Evaluate #int \ e^(-st)sint \ dt #?
1 Answer
# int \ e^(-st)sint \ dt = -(e^(-st)(s sint +cost ))/(s^2+1) + C #
Explanation:
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^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 #