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Mathematical Methods for Physics and Engineering - Matematica.NET

Mathematical Methods for Physics and Engineering - Matematica.NET

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3.6 APPLICATIONS TO DIFFERENTIATION AND INTEGRATION3.6 Applications to differentiation <strong>and</strong> integrationWe can use the exponential <strong>for</strong>m of a complex number together with de Moivre’stheorem (see section 3.4) to simplify the differentiation of trigonometric functions.◮Find the derivative with respect to x of e 3x cos 4x.We could differentiate this function straight<strong>for</strong>wardly using the product rule (see subsection2.1.2). However, an alternative method in this case is to use a complex exponential.Let us consider the complex numberz = e 3x (cos 4x + i sin 4x) =e 3x e 4ix = e (3+4i)x ,where we have used de Moivre’s theorem to rewrite the trigonometric functions as a complexexponential. This complex number has e 3x cos 4x as its real part. Now, differentiatingz with respect to x we obtaindzdx =(3+4i)e(3+4i)x =(3+4i)e 3x (cos 4x + i sin 4x), (3.36)where we have again used de Moivre’s theorem. Equating real parts we then findd (e 3x cos 4x ) = e 3x (3 cos 4x − 4sin4x).dxBy equating the imaginary parts of (3.36), we also obtain, as a bonus,d (e 3x sin 4x ) = e 3x (4 cos 4x +3sin4x). ◭dxIn a similar way the complex exponential can be used to evaluate integralscontaining trigonometric <strong>and</strong> exponential functions.◮Evaluate the integral I = ∫ e ax cos bx dx.Let us consider the integr<strong>and</strong> as the real part of the complex numbere ax (cos bx + i sin bx) =e ax e ibx = e (a+ib)x ,where we use de Moivre’s theorem to rewrite the trigonometric functions as a complexexponential. Integrating we find∫e (a+ib)x dx = e(a+ib)xa + ib + c(a − ib)e(a+ib)x=(a − ib)(a + ib) + c= eax (ae ibx − ibe ibx) + c, (3.37)a 2 + b 2where the constant of integration c is in general complex. Denoting this constant byc = c 1 + ic 2 <strong>and</strong> equating real parts in (3.37) we obtain∫I = e ax cos bx dx =eaxa 2 + b (a cos bx + b sin bx)+c 1,2which agrees with result (2.37) found using integration by parts. Equating imaginary partsin (3.37) we obtain, as a bonus,∫J = e ax sin bx dx =eaxa 2 + b (a sin bx − b cos bx)+c 2. ◭2101

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