492 UNIVERSITY OF VIRGINIA PUBLICATIONS 



Put t = 0, then will the above equation become, 



u = f {x) = oo cos + ai cos wx/r + ai cos 2irx/r + • • • 



Multiply both sides by cos mrx/r ■ dx, and integrate between the hmits 

 X = and x = r. 



r-x=T rx^r 



I f (x) ■ cos riTx/r ■ dx = \ Uq ■ cos riTx/r ■ dx + 



Ji=0 Ji=0 



I fli cos • Tx/r ■ cos TiTX'/r ■ dx -\- \ a2 cos I-Kxjr ■ cos mrx/r ■ dx 

 Ji=o • Ji=0 



x: 



a„ cos^ n-Kx/r • dx + 



Make use of the relation that, 



2 cos A ■ cos B = cos (A + 5) + cos (A — 5) 



The first term may be written, 



f ''"'' , / J 1 f f''^'" a; (x + nx) , 

 I ai cos TTx/j' ■ cos TTiix r ■ dx = ^ aii \ cos dx + 



a; (tt — wir) 



i: 



• drc 



Evaluating the above integral between the assigned limits we obtain zero. 

 In the same way all terms except the nth fall out. 

 Making use of the relations, 



sin^ A = J (1 — cos 2A), and cos^ A = ■ 1 — sin^ A, and hence, 

 cos^ A = ^ (cos 2 A + 1), and putting nirx/r = A, 



and substituting in the nth term we have, 



r'x=T r-x^T 



ar/n-K I cos A • dA= ar/2mr \ cos 2 A ■ dA -\- ar • A/2n-K = 



Ji=0 J 1=0 



ar/An-K ■ sm A -\- ar ■ A/2mr. 



Within the assigned limits the sine of A is always zero (being multiple 

 of x) : Heuce, replacing A by its value UTx/r, and solving for a„, we 

 have, 



a„ = 2/r ■ \ f {x) ■ cos nirx/r • dx 



