we know that the function
x0
(b) (0.5 points) Find an antiderivative of |x|.
x
Z
|t| dt =
(b)
0
1 2
2x ,
− 12 x2 ,
if x ≥ 0,
if x < 0.
Z
Answer.
0
tiderivative of |t|. Notice that
Z
x
|t| dt =
0
1
(c) (0.5 points) Compute lim
x→0 x
1 2
2x ,
− 12 x2 ,
Answer.
cos(t2 ) dt.
x
-1
x2
cos(t2 ) dt = 0 and
cos(t ) dt, then f (0) =
x
f 0 (x) =
0
Z
2
Let f (x) =
if x ≥ 0,
if x < 0.
x2
Z
(c)
Z
0
x2
Z
d
dx
!
cos(t2 ) dt
= cos(x4 ) · 2x − cos(x2 ) = 2x cos(x4 ) − cos(x2 ).
x
Approach I: Notice that
lim
x→0
1
x
x
|t| dt is an an-
By the Fundamental Theorem of Calculus, we know that the function
Z
x2
cos(t2 ) dt = lim
x→0
x
f (x)
f (x) − f (0)
= lim
= f 0 (0) = −1.
x→0
x
x−0
Approach II: By l’Hospital’s rule, we have
lim
x→0
1
x
Z
x2
cos(t2 ) dt = lim
x
x→0
f (x)
f 0 (0)
=
= f 0 (0) = −1.
x
1
Z r q
√
(d) (0.5 points) Evaluate
x x x dx.
(d)
8 15
x8 +C
15
Answer.
In fact, since
r q
r q
√
1
x x x =
x x · x2
r q
3
=
x x2
q
3
x · x4
=
q
7
=
x4
7
= x8
So we get
Z
Z r q
√
7
x x x dx = x 8 dx =
Z
(e) (0.5 points) Evaluate
10x · 32x dx. (Hint: ax = ex ln a ).
1
1
e(ln 10+2 ln 3)x + C or
90x + C
ln 10 + 2 ln 3
ln 90
(e)
Answer.
7
1
8 15
x 8 +1 = x 8 + C.
15
+1
7
8
Approach I: Notice that
10x · 32x = ex ln 10 · e2x ln 3 = e(ln 10+2 ln 3)x .
So we get
Z
10x · 32x dx =
=
Z
e(ln 10+2 ln 3)x dx
1
e(ln 10+2 ln 3)x + C.
ln 10 + 2 ln 3
Approach II: Notice that
10x · 32x = 10x · 9x = 90x = ex ln 90 .
So we get
Z
10x · 32x dx =
Z
(f) (0.5 points) Evaluate
Z
ex ln 90 dx =
cos(2x)
dx (Hint: cos(2x) = 2 cos2 (x) − 1 and sin2 (x) + cos2 (x) = 1).
cos(x) − sin(x)
sin(x) − cos(x) + C
(f)
Answer.
1
1 x ln 90
e
+C =
10x + C.
ln 90
ln 90
Notice that
cos(2x) = 2 cos2 (x) − 1 = 2 cos2 (x) − [cos2 (x) + sin2 (x)] = cos2 (x) − sin2 (x).
Then
cos(2x)
cos2 (x) − sin2 (x)
=
= cos(x) + sin(x).
cos(x) − sin(x)
cos(x) − sin(x)
So we get
Z
cos(2x)
dx =
cos(x) − sin(x)
Z
[cos(x) + sin(x)] dx = sin(x) − cos(x) + C.
Z
cos(2x)
dx.
· sin2 (x)
(g) (0.5 points) Evaluate
cos2 (x)
− cot(x) − tan(x) + C or −
(g)
Answer.
2
+C
sin(2x)
Approach I: Notice that
cos(2x) = 2 cos2 (x) − 1 = 2 cos2 (x) − [cos2 (x) + sin2 (x)] = cos2 (x) − sin2 (x).
Then
cos2 (x) − sin2 (x)
1
1
cos(2x)
=
=
−
= csc2 (x) − sec2 (x).
2
2
2
2
2
2
cos
(x)
cos (x) sin (x)
cos (x) sin (x)
sin (x)
So we get
Z
cos(2x)
dx =
2
cos (x) sin2 (x)
Z
[csc2 (x) − sec2 (x)] dx = − cot(x) − tan(x) + C.
