HW 1
(1) Using the concept of Riemann sum to evaluate
Z 1
n
1X
kπ
sin πxdx.
(a) lim
sin
=
n→∞ n
n
0
k=1
Z 1
n
X
n
1
(b) lim
=
dx.
2
2
2
n→∞
n +k
0 1+x
k=1
Z 1
n r
√
1X
k
(c) lim
1+ =
1 + xdx.
n→∞ n
n
0
k=1
Z 1
e1/n + e2/n + · · · + en/n
(d) lim
=
ex dx.
n→∞
n
0
r
Z 1
n
X
k
n
(e) lim
ln 1 + =
ln(1 + x)dx.
n→∞
n
0
k=1
Z 1
1
1
1
dx
1
√
√ + ··· + √
√
√ .
(f) lim √
+√
=
n→∞
n
n+1
n+ n
x
n+ 2
0 1+
For
choice of function f ∈ [0, 1], sums in the above limits can be written as
Pn a suitable
1
k
f
f with respect to partition Pn = {k/n : 0 ≤ k ≤ n}.
n=1 n
n which is a Riemann sum of
R1
The limit is thus given by the integral 0 f (x)dx.
(2) Evaluate
Z
(a)
xn ln xdx.
Solution: This can be solved by integration by parts.
Z
Z
(b)
Z n+1
xn+1
xn+1 ln x
x
−
d ln x
=
n+1
n+1
n+1
Z
Z
1
xn+1 ln x
1
xn+1 ln x
dx
−
=
−
=
xn+1 ·
xn dx
n+1
n+1
x
n+1
n+1
xn+1 ln x
xn+1
=
−
+ C.
n+1
(n + 1)2
xn ln xdx =
Z
ln xd
dx
.
sin x + 4 cos2 x
2
Solution:
1
2
Let t = tan x. Then cos x = √
1
and sin x =
1 + t2
√ t
1+t2
and dx =
dt
1+t2 .
The integral
becomes
Z
dx
=
2
sin x + 4 cos2 x
Z
t
1+t2
1
+
4
1+t2
·
dt
1 + t2
Z
dt
t+4
= ln |t + 4| + C
=
= ln | tan x + 4| + C.
Z
(c)
tan−1
1
dx.
x
Solution: We use integration by parts:
Z
Z
1
dx
1
1
1
1
·− 2
tan−1 dx = x tan−1 − x · d tan−1 = x tan−1 − x ·
x
x
x
x
1 + (1/x)2
x
Z
xdx
1
= x tan−1 +
x
1 + x2
1 1
= x tan−1 + ln(1 + x2 ) + C.
x 2
Z √
x−1
(d)
dx.
x+3
Z
Solution:
√
Let u = x − 1. Then x = 1 + u2 and dx = 2udu. Thus
Z √
Z
Z
u
u2
x−1
dx =
·
2udu
=
2
du
2
2
x+3
4+u
u +4
Z
4
=2
1− 2
du
u +4
Z
1
du
= 2u − 4
(u/2)2 + 1 2
u
= 2u − 4 tan−1 + C
2
√
√
x−1
= 2 x − 1 − 4 tan−1
+ C.
2
Z
x
(e)
dx.
(x + 1)(x + 2)(x + 3)
Solution:
Use
fraction expansion.
Z Partial
8x2 + 4x − 11
(f)
dx.
(x + 3)(x − 1)2
3
Solution:
Use partial fraction expansion.
Z
(g)
x2 tan−1 xdx.
Solution:
We use integration by parts:
Z
Z
1
tan−1 xdx3
x2 tan−1 xdx =
3
Z
x3
1
1
dx
=
tan−1 x −
x3 ·
3
3
1 + x2
Z
x3
x3
1
=
dx.
tan−1 x −
3
3
1 + x2
Z
x3
The use partial fraction expansion to evaluate
dx.
1 + x2
(3) Evaluate
Z π
√
(a)
1 − sin xdx.
0
Write sin x = 2 sin x2 cos x2 . Then
x
x 2
1 − sin x = cos − sin
.
