Answer Key for Quiz 1 (section A)
2
1. If we substitute u = x4 + x2 + 1 then du = (4x3 + 2x)dx = 2x(2x2 + 1)dx, so du
2 = (2x + 1)x dx and we
have
Z
Z
Z
1
4
2
7
2
7 du
(x + x + 1) (2x + 1)x dx = u
=
u7 du
2
2
1 u8
1 4
=
+C =
(x + x2 + 1)8 + C.
2 8
16
The substitution v = x2 isn’t powerful enough to knock off the original integral, but it does simplify it,
which is always worthwhile: dv = 2x dx, so dv
2 = x dx and we have
Z
Z
1
(v 2 + v + 1)7 (2v + 1)dv.
(x4 + x2 + 1)7 (2x2 + 1)x dx =
2
Now letting w = v 2 + v + 1, in which case dw = (2v + 1)dv, gives us
Z
Z
1
4
2
7
2
(v 2 + v + 1)7 (2v + 1)dv
(x + x + 1) (2x + 1)x dx =
2
Z
1
w8
1 4
=
w7 dw =
+C =
(x + x2 + 1)8 + C
2
16
16
as before.
2. (i) Because sin2 θ + cos2 θ = 1, we can write
Z
Z
Z
dθ
1 · dθ
(sin2 θ + cos2 θ)dθ
=
=
.
sin θ cos θ
sin θ cos θ
sin θ cos θ
Then we can split this up and simplify:
Z
Z
Z
(sin2 θ + cos2 θ)dθ
sin2 θ dθ
cos2 θ dθ
=
+
sin θ cos θ
sin θ cos θ
sin θ cos θ
Z
Z
sin θ dθ
cos θ dθ
=
+
.
cos θ
sin θ
(ii) We can do these last two integrals by substituting u = cos θ in the first one, and v = sin θ in the second
one. We have du = − sin θ dθ and dv = cos θ dθ respectively, so
Z
Z
sin θ dθ
−du
=
= − ln |u| + C = − ln | cos θ| + C
cos θ
u
and
Z
Z
cos θ dθ
dv
=
= ln |v| + C = ln | sin θ| + C.
sin θ
v
Therefore
Z
dθ
= ln | sin θ| + C − ln | cos θ| + C,
sin θ cos θ
and if we want to we can also write this as
Z
dθ
= ln | tan θ| + C.
sin θ cos θ
There are several other ways to do the integral, of which I’ll show you only one:
Z
Z
Z
dθ
dθ
sec2 θ dθ
=
=
sin θ
2
sin θ cos θ
tan θ
cos θ cos θ
where sec θ is the reciprocal of cos θ. (One would not learn from our textbook that there is such a function
as sec θ, but some of you may have picked it up on the street somewhere.) If we now let w = tan θ, then
dw = sec2 θ dθ and we have
Z
Z
Z
sec2 θ dθ
dw
dθ
=
=
= ln |w| + C = ln | tan θ| + C
sin θ cos θ
tan θ
w
as before.