104. Circular functions may always be reduced to algebraic functions by assuming sin x or cos x = z; thus the expression sin "x cos "x dx may be reduced to a binomial differential by assuming sin x = z; since f sin "x cos "x dx = ƒ zTM (1 − z2) * dz. But the integrals of circular functions will be obtained more elegantly by the usual process of integration by parts, or by means of the following trigonometrical values of the powers of sin x and cos x. To exemplify the use of these values of the powers of sin x and cos x, let f cos xdx be required. We have 8 5 ...f cos x dx = fcosx dx+fos 3x dx + cos 5z dz 8 = cos 3 x + cos 5 x; 16 16 5 1 16 16 dx COS X = = = S sin (†π+x) sin(x) 106. To integrate the forms tan x dx and cot x dx. S d ( n + x) = log tan(+x). sin x dx f cot x dx = COS X = S= cos x dx sin x = d cos x COS X dx = d sin x sin x cos x 2 dx sin 2 x = = 108. To integrate the forms du = = = = u = Writing n sin "x dx and du cos "x dx. f sin x cos x + (n − 1) ƒ sin "-2x cos sin -1x cos x + (n sin − 1) ƒ sin "~2x (1 − n-2 x dx sin 2x) d x n- - 1 2 for n in (A), we get In this manner the index may be reduced to 1 or 0, according as n is odd or even, and the final integral will be either f sin x dx cos x, or f dx = Again, u = f cos = x. -'x cos x dx = f cos "-1x d sin x = cos -1 sin x + (n − 1) ƒ cos - sin x dx and the final integral is either ƒ cos x d x = sin x or f dx = x. 109. To integrate the forms du = which reduces the integral to fo 110. To integrate the form du sin x cos x dx. If m be an odd positive integer of the form 2p+ 1, then we have cos x)" sin x dx = u = f cos "x (1 fcos "x (1-cos x)" d cos x, which may be expanded by the binomial theorem and integrated, the general term being of the form cos "x d cos x, whose integral is COS "+1 n+1 If n be an odd positive integer of the form 2 p + 1, then we have u = ƒ sin "x (1 sin x)" cos x dx = which by expansion is easily integrated. f sin "x (1 - sin x)” d sin x, - 2 p; then If m+n be an even negative integer of the form = -1 Stan "x (1 + tan x)" ̄1 sec 2x dx m -1 = Stan 'x (1 + tan x)" -1 d tan x, = which by expansion is integrable, each term being of the form a dx. But when neither of these conditions is satisfied, we must proceed by reducing the indices m and n successively. 111. To reduce m or n in f sin "x cos "x dx. Since d u = sin x cos "x dx = sin 7. = 1 2 x dx 1 x. x cos x + 2x) = 1 + sin 2 x COS 1 (cos x 1 2 4 + x. sin 3x 2 sin+log tanz dx 2 2 1 15 2 15 sin x cos x de sin - sin x) con = = cos xsec x. X. X. dx 5. sin x dx 6. COS 2x 8. 9. 10. 11. 12. 1 S S dx sin x cos x dx sin x cos 2x tan x = d x (a + b cos x)2 b sin x + APPLICATION OF THE INTEGRAL CALCULUS TO GEOMETRY. 1. TO FIND THE AREAS OF CURVES. 116. The determination of the quadrature and rectification of curves, as well as that of the volumes and surfaces of solids, will afford both useful and interesting applications of the Integral Calculus; but before proceeding to these, it will be necessary to premise the following lemma. be three series such that, however small h may be, the value of the second is less than the first and greater than the third; then if A = C, we shall have A B = C. = |