Solution to bonus problem 5 (based on the solutions by Eric Teacher and NinaSimone Lobban)
√ Z
The problem: Let I = 3 2
0
is x?
x
√
1 + cos t
2
dt. If 0 < x < π and tan I = √ , what
17 − 8 cos t
3
Solution: Begin with the substitution u = cos t, or t = arccos u, so that dt = − √
Then we have
du
.
1 − u2
√
√
√ Z cos x 1 + u
1 + cos t
du
√
dt = −3 2
17 − 8u 1 − u2
0 17 − 8 cos t
1
Z
cos
x
√
du
√
= −3 2
(17 − 8u) 1 − u
1
√ Z
I=3 2
x
Note that we have converted the original integral, which was not improper, into an
improper integral (but this is not going to cause any problem). Next, make the
√
substitution v = 1 − u, so u = 1 − v 2 and −du = 2v dv.Therefore:
√
√
√ Z 1−cos x dv
√ Z 1−cos x
v dv
=6 2
I=6 2
(9 + 8v 2 )v
9 + 8v 2
0
0
√ √
√
√ Z 1−cos x
Z arctan 2 2 1−cos x
3
2 2
dv
=
dθ
8 2 =
3 0
1 + 9v
0
!
√
2 2√
= arctan
1 − cos x
3
√
2 2v
3
after the trig substitution tan θ =
, or v = v = √ tan θ, so that dv =
3
2 2
3
2
√ sec θ. The reverse of the substitution (to get the limits of integration) is
2 2
√ !
2 2v
θ = arctan
. So now we now the value of I. And so
3
√
2 2√
2
tan I =
1 − cos x = √
3
3
or
√
√
3
1 − cos x = √
2
or
3
1 − cos x = .
2
2
Finally, we conclude that cos x = − 12 , and since 0 < x < π we have that x =
2π
.
3