Download Calculus II, Section 7.3, #8 Trigonometric Integrals Evaluate the

Calculus II, Section 7.3, #8
Trigonometric Integrals
Evaluate the integral.1
Z
dt
√
2
t t2 − 16
Z
dt
√
=
t2 t2 − 16
The radical has the form
Z
1
√
dt
t2 t2 − 4 2
√
x2 − a2 so we make the subsitution
Let t = 4 sec (θ)
so
dt = 4 sec (θ) tan (θ) dθ
Then we have
t2 = 16 sec2 (θ)
and
t2 − 16 = 16 sec2 (θ) − 16
= 16 sec2 (θ) − 1
Now 1 + tan2 (θ) = sec2 (θ), so sec2 (θ) − 1 = tan2 (θ), and we have
t2 − 16 = 16 tan2 (θ)
so
q
p
t2 − 16 = 16 tan2 (θ)
p
= 4 tan2 (θ)
= 4 |tan (θ)|
= 4 tan (θ) ,
θ ∈ QI or III.
Since
√
For |tan (θ)| to equal tan (θ), we must
have tan (θ) nonnegative, which means
we must have θ in Quadrant I or III.
We substitute all these pieces in terms of θ into the integral
Z
Z
1
1
√
dt =
· 4 sec (θ) tan (θ) dθ
16 sec2 (θ) · 4 tan (θ)
t2 t2 − 16
Z
1
dθ
=
16 sec (θ)
Z
1
1
·
dθ
=
16
sec (θ)
Z
1
· cos (θ) dθ
=
16
1
=
· (sin (θ) + C)
16
1
1
sin (θ) + C
=
16
16
1 Stewart,
Calculus, Early Transcendentals, p. 491, #8.
a2 = |a|.
Calculus II
Trigonometric Integrals
Since C is an arbitrary constant,
=
1
16 C
is also an arbitrary constant, so we’ll just write
1
sin (θ) + C
16
We have succeeded in integrating our function, but the result is in terms of θ rather than t, so we’ll try to
substitute back to get the result in terms
√ of t. If we look at our initial substitution, t = 4 sec (θ), and other
calculations, t2 − 16 = 16 tan2 (θ) or t2 − 16 = 4 tan (θ), none of them involve the sin (θ) that we got from
the integration. In this case, we use t = 4 sec (θ) and a right triangle to express sin (θ) in terms of t.
Since t = 4 sec (θ), we have
t=4·
1
cos (θ)
or
t
4
cos (θ) =
t
and we use this relationship to sketch the triangle
showun in Figure 1. (We could’ve used 4t = sec (θ),
but most of us are more comfortable with the cosine
function.)
opp.
θ
4
Figure 1
From Pythagorean Theorem, we get
42 + opp.2 = t2
16 + opp.2 = t2
opp.2 = t2 − 16
And since the opposite side is a length and thus nonnegative,
p
opp. = t2 − 16
In Figure 2, the three sides of the triangle are labeled.
From the figure, we see
√
t2 − 16
sin (θ) =
t
√
t2 − 16
t
θ
4
Figure 2
Finally, substituting gives us
Z
1
1
√
dt =
sin (θ) + C
16
t2 t2 − 16
√
1
t2 − 16
=
·
+C
16
t
√
t2 − 16
=
+C
16t
Thus,
Z
1
√
dt =
2
2
t t − 16
√
t2 − 16
+C
16t