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MAS113 CALCULUS II, TUTORIAL 4 SOLUTIONS
Section 7.5
π
π
2. (a) arctan(−1) = − . (b) csc−1 2 = .
4
6
√
π
π
−1
2 = . (b) arcsin 1 = .
4. (a) sec
4 ¶
2µ
¶ √
µ
3
3π
3π
1
=
=
6. (a) tan−1 tan
. (b) cos arcsin
.
4
4
2
2
−x arcsin x
+ 1.
24. h′ (x) = √
1 − x2
x
26. f ′ (x) = ln(arctan x) +
.
2
(1 + x ) arctan x
−1
esec t
′
.
28. h (t) = √
2−1
x
x
Z 1
4
60.
dt = (4 arctan t)10 = π.
2
Z0 t + 1
dt
1
√
= arcsin(2t) + C.
62.
2
2
Z π/21 − 4t
π
sin x
π/2
dx
=
(−
arctan(cos
x))
.
=
64.
0
1 + cos2 x
4
Z0
arctan x
1
66.
dx
=
arctan2 x + C.
2
2 Z
Z 1+x
³x´
1
1
1
1
√
p
dx =
+ C.
dx = sec−1
68.
4
2
2
x x2 − 4
(x/2) (x/2)2 − 1
70. Making a change of variables by setting u = e2x , we have
Z
Z
e2x
1
1
1
1
√
√
dx =
du = arcsin u = arcsin e2x .
2
2
2
1 − e4x
1 − u2
Section 7.6
5
e3 + e−3
≈ 10.067661996. (b) cosh(ln 3) = .
4. (a) cosh 3 =
2
3
√
e + 1/e
−1
≈ 1.175201194. (b) sinh 1 = ln(1 + 2) ≈ 0.881373587.
6. (a) sinh 1 =
2
µ x
¶2 µ x
¶2
e + e−x
e2x + e−2x
e − e−x
2
2
16. cosh x + sinh x =
+
=
= cosh 2x.
2
2
2
x −e−x
1 + eex +e
2ex
1 + tanh x
−x
=
18.
=
= e2x .
x
−x
−x
1 − tanh x
2e
1 − eex −e
+e−x
32. g ′ (x) = 2 sinh x cosh x = sinh 2x.
¡
¢
34. F ′ (x) = cosh x tanh x + sinh xsech2 x = sinh x 1 + sech2 x .
1
MAS113 CALCULUS II, TUTORIAL 4 SOLUTIONS
2
36. Z
f (t) = e secht + et secht tanh t = et secht (1 − tanh t).
1
56.
sinh(1 + 4x)dx = cosh(1 + 4x) + C.
4
Z
′
tanh xdx = ln | cosh x| + C.
58.
60.
t
Z
sech2 x
dx = ln |2 + tanh x| + C.
2 + tanh x
Section
8.2
Z
Z
1
1
6
3
2.
sin x cos xdx = sin6 x(1 − sin2 x) cos xdx = sin7 x − sin9 x + C.
7
9
1
10. Applying the formula cos2 x = (1 + cos 2x) twice, we get that the integrand is
2
µ
¶
3
1
3
1 + 3 cos 2x + ((1 + cos 4x) + cos 2x .
8
4
Moreover,
¶
µ
Z
Z
1
1 3
3
2
cos 2xdx = (1 − sin 2x) cos 2xdx =
sin 2x − sin 2x + C.
2
3
Therefore, we have
¶π
µ
Z π
1
3
1
5π
5
3
6
θ + sin 2θ +
sin 4θ −
sin 2θ
.
=
cos θdθ =
16
4
128
48
16
0
0
1
14. Using the formulas sin 2x = 2 sin x cos x and sin2 x = (1 − cos 2x) we have
2
µ
¶π/2
Z π/2
Z π/2
1
1
1
π
2
2
sin x cos xdx =
x − sin 4x
1 − cos 4xdx =
= .
8 0
8
4
16
0
0
18. First we have
(1 − sin2 θ)2
cos5 θ
=
cos θ.
sin θ
sin θ
Therefore, with the change of variables u = sin θ, we have
Z
Z
Z
(1 − u2 )2
1
1
5
4
cot θ sin θdθ =
du =
− 2u + u3 du = ln | sin θ| − sin2 θ + sin4 θ + C.
u
u
4
cot5 θ sin4 θ =
26. The integrand can be re-written as (1 + tan2 θ) tan4 θ sec2 θ. Therefore, with a change
of variables u = tan θ, we get
Z π/4
Z 1
12
4
4
sec θ tan θdθ =
(1 + u2 )u4 du = .
35
0
0
28. The integrand can be re-written as (sec2 (2x) − 1) sec4 (2x) tan(2x) sec(2x). Therefore,
with a change of variables u = sec(2x) we get
Z
Z
1
1
1
3
5
u6 + u4 du =
sec7 (2x) +
sec5 (2x) + C.
tan (2x) sec (2x)dx =
2
14
10
MAS113 CALCULUS II, TUTORIAL 4 SOLUTIONS
2
2
Z
tan6 (ay)dy =
3
32. Using tan x = sec x − 1 we get that
¡
¢
tan6 (ay) = sec2 (ay) tan4 (ay) − tan2 (ay) + 1 − 1.
Therefore,
1
1
1
tan5 (ay) −
tan3 (ay) + tan(ay) − y + C.
5a
3a
a
38. The integrand can be re-written as (1 + cot2 x) cot6 x csc2 x. Therefore, we have
Z
Z
1
1
4
6
csc x cot xdx = (1 + cot2 x) cot6 x csc2 xdx = − cot7 x − cot9 x + C.
7
9
44. Using the formula sin 2x = 2 sin x cos x we get
Z
Z
1
1
cos x + sin x
dx =
(csc x + sec x) dx = ln |(sec x + tan x)(csc x − cot x)| + C.
sin 2x
2
2
48. Using the same method as in exercise 32, we have
Z
Z
1
8
7
tan xdx = tan x − tan6 xdx.
7
Let this last integral on the right-hand side of the above be I ′ . We have
¶
¶
µ
Z µ
Z
1
1
7
′
8
′
8
tan x − I sec x −
tan x sec x − I tan x sec x dx.
tan x sec xdx =
7
7
We therefore have
¶
µ
Z
Z
1
8
7
′
8
tan x − I sec x + I ′ tan x sec xdx.
tan x sec xdx =
7
7
By a similar computation, we have
Z
Z
6
′
tan x sec xdx = I sec x − I ′ tan x sec xdx.
From these we infer that
√
µ
¶π/4
Z π/4
1
2 7
7
8
7
tan x sec xdx =
tan x sec x
− I.
− I=
8
8
8
8
0
0
1
66. The integrand can be re-written as (cos((n − m)x) − cos((m + n)x)). If m − n 6= 0,
2
then
µ
¶π
Z π
1
1
1
sin mx sin nxdx =
sin((n − m)x) −
sin((m + n)x)
= 0.
2 n−m
n+m
−π
−π
On the other hand, if m − n = 0, then
µ
¶π
Z
Z π
1
1
1 π
x−
sin 2mx
(1 − cos 2mx) dx =
= π.
sin mx sin nxdx =
2 −π
2
2m
−π
−π
4
MAS113 CALCULUS II, TUTORIAL 4 SOLUTIONS
68. We have
1
π
Z π
N
1X
f (x) sin mxdx =
an
sin nx sin mxdx.
π n=1
−π
−π
Z
π
The integral above vanishes unless m = n in which case it is π. Hence we have the desired
result.