Approach II: Since sin(2x) = 2 sin(x) cos(x), then
Z
Z
cos(2x)
cos(2x)
dx =
1
2 dx
2
2
cos (x) · sin (x)
2 sin(2x)
Z
cos(2x)
dx
= 4
sin2 (2x)
Z
1
= 2
du let u = sin(2x), then du = 2 cos(2x) dx
u2
2
= − +C
u
2
+ C Since u = sin(2x).
= −
sin(2x)
r
Z
(h) (0.5 points) Evaluate
1+x
+
1−x
r
1−x
1+x
!
dx (Hint: Simplify the integrand).
2 sin−1 (x) + C
(h)
Answer.
Notice that
r
1+x
+
1−x
r
1−x
1+x
=
=
=
=
√
√
1+x
1−x
√
+√
1−x
1+x
√
√
√
√
1+x· 1+x
1−x· 1−x
√
√
√
+√
1−x· 1+x
1+x· 1−x
1+x
1−x
√
+√
2
1−x
1 − x2
2
√
.
1 − x2
So we get
Z r
1+x
+
1−x
r
1−x
dx =
1+x
Z
√
2
dx = 2 sin−1 (x) + C.
2
1−x
Z
(i) (0.5 points) Evaluate
√
dx
√ (Hint: Let u = 6 x).
x+ 3x
1
1
1 1 1 1
6 x 2 − x 3 + x 6 − ln x 6 + 1 + C
(i)
3
2
√
√
Let u = 6 x, then x = u6 and dx = 6u5 du. So we get
Z
Z
dx
6u5
√
√
du
=
u3 + u2
x+ 3x
Z
u3
= 6
du
u+1
Z 2
u (u + 1) − u2
= 6
du
u+1
Z u2
2
du
u −
= 6
u+1
Z u(u + 1) − u
= 6
u2 −
du
u+1
Z u
2
= 6
u −u+
du
u+1
Z u+1−1
2
= 6
u −u+
du
u+1
Z 1
= 6
u2 − u + 1 −
du
u+1
1 3 1 2
= 6 u − u + u − ln |u + 1| + C
3
2
1
1
1 1 1 1
= 6 x 2 − x 3 + x 6 − ln x 6 + 1 + C
3
2
Z
(j) (0.5 points) Evaluate
tan−1 (x) dx.
Answer.
x tan−1 (x) −
(j)
Answer.
1
ln |x2 + 1| + C
2
By ‘ILATE’ rule, let
1
u = tan−1 (x) =⇒ du =
dx
1
+
x2
Z
dv = dx =⇒ v = dx = x.
Then we get
Z
−1
tan
Z
(x) dx =
u dv
Z
= uv − v du
Z
x
dx
+1
Z
1
dw
= x tan−1 (x) −
Let w = x2 + 1
2
w
1
= x tan−1 (x) − ln |w| + C
2
1
−1
= x tan (x) − ln |x2 + 1| + C Since w = x2 + 1.
2
−1
= x tan
(x) −
x2
1
Since u = x 6
Z 1
(k) (0.5 points) Evaluate
ln(ln x) +
dx.
ln x
x ln(ln x) + C
(k)
Answer.
Let
1
1
u = ln(ln x) =⇒ du =
· dx
Z ln x x
dv = dx =⇒ v = dx = x.
Then
Z
Z
Z 1
1
dx =
ln(ln x) dx +
ln(ln x) +
dx
ln x
ln x
Z
Z
1
=
u dv +
dx
ln x
Z
Z
1
= uv − v du +
dx
ln x
Z
Z
1
1
1
= x ln(ln x) − x ·
· dx +
dx
ln x x
ln x
Z
Z
1
1
= x ln(ln x) −
dx +
dx
ln x
ln x
= x ln(ln x) + C.
(l) (0.5 points) Find the area of the region bounded by the graph of ln x and x-axis between x = e−1 and
x = e.
2 − 2e−1
(l)
Answer.
Notice that the question is to find the are not the net area. So the area is:
Z e
Z 1
Z e
| ln x| dx =
| ln x| dx +
| ln x| dx
e−1
e−1
1
Z 1
Z e
ln x dx.
= −
ln x dx +
e−1
1
Notice that
Z
Z
ln x dx = x ln x −
x·
1
dx
x
Z
= x ln x −
1 dx
= x ln x − x + C.
