How Can You Solve the Integral of exp(-t) times cos^2(t)?

[Somefantastik]

$$\int^{t}_{0}e^{-s}cos^2(s)ds$$

let $$u = e^{-s}, \ du = -e^{-s}, \ dv = cos^{2}(s)ds, \ v = \frac{1}{2}s + \frac{1}{4}sin(2s)$$

then

$$\int^{t}_{0}e^{-s}cos^2(s)ds =\left[e^{-s}(\frac{1}{2}s - \frac{1}{4}sin(2s)\right]^{t}_{0} + \int^{t}_{0}e^{-s}(\frac{1}{2}s-\frac{1}{4}sin(2s)ds$$

let $$u = e^{-s}, \ du = -e^{-s}ds, \ dv = \frac{1}{2}s-\frac{1}{4}sin(2s)ds, \ v = \frac{1}{4}s^{2} + \frac{1}{8}cos(2s)$$

then

$$= e^{-t}\left(\frac{1}{2}t - \frac{1}{4}sin(2t)\right) + \left[e^{-s}\left(\frac{1}{4}s^{2} + \frac{1}{8}cos(2s)\right)\right]^{t}_{0} + \int^{t}_{0}e^{-s}\left(\frac{1}{4}s^{2} + \frac{1}{8}cos(2s)\right)$$

How to proceed? I can't seem to get this

$$\int^{t}_{0}e^{-s}\left(\frac{1}{4}s^{2} + \frac{1}{8}cos(2s)\right)$$

to equal this

$$\int^{t}_{0}e^{-s}cos^2(s)ds$$

So I'm not sure what to do next.

Any suggestions?

 

[HallsofIvy]

I would be inclined to do it the other way around: let u= cos²(s), dv= e^-s.

That, together with the fact that sin(2s)= 2 sin(s)cos(s), should make the second integration by parts easy.

 

[Somefantastik]

Ok thanks, I'll give that a try.