Evaluate #int \ e^(-st)sint \ dt #?

Question

Ezekiel Alexander · Accepted Answer

# int \ e^(-st)sint \ dt = -(e^(-st)(s sint +cost ))/(s^2+1) + C #

We seek the integral: # I =int \ e^(-st)sint \ dt # We can then apply Integration By Parts: Let # { (u,=sint, => (du)/dt,=cost), ((dv)/dt,=e^(-st), => v,=-1/se^(-st) ) :}# Then plugging into the IBP formula: # int \ (u)((dv)/dt) \ dt = (u)(v) - int \ (v)((du)/dt) \ dt # We have: # int \ (sinmt)(e^(-st)) \ dt = (sin t)(-1/se^(-st)) - int \ (-1/se^(-st))(cos t) \ dt # # :. I = -1/s e^(-st)sint + 1/s int \ e^(-st)cos t \ dt # Now consider the integral given by: # I_2 = int \ e^(-st)cos t \ dt # We will now need to apply IBP again: Let # { (u,=cost, => (du)/dt,=-sint), ((dv)/dt,=e^(-st), => v,=-1/se^(-st) ) :}# Then plugging into the IBP formula we have:: # int \ (cost)(e^(-st)) \ dt = (cost)(-1/se^(-st)v) - int \ (-1/se^(-st))(-sint) \ dt # # I_2 = -1/s e^(-st)cost - 1/s \ int \ e^(-st)sint \ dt # # I_2 = -1/s e^(-st)cost - 1/s I # And so combining the results we find that: # I = -1/s e^(-st)sint + 1/s {-1/s e^(-st)cost - 1/s I} # As #s != 0#, then by Multiplying by #s^2# we get: # s^2I = -s e^(-st)sint - e^(-st)cost - I # # :. s^2I +I = -s e^(-st)sint - e^(-st)cost # # :. (s^2+1)I = -e^(-st)(s sint +cost ) # # :. I = (-e^(-st)(s sint +cost ))/(s^2+1) # And not forgetting the constant of integration, # I = (-e^(-st)(s sint +cost ))/(s^2+1) + C #

Scarlett Allison · Answer

undefined The integral of $ e^{-st} \cdot \sin(t) \, dt $ is $ \frac{s}{s^2 + 1} $.