So we get
Z
e
| ln x| dx =
e−1
− [x ln x − x]|1e−1 + [x ln x − x]|e1
= −[1 ln 1 − 1] + [e−1 ln e−1 − e−1 ] + [e ln e − e] − [1 ln 1 − 1]
= 1 − e−1 − e−1 + e − e + 1
= 2 − 2e−1
1
3
3
1
+
2
+
·
·
·
+
n
.
n4
(m) (0.5 points) Compute lim
n→∞
1
4
(m)
Answer.
Write
1
1 + 23 + · · · + n3 as a Riemann sum:
4
n
n
n
n 1 X 3
1 X k3
1X k 3
1
3
3
1 + 2 + ··· + n = 4
k =
=
.
n4
n
n
n3
n
n
k=1
k=1
k=1
So we know that
n
1
lim
1 + 2 3 + · · · + n3 =
4
n→∞ n
1X
lim
n→∞ n
k=1
Z 1
x3 dx
=
0
1 4 1
=
x
4 0
1
=
.
4
(n) (0.5 points) Compute lim n
n→∞
1
1
1
+
+ ··· + 2 .
n2 + 1 n2 + 2 2
2n
π
4
(n)
Answer.
Write n
n
3
k
n
1
1
1
+ 2
+ ··· + 2
2
2
n +1 n +2
2n
1
1
1
+
+ ··· + 2
n2 + 1 n2 + 2 2
2n
=n
n
X
k=1
as a Riemann sum:
n
n
k=1
k=1
lim
1X
1
2
n
1+ k
1
1 X n2
1X
1
=
=
2 .
n2 + k 2
n
n2 + k 2
n
1+ k
So we get
lim n
n→∞
1
1
1
+
+ ··· + 2
n2 + 1 n2 + 2 2
2n
n
=
n→∞
Z
=
=
k=1
n
1
1
dx
2
0 1+x
1
tan−1 (x)
0
= tan−1 (1) − tan−1 (0)
π
=
.
4
1
(o) (0.5 points) Compute lim
n→∞ n
(o)
π
2π
n−1
sin + sin
+ · · · + sin
π .
n
n
n
2
π
n
Answer.
Since sin(π) = 0, then
1
n
n−1
n
π
2π
n−1
1X
k
1X
k
sin + sin
+ · · · + sin
π =
sin
π =
sin
π .
n
n
n
n
n
n
n
k=1
k=1
Then
1
lim
n→∞ n
π
2π
n−1
sin + sin
+ · · · + sin
π
=
n
n
n
=
=
=
=
=
Z
(p) (0.5 points) Evaluate
n
1X
k
lim
sin
π
n→∞ n
n
k=1
Z 1
sin(xπ) dx
0
1
1
− cos(πx)
π
0
1
1
− cos(π) + cos(0)
π
π
1
1
+
π π
2
.
π
sin3 (x) cos2 (x) dx.
1
1
sin4 (x) cos(x) sin2 (x) cos(x) 2 cos(x)
−
−
+C
(p) − cos3 (x) + cos5 (x) + C or
3
5
5
15
15
Answer.
Approach I: In fact, we have
Z
Z
3
2
sin (x) cos (x) dx =
sin2 (x) cos2 (x) · sin(x) dx
Z
=
[1 − cos2 (x)] cos2 (x) · sin(x) dx
Z
= − [cos2 (x) − cos4 (x)] d cos(x)
1
1
= − cos3 (x) + cos5 (x) + C.
3
5
Approach II: Since sin2 (x) + cos2 (x) = 1, then
Z
Z
3
2
sin (x) cos (x) dx =
sin3 (x)[1 − sin2 (x)] dx
Z
Z
3
=
sin (x) dx − sin5 (x) dx.
Recall the reduction formula:
Z
Z
sinn−1 (x) cos(x) n − 1
n
sin (x) dx = −
+
sinn−2 (x) dx.
n
n
Then
Z
sin4 (x) cos(x) 4
sin (x) dx = −
+
5
5
5
Z
sin3 (x) dx.
So we get
Z
sin3 (x) cos2 (x) dx =
=
=
Z
(q) (0.5 points) Evaluate
Z
sin4 (x) cos(x) 1
+
sin3 (x) dx
5
5
Z
sin2 (x) cos(x) 2
sin4 (x) cos(x) 1
−
+
+
sin(x) dx + C
5
5
3
3
By the reduction formula again
4
sin (x) cos(x) sin2 (x) cos(x) 2 cos(x)
−
−
+ C.
