Mathematical Methods for Physics and Engineering - Matematica.NET

2.2 INTEGRATIONRearranging this expression to obtain I explicitly and including the constant of integrationwe findI =eax(b sin bx + a cos bx)+c. (2.37)a 2 + b2 Another method of evaluating this integral, using the exponential of a complex number,is given in section 3.6. ◭2.2.9 Reduction formulaeIntegration using reduction formulae is a process that involves first evaluating asimple integral and then, in stages, using it to find a more complicated integral.◮Using integration by parts, find a relationship between I n and I n−1 whereI n =∫ 10(1 − x 3 ) n dxand n is any positive integer. Hence evaluate I 2 = ∫ 10 (1 − x3 ) 2 dx.Writing the integrand as a product and separating the integral into two we findI n ==∫ 10∫ 10(1 − x 3 )(1 − x 3 ) n−1 dx(1 − x 3 ) n−1 dx −∫ 10x 3 (1 − x 3 ) n−1 dx.The first term on the RHS is clearly I n−1 and so, writing the integrand in the second termon the RHS as a product,Integrating by parts we findI n = I n−1 +I n = I n−1 −∫ 10(x)x 2 (1 − x 3 ) n−1 dx.[ x] 1∫ 13n (1 − x3 ) n −0 013n (1 − x3 ) n dx= I n−1 +0− 13n I n,which on rearranging givesI n =3n3n +1 I n−1.We now have a relation connecting successive integrals. Hence, if we can evaluate I 0 ,wecan find I 1 , I 2 etc. Evaluating I 0 is trivial:I 0 =∫ 10(1 − x 3 ) 0 dx =∫ 10dx = [x] 1 0 =1.Hence(3 × 1)I 1 =(3 × 1) + 1 × 1= 3 4 , I (3 × 2)2 =(3 × 2) + 1 × 3 4 = 9 14 .Although the first few I n could be evaluated by direct multiplication, this becomes tediousfor integrals containing higher values of n; these are therefore best evaluated using thereduction formula. ◭69