Derive the general reduction formula?

A simple way to do this (other than using integration by parts) is to note that:

∫ e^(ax)*cos(bx) dx = Re{∫ e^(ax)*e^(bix) dx}

We have:

∫ e^(ax)*e^(bix) dx

=> ∫ e^(ax + bix) dx

= ∫ e^[x(a + bi)] dx

= e^[x(a + bi)]/(a + bi) + C.

Then:

∫ e^(ax)*cos(bx) dx

=> Re{e^[x(a + bi)]/(a + bi)} + C

= e^(ax)*Re{e^(bix)/(a + bi)} + C

= e^(ax)*Re{(a - bi)e^(bix)/(a^2 + b^2)} + C

= e^(ax)*[a*cos(bx) + b*sin(bx)]/(a^2 + b^2) + C.

I hope this helps!

∫ [e^(ax)cos(bx)] dx =

let I =∫e^(ax) cos(bx) dx

integrate by taking e^(ax) as a first function

I=1/b e^(ax) sin(bx) - a/b∫e^(ax) sin(bx) dx

now again integrate by taking e^(ax) as first function

I=1/b e^(ax) sin(bx) + a/b^2e^(ax) cos(bx) - a^2/b^2∫e^(ax)cos(bx) dx

I=1/b e^(ax)sin(bx) +a/b^2 e^(ax)cos(bx) -Ia^2/b^2

I= e^(ax)/(a^2+b^2) {a cos(bx) +b sin(bx)} +c

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