141 polynomials in cosines and sines of multiples of

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141. Polynomials in cosines and sines of multiples of x . We can always integrate any function which is the sum of a finite number of terms such as A cos m ax sin m 0 ax cos n bx sin n 0 bx . . . , where m , m 0 , n , n 0 , . . . are positive integers and a , b , . . . any real numbers whatever. For such a term can be expressed as the sum of a finite number of terms of the types α cos { ( pa + qb + . . . ) x } , β sin { ( pa + qb + . . . ) x } and the integrals of these terms can be written down at once. Examples LI. 1. Integrate sin 3 x cos 2 2 x . In this case we use the for- mulae sin 3 x = 1 4 (3 sin x - sin 3 x ) , cos 2 2 x = 1 2 (1 + cos 4 x ) . Multiplying these two expressions and replacing sin x cos 4 x , for example, by 1 2 (sin 5 x - sin 3 x ), we obtain 1 16 Z (7 sin x - 5 sin 3 x + 3 sin 5 x - sin 7 x ) dx = - 7 16 cos x + 5 48 cos 3 x - 3 80 cos 5 x + 1 112 cos 7 x. The integral may of course be obtained in different forms by different meth- ods. For example Z sin 3 x cos 2 2 x dx = Z (4 cos 4 x - 4 cos 2 x + 1)(1 - cos 2 x ) sin x dx, which reduces, on making the substitution cos x = t , to Z (4 t 6 - 8 t 4 + 5 t 2 - 1) dt = 4 7 cos 7 x - 8 5 cos 5 x + 5 3 cos 3 x - cos x.
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[VI : 143] DERIVATIVES AND INTEGRALS 299 It may be verified that this expression and that obtained above differ only by a constant. 2. Integrate by any method cos ax cos bx , sin ax sin bx , cos ax sin bx , cos 2 x , sin 3 x , cos 4 x , cos x cos 2 x cos 3 x , cos 3 2 x sin 2 3 x , cos 5 x sin 7 x . [In cases of this kind it is sometimes convenient to use a formula of reduction ( Misc. Ex. 39).] 142. The integrals Z x n cos x dx , Z x n sin x dx and associated integrals. The method of integration by parts enables us to generalise the preceding results. For Z x n cos x dx = x n sin x - n Z x n - 1 sin x dx, Z x n sin x dx = - x n cos x + n Z x n - 1 cos x dx, and clearly the integrals can be calculated completely by a repetition of this process whenever n is a positive integer. It follows that we can always calculate Z x n cos ax dx and Z x n sin ax dx if n is a positive integer; and so, by a process similar to that of the preceding paragraph, we can calculate Z P ( x, cos ax, sin ax, cos bx, sin bx, . . . ) dx, where P is any polynomial. Examples LII. 1. Integrate x sin x , x 2 cos x , x 2 cos 2 x , x 2 sin 2 x sin 2 2 x , x sin 2 x cos 4 x , x 3 sin 3 1 3 x . 2. Find polynomials P and Q such that Z { (3 x - 1) cos x + (1 - 2 x ) sin x } dx = P cos x + Q sin x. 3. Prove that Z x n cos x dx = P n cos x + Q n sin x , where P n = nx n - 1 - n ( n - 1)( n - 2) x n - 3 + . . . , Q n = x n - n ( n - 1) x n - 2 + . . . .
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[VI : 143] DERIVATIVES AND INTEGRALS 300 143. Rational Functions of cos x and sin x . The integral of any rational function of cos x and sin x may be calculated by the substitution tan 1 2 x = t . For cos x = 1 - t 2 1 + t 2 , sin x = 2 t 1 + t 2 , dx dt = 2 1 + t 2 , so that the substitution reduces the integral to that of a rational function of t . Examples LIII. 1. Prove that Z sec x dx = log | sec x + tan x | , Z cosec x dx = log | tan 1 2 x | . [Another form of the first integral is log | tan( 1 4 π + 1 2 x ) | ; a third form is 1 2 log | (1 + sin x ) / (1 - sin x ) | .] 2.
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