SECT. VI.] VERIFICATION OF THE FORMULA. 187 



x = to x = TT ; and then that we have changed these curves by 

 multiplying all their ordinates by the corresponding ordinates of 

 a curve whose equation is y = &amp;lt;f&amp;gt;(x). The equations of the re 

 duced curves are 



y = sin x cf&amp;gt; (x), y = sin 2x &amp;lt;/&amp;gt; (x), y = sin 3x &amp;lt;/&amp;gt; (x), &c. 



The areas of the latter curves, taken from x = to x TT, 

 are the values of the coefficients a, 6, c, d, &c., in the equation 



I 



~ TT &amp;lt;f&amp;gt; (x) = a sin x + b sin 2a? + c sin 3x + d sin 4# + &c. 



221. We can verify the foregoing equation (D), (Art. 220), 

 by determining directly the quantities a lt 2 , a 3 , ... a. y &c., in the 

 equation 



&amp;lt; (a?) = a : sin a? + a 2 sin 2# + a 3 sin 3x + . . . a, sin Jic + &e. ; 



for this purpose, we multiply each member of the latter equation 

 by sin ix dx, i being an integer, and take the integral from x = 

 to X = TT, whence we have 



I &amp;lt;f)(x) sin ix dx = a x I sin x sin ix dx + 2 (sin 2# sm ix dx + &c. 



+ aj I sinjx sin ix dx + ... &c. 



Now it can easily be proved, 1st, that all the integrals, which 

 enter into the second member, have a nul value, except only the 



term a L \ sin ix sin ixdx ; 2nd, that the value of Ismixsmixdx is 

 i-TT ; whence we derive the value of a i} namely 



2 r 



- I (f&amp;gt; (a?) sin ix dx. 



The whole problem is reduced to considering the value of the 

 integrals which enter into the second member, and to demon- -i 

 strating the two preceding propositions. The integral 



2 I sin jjc si 11 ixdx, 