2
2
Hence the integral becomes
Z π
x
x
| cos − sin |dx.
2
2
0
Z π/2
cos θ
p
(b)
dθ.
0
2 − sin2 θ
Solution:
Let u = sin θ. The integral becomes
Z 1
Z 1
du
du
1
√
q
√ .
=
√
2
2
−
u
0
0
1 − (u/ 2)2 2
Z π4
√
π
Let u = 2 sin t. Then the integral becomes
1dt = .
4
0
Z 1 2
x +1
(c)
dx.
4
0 x +1
Solution:
Write
x4 + 1 = x4 + 2x2 + 1 − 2x2
√
= (x2 + 1) − ( 2x)2
√
√
= (x2 − 2x + 1)(x2 + 2x + 1).
4
Then
partial fraction expansion.
√
Z 1 use
4
x
√ dx.
(d)
x
0 1+
Solution:
Z π/2
(e)
| cos 2x − sin x|dx.
0
Solution:
Find x in [0, π/2] such that cos 2x > sin x and cos 2x < sin x.
Z
(f)
2
10
x+1
√
dx.
x x−1
Solution:
Z 1
2
x3 ex
(g)
dx.
2
2
0 (x + 1)
Solution:
Z 3
(h)
|x2 − 4|dx.
1
Solution:
Z 1
dt
√
.
(i)
2t
− t2
1/2
Solution:
Z 1
ln(1 + x)
(j)
dx.
ln(1
+
x) + ln(2 − x)
0
Solution:
Z π/2
(k)
sin4 x cos5 xdx.
0
Solution:
Z π
(l)
cosn x sin2 (n + 1)xdx.
0
Solution:
Z π/2
(m)
sin x sin 2x sin 3xdx.
0
5
Solution:
Z 2π
(n)
0
dx
.
(2 + cos x)(3 + cos x)
Solution:
(4) Evaluate
Z
2
x · 3x
(a)
dx.
3x2 − 2
Solution:
2
2
Let u = 3x − 2. Then du = 3x · ln 3 · 2xdx.
Z
cos x
√
dx.
(b)
4 − cos2 x
Solution:
Write 4 − cos2 x = 3 + sin2 x. The integral becomes
Z
cos x
p
dx.
3 + sin2 x
√
Let u = 3 sin x.
Z
x
√
(c)
dx.
2
x + 2x
Solution:
Write x2 + 2x = (x + 1)2 − 1. Then the integral becomes
Z
Z
Z
Z
x
x
x+1
1
√
p
p
p
dx =
dx =
dx −
dx.
2
2
2
x + 2x
(x + 1) − 1
(x + 1) − 1
(x + 1)2 − 1
For the first integral, we let u = (x + 1)2 . Then du = 2(x + 1)dx. Thus
Z
Z
p
√
x+1
du
p
√
= u − 1 + C = x2 + 2x + C.
dx =
2 u−1
(x + 1)2 − 1
Z
(d)
sech−1 (x)dx.
Solution:
Z
u
(e)
du.
6
u −8
Solution:
Z
(f)
sin6 xdx.
Solution:
6
Z
(g)
1
dx.
1 + ex
Solution:
Z
(h)
sin(ln x)dx.
Solution:
(5) Evaluate
Z 4
√
(a)
sec−1 xdx.
1
Solution:
√
Z 2
x
√
dx.
(b)
√
x+ 2−x
0
(c)
Solution:
Z π
p
3
(x2 x3 + 1 + | cos x|)dx.
−π
Solution:
Z 1
tan−1 x
dx.
(d)
1+x
0
Solution:
Z 2
3
(e)
x5 e−x dx.
0
Solution:
Write
3
x5 e−x =
−
x3
3
−3x2 e−x
3
.
Thus we may use integration by parts to solve the problem:
Z 2
Z 2 3
3
x
5 −x3
x e
dx =
−
de−x
3
0
0
Z 2
3
x3 −x3 2
=− e
|0 −
−x2 e−x dx
3
0
x3 −x3 2 1 x3 2
=− e
|0 − e |0 .
