Question

Evaluate:

`int e^(ax)*cos(bx + c)dx`

Answer 1

Let I = `int e^(ax)cos(bx + c)dx`   ....(1)

Integrating by parts

I = `cos(bx + c) int e^(ax)dx - int(int e^(ax)dx*d/dx cos(bx + c))dx`

= `cos(bx + c)*e^(ax)/a - int e^(ax)/a(-sin(bx + c)b)dx`

= `e^(ax)/a cos(bx + c) + b/a int e^(ax)*sin(bx + c)dx`

= `e^(ax)/a cos(bx + c) + b/a[sin(bx + c) int e^(ax)dx - int(int e^(ax)dx d/dx sin(bx + c))dx]`

= `e^(ax)/acos(bx + c) + b/a[e^(ax)/a sin(bx + c) - int e^(ax)/a (cos(bx + c)b)dx]`

= `e^(ax)/a cos(bx + c) + b/a^2 e^(ax) sin(bx + c) - b^2/a^2 int e^(ax) cos(bx + c)dx`

∴ I = `e^(x)/a cos(bx + c) + b/a^2 e^(ax) sin(bx + c) - b^2/a^2*I`  ....[From (1)]

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

∴ `I((a^2 + b^2))/a^2 = e^(ax) [(cos(bx + c))/a + (bsin(bx + c))/a^2]`

∴ I = `e^(ax)/(a^2 + b^2)[acos(bx + c) + bsin(bx + c)] + c`