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# 113 it follows at once from theorem 3 of 113 that z f

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113 . It follows at once from Theorem (3) of § 113 that Z f 0 ( x ) F ( x ) dx = f ( x ) F ( x ) - Z f ( x ) F 0 ( x ) dx. It may happen that the function which we wish to integrate is expressible in the form f 0 ( x ) F ( x ), and that f ( x ) F 0 ( x ) can be integrated. Suppose, for example, that φ ( x ) = ( x ), where ψ ( x ) is the second derivative of a known function χ ( x ). Then Z φ ( x ) dx = Z 00 ( x ) dx = 0 ( x ) - Z χ 0 ( x ) dx = 0 ( x ) - χ ( x ) . We can illustrate the working of this method of integration by applying it to the integrals of the last section. Taking f ( x ) = ax + b, F ( x ) = p ax 2 + 2 bx + c = y, we obtain a Z y dx = ( ax + b ) y - Z ( ax + b ) 2 y dx = ( ax + b ) y - a Z y dx + ( ac - b 2 ) Z dx y , so that Z y dx = ( ax + b ) y 2 a + ac - b 2 2 a Z dx y ; and we have seen already ( § 135 ) how to determine the last integral. Examples XLIX. 1. Prove that if a > 0 then Z p x 2 + a 2 dx = 1 2 x p x 2 + a 2 + 1 2 a 2 log { x + p x 2 + a 2 } , Z p x 2 - a 2 dx = 1 2 x p x 2 - a 2 - 1 2 a 2 log | x + p x 2 - a 2 | , Z p a 2 - x 2 dx = 1 2 x p a 2 - x 2 + 1 2 a 2 arc sin( x/a ) .

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[VI : 138] DERIVATIVES AND INTEGRALS 292 2. Calculate the integrals Z dx a 2 - x 2 , Z p a 2 - x 2 dx by means of the substitution x = a sin θ , and verify that the results agree with those obtained in § 135 and Ex. 1. 3. Calculate Z x ( x + a ) m dx , where m is any rational number, in three ways, viz. (i) by integration by parts, (ii) by the substitution ( x + a ) m = t , and (iii) by writing ( x + a ) - a for x ; and verify that the results agree. 4. Prove, by means of the substitutions ax + b = 1 /t and x = 1 /u , that (in the notation of §§ 130 and 138 ) Z dx y 3 = ax + b Δ y , Z x dx y 3 = - bx + c Δ y . 5. Calculate Z dx p ( x - a )( b - x ) , where b > a , in three ways, viz. (i) by the methods of the preceding sections, (ii) by the substitution ( b - x ) / ( x - a ) = t 2 , and (iii) by the substitution x = a cos 2 θ + b sin 2 θ ; and verify that the results agree. 6. Integrate p ( x - a )( b - x ) and p ( b - x ) / ( x - a ). 7. Show, by means of the substitution 2 x + a + b = 1 2 ( a - b ) { t 2 +(1 /t ) 2 } , or by multiplying numerator and denominator by x + a - x + b , that if a > b then Z dx x + a + x + b = 1 2 a - b t + 1 3 t 3 . 8. Find a substitution which will reduce Z dx ( x + a ) 3 / 2 + ( x - a ) 3 / 2 to the integral of a rational function. ( Math. Trip. 1899.) 9. Show that Z R { x, n ax + b } dx is reduced, by the substitution ax + b = y n , to the integral of a rational function. 10. Prove that Z f 00 ( x ) F ( x ) dx = f 0 ( x ) F ( x ) - f ( x ) F 0 ( x ) + Z f ( x ) F 00 ( x ) dx
[VI : 139] DERIVATIVES AND INTEGRALS 293 and generally Z f ( n ) ( x ) F ( x ) dx = f ( n - 1) ( x ) F ( x ) - f ( n - 2) ( x ) F 0 ( x ) + · · · + ( - 1) n Z f ( x ) F ( n ) ( x ) dx. 11. The integral Z (1 + x ) p x q dx , where p and q are rational, can be found in three cases, viz. (i) if p is an integer, (ii) if q is an integer, and (iii) if p + q is an integer. [In case (i) put x = u s , where s is the denominator of q ; in case (ii) put 1+ x = t s , where s is the denominator of p ; and in case (iii) put 1+ x = xt s , where s is the denominator of p .] 12. The integral Z x m ( ax n + b ) q dx can be reduced to the preceding integral by the substitution ax n = bt . [In practice it is often most convenient to calculate a particular integral of this kind by a ‘formula of reduction’ (cf. Misc. Ex. 39).] 13. The integral Z R { x, ax + b, cx + d } dx can be reduced to that of a rational function by the substitution 4 x = - ( b/a ) { t + (1 /t ) } 2 - ( d/c ) { t - (1 /t ) } 2 .

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