Mathematical Methods for Physics and Engineering - Matematica.NET

2.2 INTEGRATION◮Evaluate the integral∫I =1x 2 +4x +7 dx.We can write the integral in the form∫I =1(x +2) 2 +3 dx.Substituting y = x + 2, we find dy = dx and hence∫1I =y 2 +3 dy,Hence, by comparison with the table of standard integrals (see subsection 2.2.3)√ ( ) √ ( ) 3 y√3 3 x +2I =3 tan−1 + c =3 tan−1 √ + c. ◭32.2.8 Integration by partsIntegration by parts is the integration analogy of product differentiation. Theprinciple is to break down a complicated function into two functions, at least oneof which can be integrated by inspection. The method in fact relies on the resultfor the differentiation of a product. Recalling from (2.6) thatd dv(uv) =udx dx + dudx v,where u and v are functions of x, we now integrate to find∫uv = u dv ∫ dudx dx + dx vdx.Rearranging into the standard form for integration by parts gives∫u dv ∫ dudx dx = uv − vdx. (2.36)dxIntegration by parts is often remembered for practical purposes in the formthe integral of a product of two functions is equal to {the first times the integral ofthe second} minus the integral of {the derivative of the first times the integral ofthe second}. Here, u is ‘the first’ and dv/dx is ‘the second’; clearly the integral vof ‘the second’ must be determinable by inspection.◮Evaluate the integral I = ∫ x sin x dx.In the notation given above, we identify x with u and sin x with dv/dx. Hence v = − cos xand du/dx = 1 and so using (2.36)∫I = x(− cos x) − (1)(− cos x) dx = −x cos x +sinx + c. ◭67