Calculus II Integration by parts

Question

Evaluate the integral: from 0 to t ∫-9essin(t-s)ds

Answer 1

I am interpreting "∫-9essin(t-s)ds" as "∫-9e^ssin(t-s)ds".

If my interpretation is correct then see the solution at http://fornerds.epizy.com/Tutoring/byParts.pdf

Answer 2

Question:

Evaluate the integral: from 0 to t ∫-9essin(t-s)ds

Solution:

∫ -9e s sin(t-s) ds from 0 to t

= -9e ∫ s * sin(t-s) ds

= -9e INT[ s * sin(t-s)] ds

Using integration by parts, choose u by LIATE:

u = s

du = ds

v = ?

dv = sin(t-s) ds

INT[u] dv = uv - INT[v] du

We now need to solve for v.

v = INT [sin(t-s)] ds

Let's do a w substitution since u is already taken.

w = t-s

Treat t like a constant.

dw = -ds

v = INT [sin(t-s)] ds

v = - INT [sin(w)] dw

v = cos(w)

v = cos(t-s)

Back to integration by parts:

u = s

du = ds

v = cos(t-s)

dv = sin(t-s) ds

INT[u] dv = uv - INT[v] du

= s cos(t-s) - INT [cos(t-s)] ds

We now need to solve for INT [cos(t-s)] ds.

Let's do a u substitution.

u = t-s

Treat t like a constant.

du = -ds

= INT [cos(t-s)] ds

= - INT [cos(u)] du

= -sin(u)

= -sin(t-s)

-9e INT[ s * sin(t-s)] ds = -9e [s * cos(t-s) - sin(t-s)] + C

-9e [s cos(t-s) - sin(t-s)] | 0 to t

= -9e [t cos(t-t) - sin(t-t)] + 9e [0 cos(t-0) - sin(t-0)]

= -9e [t cos(0) - sin(0)] + 9e [0 cos(t) - sin(t)]

= -9e [t*1 - 0] + 9e [0 - sin(t)]

= -9e [t] + 9e [- sin(t)]

= -9e [t + sin(t)]

THEREFORE, ∫ -9 e s sin(t-s) ds from 0 to t = -9e [t + sin(t)]