REDUCTION FORMULAS 
39 
In particular we obtain on setting m = 0 
(4) 
/ sin” -1 * COS X , n — 1 Г . „ 
sin" x dx —I I sin" 
n ' n J 
2 x dx. 
Thus for n = 2, 4 we have : 
J sin 2 ж dx = — 1 sin x cos ф + 1 x ; 
sin 4 *d* = — £sin 3 * cos x -f f^ sin 2 *d* 
= — ^ sin 3 x cos x — f sin x cos x + | x, 
and by setting n = 6 the student can now verify the result of Ex. 5 
in Chap. IX, § 8, of the Introduction to the Calculus. 
We must warn the student, however, against the stupidity of 
applying any of these reduction formulas when there is an obvious 
short cut. Thus 
I 
sin 3 xdx— — 
cos 2 x) d cos x = — cos x + ^ cos 3 x, 
and to use Formula (4), n = 3, to evaluate this integral would be 
much like multiplying 700 and 800 together by logarithms. 
If n is negative, we wish to increase it, and so Formulas (3) and 
(4) should be used backwards, i.e. solved for the integral on the 
right-hand side. A neater form for the result is obtained by going 
back to (2) and setting 
v — 1 = — n, 
¡x — 1 = m : 
( 5 ) 
cos m xdx 
sin”* 
COS m+1 X 
(n — 1) sin” -1 * 
+ 
n — m 
n — 
cos’” x dx 
sin” -2 x 
On setting m = 0, we have : 
(6) 
dx cos* n — 2 dx 
sin"* (n —1)sin" -1 * n — lj sin” -2 * 
If in (5) ft and m are equal, there is a simpler formula. 
Here, 
(0S x = cot"* = cot" -2 *(esc 2 * — 1). 
sin"* 
Hence 
(7) j cot" xdx = — ^—y ~ J* cot" -2 * dx. 
In all these formulas, n and m may be fractional or incommen 
surable. But in that case the given integral cannot in general be