454 Trigonometric functions
Instead of inviting attention to problems of this nature, we present prob-
lems more likely to promote scientific competence.
Table 8.28
Jsinududx=-cosu+c
f tanudxdx=logIsecul+c
f
secudxdx=log Isecu+tanul+c
f
dduxdx
n+1+c'f
uddu
Jcosudxdx=sinu+c
Jcot u du dx = log sin ul + c
I csc is du dx = log Icsc u - cot ul + c
x
du
+ c, f eud dr= eu}c.
dx
Problems 8.29
1 Make all of the calculations necessary to show that
(a)
(c)
(e)
(g)
Iosin x dx = 2
b
lim J sin wx dx = 0a
1 r
lim
J
sin2 wt dt =
xdx = log 2
(b)
fox"
cos x dx = 1
x
?/2.
(h) fo tan x dx
(d) lim f cos wx dx = 0
(f) lim
1
- fo wt dt =
-m x o
x
= 00
2 Recall that, when is is a differentiable function of x and is 0 0,
(1) dxlog Jul u dx'
the absolute-value signs, which are superfluous when u > 0, need not bother us.
Supposing that x is not an odd multiple of a/2 and that
f(x) = log sec x + tan xl, g(x) = log I tan 0 + 4) I,
show that f' (x) = sec x and g'(x) = sec x.
Remark: This proves that the two formulas
A4)
fsee x dx = log sec x + tan xl + Cl
f secxdx=logItan(2+7rf see x dx = log I tan 0 )I+s
are both correct. Some integral tables contain both of them. These things
imply that, over each interval containing no odd multiple of 7r/2, f(x)- g(x)