Workshop_Solutions_5

Workshop_Solutions_5

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Unformatted text preview: e the indefinite integrals 2 J= sin(t) dt and 2 K= cos(t) dt. Solution: (a) Making the substitution dy =2 dt y = y (t) = 2 t, ⇒ 1 dy = dt 2 gives that cos(2t) dt = cos(y ) 1 1 dy = 2 2 cos(y ) dy = 1 1 sin(y ) = sin(2t). 2 2 (2) (b) Since cos(t + s) = cos(t) cos(s) − sin(t) sin(s), we have that cos(2t) = cos(t + t) = cos(t) cos(t) − sin(t) sin(t) = cos(t) Using sin(t) 2 2 + cos(t) 2 2 − sin(t) . = 1 yields that cos(2t) = cos(t) 2 − sin(t) 2 = 1 − sin(t) cos(2t) = cos(t) 2 − sin(t) 2 = cos(t) 2 2 2 − sin(t) − 1 − cos(t) 2 2 = 1 − 2 sin(t) , = 2 cos(t) 2 − 1. (c) From (b) we have cos(2t) = 1 − 2 sin(t) 2 and cos(2t) = 2 cos(t) 2 −1 and it follows that sin(t) 2 = 1 1 − cos(2t) 2 and cos(t) 2 = 1 1 + cos(2t) . 2 (3) Substituting (3) into the integrals J and K , respectively, yields 2 1 2 J= sin(t) dt = K= cos(t) dt = 2 1 2 1 − cos(2t) dt = 1 1 1 1 t − sin(2t) + c = t − sin(2t) + c, 2 2 2 4 1 + cos(2t) dt = 1 1 1 1 t + sin(2t) + c = t + sin(2t) + c, 2 2 2 4 where we have used (2) to evaluate the integral in the thi...
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This document was uploaded on 01/31/2014.

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