5
15
15
tan(x) sec2 (x) dx.
1
1
tan2 (x) + C or sec2 (x) + C
2
2
(q)
Answer.
Approach I: Notice that d tan(x) = sec2 (x) dx, then
Z
Z
2
tan(x) sec (x) dx =
tan(x) d tan(x)
=
1
tan2 (x) + C.
2
Approach II: Notice that d sec(x) = sec(x) tan(x) dx, then
Z
Z
2
tan(x) sec (x) dx =
sec(x) d sec(x)
=
1
sec2 (x) + C.
2
Approach III: Notice that sec2 (x) = 1 + tan2 (x), then
Z
Z
2
tan(x) sec (x) dx =
tan(x)[1 + tan2 (x)] dx
Z
Z
=
tan(x) dx + tan3 (x) dx.
Recall the reduction formula:
Z
tann−1 (x)
−
n−1
Z
tan2 (x)
tan (x) dx =
−
2
Z
tann (x) dx =
Then
Z
3
tann−2 (x) dx.
tan(x) dx.
So we get
Z
Z
(r) (0.5 points) Evaluate
tan(x) sec2 (x) dx =
tan2 (x)
+ C.
2
tan2 (x) sec3 (x) dx.
(r)
sec3 (x) tan(x) sec(x) tan(x) 1
−
− ln | sec(x) + tan(x)| + C
4
8
8
Notice that 1 + tan2 (x) = sec2 (x), then
Z
Z
Z
Z
2
3
2
3
5
tan (x) sec (x) dx = [sec (x) − 1] sec (x) dx = sec (x) dx − sec3 (x) dx.
Answer.
Z
Recall the reduction formula for
Z
secn (x) dx:
secn−2 (x) tan(x) n − 2
sec (x) dx =
+
n−1
n−1
n
Z
secn−2 (x) dx.
Then we have
Z
5
sec (x) dx =
Z
tan2 (x) sec3 (x) dx =
=
=
Recall
sec3 (x) tan(x)
4
3
sec (x) tan(x)
4
3
sec (x) tan(x)
4
3
sec (x) tan(x)
4
+
+
−
−
Z
3
sec3 (x) dx
4
Z
Z
3
3
sec (x) dx − sec3 (x) dx
4
Z
1
sec3 (x) dx
4
Z
1 sec(x) tan(x) 1
+
sec(x) dx .
4
2
2
Z
sec(x) dx = ln | sec(x) + tan(x)| + C.
Then
Z
tan2 (x) sec3 (x) dx =
Z
2. (1 point) For any m, n, let I(m, n) =
sec3 (x) tan(x) sec(x) tan(x) 1
−
− ln | sec(x) + tan(x)| + C.
4
8
8
cosm (x) sinn (x) dx, then if m + n 6= 0, show that
cosm−1 (x) sinn+1 (x) m − 1
+
I(m − 2, n)
m+n
m+n
cosm+1 (x) sinn−1 (x)
n−1
= −
+
I(m, n − 2).
m+n
m+n
I(m, n) =
Answer. Approach I: By the product rule, we have
d cosm−1 (x) sinn+1 (x)
m−1
n+1
= −
cosm−2 (x) · sin(x) · sinn+1 (x) +
cosm−1 (x) sinn (x) · cos(x)
dx
m+n
m+n
m+n
m−1
n+1
= −
cosm−2 (x) · sinn (x) · sin2 (x) +
cosm (x) sinn (x)
m+n
m+n
m−1
n+1
= −
cosm−2 (x) · sinn (x) · [1 − cos2 (x)] +
cosm (x) sinn (x)
m+n
m+n
Since sin2 (x) + cos2 (x) = 1
m−1
m−1
n+1
= −
cosm−2 (x) sinn (x) +
cosm (x) sinn (x) +
cosm (x) sinn (x)
m+n
m+n
m+n
m−1
= −
cosm−2 (x) sinn (x) + cosm (x) sinn (x)
m+n
That is,
cosm−1 (x) sinn+1 (x)
m+n
Z
Z
m−1
m−2
n
= −
cos
(x) sin (x) dx + cosm (x) sinn (x) dx
m+n
m−1
= −
I(m − 2, n) + I(m, n).
