(^476) Trigonometric functions
4 In Problem 2 of Problems 4.39, we called attention to the important non-
elementary beta integral formula
(1)
f0l
x2l(1 - x)q dx p!q!
(p+q+1)!
which is correct when the integral is a Riemann integral and p and q are non-
negative; we can now report that it is correct when the integral is a Riemann-
Cauchy integral and p and q exceed -1. Show that putting x = sin2 0, where
0 5 0 5 a/2 when 0 < x 5 1, yields the formula
(2)
=/z ''nti 0 1
f i 'gtl a dB
p!g!
o sn cos =2 (p + q + 1
Be sure to observe the fact that the limits of integration are correctly determined;
setting x = 0 in formulas involving x is equivalent to setting 0 = 0 in formulas
involving 0, and setting x = I in formulas involving x is equivalent to setting
0 = 7/2 in formulas involving 8. Finally, replace 0 by x and let 2p + 1 = a,
2q + I = 13 to obtain the formula
(3)
a-1
1/2
sins x cosA x dx =
2 > ( 2 )
fo (a+13'\
which is valid when a and 0 exceed -1. Remark: If we happen to know that
(--f)! _ , we can conclude from (3) that
(4) fo /2 sin"x dx = I2 ``
\I 2/!al
5 Replace x by y2 in the first formula of the preceding problem and then
replace y by x to obtain the formula
i x1 p!q!
2p+'(1- x2)q
o
dx=2(p+q+1)
Tell why the limits of integration are correct.
6 When we proved the formulas for derivatives of trigonometric functions,
we were unwilling to base the proofs upon the formula R = -'r26 for the area of
a sector of a circle having central angle 8 and radius r. Supposing that 0 <
0 < 7/2, this problem requires that the formula be proved. With the aid of an
appropriate figure, show that
.Q= forf(x)dx
where
f(x) = x tan 0 (0<x5rcos0)
f(x) = r' - xz (r cos 9 5 x <- r).
Then find 4 with the aid of the formula (8.455) which has been proved.
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(lu)
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