please help me in this hard question

Question

let c = (0 to π ) ∫ e^(-2x) * cos x dx and s = (0 to π ) ∫ e^(-2x) * sin x dx

a) show by integration by parts , or otherwise that c = 2s and s=-2c+1+e^(-2π)

b) hence, or otherwise find the values of c and s

c) the finite region bounded by the axes and the curve y=e^(-x) * cos(0.5x) , for 0<=x<=π is rotated through four right angles about the x-axis find the volume of the solid of revolution

Answer 1

a). Integrate the first integral by parts:  u = e^-2x    du = -2e^-2xdx

                                                       dv = cosxdx    v = sinx

 

So,

∫_{(from 0 to π)}e^-2xcosxdx=(e^-2xsinx)_{(from 0 to π)}+2∫_{(from 0 to π)}e^-2xsinxdx

 

   Thus, c = 0 + 2s.  So, c = 2s

 

If instead we first integrate e^-2xsinx letting u = e^-2x         du = -2e^-2x

                                                            dv = sinxdx      v = -cosx

 

We have:

 

∫_{(from 0 to π)}e^-2xsinxdx=-(e^-2xcosx)_{(from 0 to π)} - 2∫_{(from 0 to π)}e^2xcosxdx 

 

Therefore, s = -( -e^-2π - 1) - 2c

 

            So, s = e^-2π + 1 - 2c

 

b). From part a), c = 2s and s = e^-2π +1 - 2c

 

    So, c = 2(e^-2π+1-2c)

 

        5c = 2(e^-2π+1)     c = (2/5)(e^-2π+1)    s = (1/2)c = (1/5)(e^-2π+1)

 

c). V = π∫_{(from 0 to π)}(e^-xcos(0.5x))²dx

 

        = π∫_{(from 0 to π)}e^-2x(√[(1+cosx)/2])²dx

 

        = π/4∫_{(from 0 to π)}e^-2xdx + π/4∫_{(from 0 to π)}e^-2xcosxdx

 

        = -π/8(e^-2π -1) + π/4(c)

 

        = -π/8(e^-2π-1) + π/4[(2/5)(e^-2π+1)]

 

        = 3π/8 + (-π/8 + π/10)e^-2π

 

        = 3π/8 - (π/40)e^-2π

 

 

 

Answer 2

To get you started, I will show how to do the first integral. Both must be done using a repeated integration by parts and some algebra at the end.

 

I will start by finding the indefinite integral,

 

∫ e^-2x cos x dx = I.

 

Use integration by parts with u = e^-2x and dv = cos x dx. Then du = -2e^-2x dx and v = sin x. The integration by parts formula is

 

∫ u dv = uv - ∫ v du

 

∫ e^-2x cos x dx = e^-2x sin x - ∫ -2e^-2x sin x dx

 

                       = e^-2x sin x + 2∫ e^-2x sin x dx .

 

Use another integration by parts with U = e^-2x and dV = sin x dx. Then dU = -2e^-2x dx and V = -cos x. We now continue the evaluation:

 

I = e^-2x sin x + 2[e^-2x (-cos x) - ∫ -2e^-2x (-cos x) dx]

 

   = e^-2x sin x - 2e^-2x cos x - 4 ∫ e^-2x cos x dx

 

   = e^-2x sin x - 2e^-2x cos x - 4I .

 

Add 4I to both sides, then divide by 5 and tack on a "+C" to get the antiderivative:

 

I = 1/5 e^-2x (sin x - 2 cos x) + C .

 

Now you can finish the problem by evaluating this from 0 to π.

 

1/5 [ e^-2π (sin π - 2 cos π) - e^-2(0) (sin 0 - 2 cos 0) ] = 1/5 [ e^-2π (2) - 1 (-2) ] = 2/5 ( e^-2π + 1 ).

 

Now that I have expended so much time doing the first integral for you, you can finish the rest of the problem.

 

If you get REALLY stuck on this again, you should post a reply and someone else or myself will come to rescue you.