3
3
Z 1/2
dx
√
(f)
.
2
x + 4x + 5
0
7
Solution:
write x2 + 4x + 5 = (x + 2)2 + 1. Taking u = x + 2, the integral becomes
3
2
Z
√
2
√
Z
π/2
(g)
0
du
.
u2 + 1
d
[sin u cos u2 ]du.
du
Solution:
By the fundamental theorem of calculus:
√
Z
π/2
0
Z
π/2
(h)
0
√
d
π/2
.
[sin u cos u2 ]du = sin u cos u2 |0
du
sin x
dx.
cos2 x + 3 cos x + 2
Solution:
Let t = cos x. Then the integral becomes
1
Z
0
t2
dt
.
+ 3t + 2
Then use partial fraction expansion.
Z π/4
dθ
(i)
.
2
(tan θ + 4 tan θ + 3) cos2 θ
0
Solution:
Write the integral as
Z
π/4
0
dθ
=
(tan2 θ + 4 tan θ + 3) cos2 θ
Z
π/4
0
sec2 θdθ
.
(tan2 θ + 4 tan θ + 3)
Let u = tan θ. Then the integral becomes
Z
0
1
du
.
u2 + 4u + 3
Then use partial fraction expansion.
Z 7π/3
(j)
sin3 xdx.
π/3
Solution:
Write sin3Zx = sin x · sin2 x = sin x(1 −Zcos2 x). Let u = cos x.
sin x
cos x
(6) Evaluate I1 =
dx and I2 =
dx.
cos x + sin x
cos x + sin x
Hint: evaluate both I1 + I2 and I1 − I2 .
8
We know
Z
I1 + I2 =
1dx = x + C
Z
I1 − I2 =
sin x − cos x
dx.
sin x + cos x
Let u = sin x + cos x. We have du = (cos x − sin x)dx. Thus
Z
du
I1 − I2 = −
= − ln |u| + C = − ln | sin x + cos x| + C.
u
This implies that I1 =
1
1
1
1
x − ln | sin x + cos x| + C and I2 = x + ln | sin x + cos x| + C
2
2
2
2
Z
(7) Suppose a < b. Evaluate
b
p
(x − a)(b − x)dx.
a
Solution:
Let x = (b − a) sin2 t + a. Then dx = 2(b − a) sin t cos tdt and
x − a = (b − a) sin2 t,
Thus
Z p
b − x = (b − a) cos2 t.
Z
(x − a)(b − x)dx =
=
=
=
=
(b − a) sin t cos t · 2(b − a) sin t cos tdt
Z
1
(b − a)2 sin2 2tdt
2
Z
1
1 − cos 2t
2
(b − a)
dt
2
2
1
1
(b − a)2 t − sin 2t + C
4
2
!
p
r
(x − a)(b − x)
1
x−a
−1
2
(b − a) sin
−
+ C.
4
b−a
b−a
(8) Suppose a > 0 ac − b2 > 0. Show that
Z π/2
π
dx
=√
.
2 x + 2b cos x sin x + c sin2 x
a
cos
ac
− b2
−π/2
Solution:
dt
= dx and
1 + t2
1
cos x = √
.
1 + t2
Let I be the integral and t = tan x. Then
sin x = √
t
,
1 + t2
Thus the integral becomes
Z
∞
−∞
dt
.
a + 2bt + ct2
Completing the square of the enumerator, we obtain
2
b
ac − b2
2
a + 2bt + ct = c t +
+
c
c
9
Then
Z
∞
dt
=
a
+
2bt
+ ct2
−∞
Let u = ct + b. Then du = cdt. Thus
Z ∞
I=
−∞
Z
∞
−∞
cdt
.
(ct + b)2 + ac − b2
du
.
u2 + ac − b2
u
du
Let v = √
. Then dv = √
2 . Hence
2
ac − b
ac − b
Z ∞
dv
1
I=√
.
ac − b2 −∞ v 2 + 1
Z ∞
dv
π
.
We know
= π. Thus I = √
2+1
v
ac − b2
−∞