m+n
So we get
I(m, n) =
By the product rule, we have
cosm+1 (x) sinn−1 (x)
d
−
=
dx
m+n
=
=
=
=
cosm−1 (x) sinn+1 (x) m − 1
+
I(m − 2, n).
m+n
m+n
m+1
n−1
cosm (x) · sin(x) · sinn−1 (x) −
cosm+1 (x) · sinn−2 (x) · cos(x)
m+n
m+n
m+1
n−1
cosm (x) sinn (x) −
cosm sinn−2 (x) · cos2 (x)
m+n
m+n
n−1
m+1
cosm (x) sinn (x) −
cosm sinn−2 (x) · [1 − sin2 (x)]
m+n
m+n
Since sin2 (x) + cos2 (x) = 1
m+1
n−1
n−1
cosm (x) sinn (x) −
cosm sinn−2 (x) +
cosm sinn (x)
m+n
m+n
m+n
n−1
cosm (x) sinn (x) −
cosm (x) sinn−2 (x).
m+n
Then we get
cosm+1 (x) sinn−1 (x)
−
m+n
n−1
=
cos (x) sin (x) dx −
m+n
n−1
= I(m, n) −
I(m, n − 2).
m+n
Z
m
n
Z
cosm (x) sinn−2 (x) dx
So we get
I(m, n) = −
cosm+1 (x) sinn−1 (x)
n−1
+
I(m, n − 2).
m+n
m+n
Approach II: Use the integration by parts, we have
Z
I(m, n) =
cosm−1 (x) sinn (x) · cos(x) dx
Z
=
cosm−1 (x) sinn (x) d sin(x)
Z
m−1
n+1
= cos
(x) sin
(x) − sin(x) −(m − 1) cosm−2 (x) sin(x) · sinn (x)
+ cosm−1 (x) · n sinn−1 (x) cos(x) dx By the product rule
Z
Z
m−1
n+1
m−2
n+2
= cos
(x) sin
(x) + (m − 1) cos
(x) sin
(x) dx − n cosm (x) sinn (x) dx
Z
m−1
n+1
= cos
(x) sin
(x) + (m − 1) cosm−2 (x) sinn (x)[1 − cos2 (x)] dx
Z
−n cosm (x) sinn (x) dx Since sin2 (x) + cos2 (x) = 1
Z
Z
m−1
n+1
m−2
n
= cos
(x) sin
(x) + (m − 1) cos
(x) sin (x) dx − (m − 1) cosm (x) sinn (x) dx
Z
−n cosm (x) sinn (x) dx
= cosm−1 (x) sinn+1 (x) + (m − 1)I(m − 2, n) − (m + n − 1)I(m, n).
Since 1 + m + n − 1 = m + n, then
(m + n)I(m, n) = cosm−1 (x) sinn+1 (x) + (m − 1)I(m − 2, n).
That is, we get
I(m, n) =
cosm−1 (x) sinn+1 (x) m − 1
+
I(m − 2, n).
m+n
m+n
Use the integration by parts, we have
Z
I(m, n) =
cosm (x) sinn−1 (x) · sin(x) dx
Z
= − sinn−1 (x) cosm (x) d cos(x)
Z
= − sinn−1 (x) cosm+1 (x) + cos(x) (n − 1) sinn−2 (x) cos(x) · cosm (x)
− sinn−1 (x) · m cosm−1 (x) sin(x) dx By the product rule
Z
Z
n−1
m+1
m+2
n−2
= − sin
(x) cos
(x) + (n − 1) cos
(x) sin
(x) dx − m sinn (x) cosm (x) dx
Z
n−1
m+1
= − sin
(x) cos
(x) + (n − 1) cosm (x) sinn−2 (x)[1 − sin2 (x)] dx
Z
−m sinn (x) cosm (x) dx Since sin2 (x) + cos2 (x) = 1
Z
Z
n−1
m+1
m
n−2
= − sin
(x) cos
(x) + (n − 1) cos (x) sin
(x) dx − (n − 1) cosm (x) sinn (x) dx
Z
−m sinn (x) cosm (x) dx
= − sinn−1 (x) cosm+1 (x) + (n − 1)I(m, n − 2) − (m + n − 1)I(m, n).
Since 1 + m + n − 1 = m + n, then
(m + n)I(m, n) = − sinn−1 (x) cosm+1 (x) + (n − 1)I(m, n − 2).
