Calculus: Analytic Geometry and Calculus, with Vectors

8.2 Trigonometric integrands 455

must be a constant, but they do not imply that this constant must be zero. The
identity

x 1+tan2 cost+sin2cos2+sin2

tanC2+ - _ 4

1 - tan cos sin cos + sin

(^22222)

1 + 2sin2cos2_1+sinx

x x cos x sec x + tan x

cost 2 - sine 2

clarifies the matter.

3 Prove that if x is not an even multiple of r/2 and

f(x) = log csc x - cot x1, g(x) = log tan 2 ,

then f' (x) = csc x and g'(x) = csc x. Remark: This proves that the two formulas

fcsc x dx = log Icsc x - cot xj + c, f csc x dx = log

are both correct.

4 When a steady (or constant) current I (measured in amperes) flows

through a wire (or resistor) having resistance R (measured in ohms), the quantity

Q (measured in watthours, or thousandths of kilowatthours) of energy converted

into heat in At hours is calculated from the formula

(1) Q = 11R At.

With this basic information, show that if 1(t) is integrable (Riemann) over the

time interval t1 < t S t2, then the quantity Q(tl,t2) of energy converged into heat

between times t1 and t2 is (or should be)

(2) Q(tl,t2) = R f,[I(t)J2 dt.

Suppose now that 1(t) is the sinusoidal (or alternating) current determined by

(3) 1(t) = lo sin (wt + 0),

where Io is a constant "maximum current," w is a constant "angular frequency,"

and 0 is a "phase angle." Show that, in this case,

2

(4) Q(t142) = t2

_

2

tl I2oR-

I R

[sin (2wt2 + 20) - sin (2wt1 + 20)].

The last term is 0 whenever 2wt2 - 2wt1 is an integer multiple of 2a, and in every

case the absolute value of the last term cannot exceed RIo/2w. In all ordinary

applications of this formula, t2 - t1 is so large in comparison to 1/w that the last

term in (4) is insignificant. In such cases, (4) is always replaced by

(5) Q(tl,t2) - 3IR(12 - tl).