Integration by Parts

Answer 1

it is one of those problems where the integral you are trying to find

shows up on the right side, so you solve for it.....

U = cos(bx) dV = exp(ax)

dU = -b sin(bx) V = (1/a) exp(ax)

the integral expands to:

(1/a) exp(ax) cos(bx) - integral [ (1/a)exp(ax) (-b sin (bx))

= (1/a) exp(ax) cos(bx) + (b/a) integral [ exp(ax) sin(bx)]

integrate the bold integral by parts:

U = sin(bx) dV = exp(ax)

du = b cos(bx) B = (1/a) exp(Ax)

the integral expands further:

= (1/a) exp(ax) cos(bx) + (b/a) [ (1/a) exp(ax) sin(bx) - integral [ (1/a)exp(ax) (b cos bx) ]

= (1/a) exp(ax)cos(bx)+ (b/a) [ (1/a) exp(Ax) sin(bx) - (b/a) integral [ exp(ax) cos(bx) ]

= (1/a) exp(ax)cos(bx) + (b/a^2) exp(Ax) sin(bx) - (b^2/a^2) integral [ exp(ax)cos(bx)]

The integral in bold is the same as the one we are trying to find, so we solve for it.

Moving it to the left side, the coefficient of it is 1 + b^2/a^2 = (a^2+b^2)/a^2

To solve for it, multiplies both sides of the reciprocal a^2/(a^2+b^2)

The raw answer is:

a^2/(a^2+b^2) (1/a) exp(ax)cos(bx) + (b/a^2) exp(Ax) sin(bx)

The final answer, upon simplifying becomes:

exp(ax) [ a cos bx + b sin bx]/(a^2+b^2)