That is, we get
I(m, n) = −
cosm+1 (x) sinn−1 (x)
n−1
+
I(m, n − 2).
m+n
m+n
Approach III: Use the integration by parts, we have
Z
I(m, n) =
cosm (x) sinn (x) dx
Z
=
cosm−1 (x) sinn (x) · cos(x) dx
Z
=
cosm−1 (x) sinn (x) d sin(x)
Z
1
cosm−1 (x) d sinn+1 (x)
=
n+1
Z
1
m−1
m−1
n+1
=
cos
(x) sin
(x) +
cosm−2 (x) · sin(x) · sinn+1 (x) dx
n+1
n+1
Z
1
m−1
m−1
n+1
=
cos
(x) sin
(x) +
cosm−2 (x) sinn (x) · sin2 (x) dx
n+1
n+1
Z
1
m−1
m−1
n+1
=
cos
(x) sin
(x) +
cosm−2 (x) sinn (x) · [1 − cos2 (x)] dx
n+1
n+1
Since sin2 (x) + cos2 (x) = 1
Z
Z
1
m−1
m−1
m−1
n+1
m−2
n
=
cos
(x) sin
(x) +
cos
(x) sin (x) dx −
cosm (x) sinn (x) dx
n+1
n+1
n+1
1
m−1
m−1
=
cosm−1 (x) sinn+1 (x) +
I(m − 2, n) −
I(m, n).
n+1
n+1
n+1
Since 1 +
m−1
m+n
=
, then
n+1
n+1
m+n
1
m−1
I(m, n) =
cosm−1 (x) sinn+1 (x) +
I(m − 2, n).
n+1
n+1
n+1
That is,
cosm−1 (x) sinn+1 (x) m − 1
+
I(m − 2, n).
m+n
m+n
I(m, n) =
Use the integration by parts, we have
Z
I(m, n) =
cosm (x) sinn−1 (x) · sin(x) dx
Z
= − sinn−1 (x) cosm (x) d cos(x)
Z
1
= −
sinn−1 (x) d cosm+1 (x)
m+1
Z
n−1
cosm+1 (x) sinn−1 (x)
+
= −
cosm+1 (x) sinn−2 (x) cos(x) dx
m+1
m+1
Z
n−1
cosm+1 (x) sinn−1 (x)
+
cosm (x) sinn−2 (x) cos2 (x) dx
= −
m+1
m+1
Z
cosm+1 (x) sinn−1 (x)
n−1
= −
+
cosm (x) sinn−2 (x)[1 − sin2 (x)] dx
m+1
m+1
Since sin2 (x) + cos2 (x) = 1
Z
Z
n−1
n−1
cosm+1 (x) sinn−1 (x)
+
cosm (x) sinn−2 (x) dx −
cosm (x) sinn (x) dx
= −
m+1
m+1
m+1
cosm+1 (x) sinn−1 (x)
n−1
n−1
= −
+
I(m, n − 2) −
I(m, n).
m+1
m+1
m+1
Since 1 +
n−1
m+n
=
, then
m+1
m+1
m+n
cosm+1 (x) sinn−1 (x)
n−1
I(m, n) = −
+
I(m, n − 2).
m+1
m+1
m+1
That is,
I(m, n) = −
Z
3. (1 point) Show that
sinn (x) dx = −
Z
Answer.
Let J(n) =
cosm+1 (x) sinn−1 (x)
n−1
+
I(m, n − 2).
m+n
m+n
sinn−1 (x) cos(x) n − 1
+
n
n
Z
sinn−2 (x) dx.
sinn (x) dx.
Approach I: Notice that J(n) = I(0, n), where I(m, n) is defined in Problem 2. By the result of Problem
2, we have
J(n) = I(0, n) = −
sinn−1 (x) cos(x) n − 1
cos(x) sinn−1 (x) n − 1
+
I(0, n − 2) = −
+
J(n − 2)
n
n
n
n
Approach II: Notice that
d
sinn−1 (x) cos(x)
(n − 1) sinn−2 (x) cos2 (x) − sinn−1 (x) sin(x)
−
= −
By the product rule
dx
n
n
n−1
sinn (x)
= −
sinn−2 (x)[1 − sin2 (x)] +
Since sin2 (x) + cos2 (x) = 1
n
n
n−1
sinn (x)
n−1
sinn−2 (x) +
sinn (x) +
= −
n
n
n
n−1
= −
sinn−2 (x) + sinn (x).
n
Then we get
n−1
sinn−1 (x) cos(x)
=−
−
n
n
Z
n−2
sin
Z
(x) dx +
sinn (x) dx.
