75.
Z
dx
Ztan x + sin x
cos x d x
=
sin x(1 + cos x)
Let z = tan(x/2),
cos x =
1 − z2
,
1 + z2
2 dz
1 + z2
2z
sin x =
1 + z2
dx =
1 − z 2 2 dz
1 + z2 1 + z2
=
2z
1 − z2
1+
1 + z2
1 + z2
Z
Z
1
1 − z2
(1 − z 2 ) dz
=
dz
=
z(1 + z 2 + 1 − z 2 )
2
z
1
z2
+C
= ln |z| −
2
4
1 x 1 x 2
= ln tan −
+ C.
tan
2
2
4
2
Remark: Since
x
sin2
x
2 = 1 − cos x ,
tan2 =
x
2
1 + cos x
cos2
2
the answer can also be written as
1 1 − cos x 1 1 − cos x
ln
− ·
+ C.
4 1 + cos x 4 1 + cos x
Z
65.
x 1/2
d x Let x = u 6
1 + x 1/3
d x = 6u 5 du
Z
u8
du
=6
u2 + 1
Z 8
u + u6 − u6 − u4 + u4 + u2 − u2 − 1 + 1
=6
du
u2 + 1
Z 1
6
4
2
=6
du
u −u +u −1+ 2
u +1
!
u5
u3
u7
−
+
− u + tan−1 u + C
=6
7
5
3
√
6
6
= x 7/6 − x 5/6 + 2 x − 6x 1/6 + 6 tan−1 x 1/6 + C.
7
5
Z
64.
Z x2
1
3
dx =
dx
1+ 2
2x 2 − 3
2
2x − 3
√ Z x
1
3
1
= +
√
√ −√
√ dx
2
4
3
2x
−
3
2x
+
√
√ √
x
3 2x − 3 = + √ ln √
√ + C.
2 4 2 2x + 3 Z
√
40.
Z
10 x+2 d x
√
x +2
√
x +2
dx
du = √
2 x +2
Z
√
2
2
= 2 10u du =
10u + C =
10 x+2 + C.
ln 10
ln 10
Let u =
40. Since ln x grows more slowly than any positive power of x, therefore we haveZln x ≤ kx 1/4
∞
1
dx
1
≥ 3/4 for x ≥ 2 and
for some constant k and every x ≥ 2. Thus, √
√
kx
x ln x
2
Z x ln x
1 ∞ dx
.
diverges to infinity by comparison with
k 2 x 3/4
34. On [0,1], √
1
1
1
1
≤ √ . On [1, ∞), √
≤ 2 . Thus,
2
2
x
x+x
x
x+x
Z
Z
1
0
∞
1
Z 1
dx
dx
≤
√
√
2
x+x
x
Z0 ∞
dx
dx
.
≤
√
2
x2
x+x
1
Since both of these integrals are convergent, therefore so is their sum
Z
∞
√
0
dx
.
x + x2
24.
y(x) = 3 +
dy
= e−y ,
dx
ey = x + C
Z
x
e−y dt
H⇒
y(0) = 3
0
i.e. e y dy = d x
H⇒
3 = y(0) = ln C
y = ln(x + e3 ).
y = ln(x + C)
H⇒
C = e3
7. The region is a quarter-elliptic disk with semi-axes a = 2 and b = 1. The area of the region
is A = π ab/4 = π/2. The moments about the coordinate axes are
Mx=0 =
Z
2
s
x 1−
0
x2
Let u = 1 −
4
x
du = − d x
2
1√
4
u du =
3
0
Z 2
2
1
x
=
1−
dx
2 0
4
1
2
x 3 2
=
= .
x−
2
12 0
3
=2
M y=0
Z
x2
dx
4
Thus x = Mx=0 / A = 8/(3π ) and y = M y=0 / A = 4/(3π ). The centroid is 8/(3π ), 4/(3π ) .