That is,
Z
sinn−1 (x) cos(x) n − 1
sin (x) dx = −
+
n
n
n
Z
sinn−2 (x) dx.
Approach III: Use the integration by parts, we have
Z
J(n) =
sinn−1 (x) sin(x) dx
Z
= − sinn−1 (x) d cos(x)
Z
n−1
= − sin
(x) cos(x) + cos(x) · (n − 1) sinn−2 (x) cos(x) dx
Z
n−1
= − sin
(x) cos(x) + (n − 1) sinn−2 (x) cos2 (x) dx
Z
= − sinn−1 (x) cos(x) + (n − 1) sinn−2 (x)[1 − sin2 (x)] dx Since sin2 (x) + cos2 (x) = 1
Z
Z
= − sinn−1 (x) cos(x) + (n − 1) sinn−2 (x) dx − (n − 1) sinn (x) dx
= − sinn−1 (x) cos(x) + (n − 1)J(n − 2) − (n − 1)J(n).
So we get
nJ(n) = − sinn−1 (x) cos(x) + (n − 1)J(n − 2).
That is,
J(n) = −
sinn−1 (x) cos(x) n − 1
+
J(n − 2).
n
n
Approach IV: Notice that
Z
J(n) =
sinn−2 (x) sin2 (x) dx
Z
=
sinn−2 (x)[1 − cos2 (x)] dx Since sin2 (x) + cos2 (x) = 1
Z
Z
n−2
=
sin
(x) dx − sinn−2 cos2 (x) dx
Z
= J(n − 2) − sinn−2 cos(x) d sin(x)
Z
n−2
= J(n − 2) − sin
cos(x) · sin(x) + sin(x) · (n − 2) sinn−3 (x) cos(x) · cos(x) − sinn−2 (x) · sin(x) dx
Use the integration by parts
n−1
= J(n − 2) − sin
n−1
= J(n − 2) − sin
Z
(x) cos(x) + (n − 2)
n−2
sin
Z
(x) cos(x) + (n − 2)
2
(x) cos (x) dx −
Z
sinn (x) dx
sinn−2 (x)[1 − sin2 (x)] dx − J(n)
Since sin2 (x) + cos2 (x) = 1
= J(n − 2) − sinn−1 (x) cos(x) + (n − 2)J(n − 2) − (n − 2)J(n) − J(n)
= (n − 1)J(n − 2) − sinn−1 (x) cos(x) − (n − 1)J(n).
Since 1 + n − 1 = n, then
nJ(n) = − sinn−1 (x) cos(x) + (n − 1)J(n − 2).
That is, we get
J(n) = −
sinn−1 (x) cos(x) n − 1
+
J(n − 2).
n
n
Z
4. (1 point) Use the integration by parts to get a reduction formula of the integral
Z
Answer.
Let In =
ex sinn (x) dx.
ex sinn (x) dx, by ‘ILATE’ rule, we have
Z
sinn (x) dex
Z
= ex sinn (x) − nex sinn−1 (x) cos(x) dx
Z
x
n
= e sin (x) − n sinn−1 (x) cos(x) dex
Z
x
n
x
n−1
x
n−2
2
n−1
= e sin (x) − n e sin
(x) cos(x) − e (n − 1) sin
cos (x) − sin
(x) sin(x) dx
Z
Z
= ex sinn (x) − nex sinn−1 (x) cos(x) + n(n − 1) ex sinn−2 cos2 (x) dx − n ex sinn (x) dx
Z
x
n
x
n−1
= e sin (x) − ne sin
(x) cos(x) + n(n − 1) ex sinn−2 [1 − sin2 (x)] dx − nIn
In =
Since sin2 (x) + cos2 (x) = 1
= ex sinn (x) − nex sinn−1 (x) cos(x) + n(n − 1)
Z
ex sinn−2 dx − n(n − 1)
Z
ex sinn (x) dx − nIn
= ex sinn (x) − nex sinn−1 (x) cos(x) + n(n − 1)In−2 − n(n − 1)In − nIn
= ex sinn (x) − nex sinn−1 (x) cos(x) + n(n − 1)In−2 − n2 In .
So we get
In =
n2
1 x n
e sin (x) − nex sinn−1 (x) cos(x) + n(n − 1)In−2 .
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