integral of sinlnxSolution sin ln x dx Let uln x then eu x

Solution

?sin (ln x) dx Let u=ln x, then e^u = x and dx = e^u du. So we have: ?sin u e^u du Integrate by parts: Let v = sin u, dw = e^u du, w=e^u, dv = cos u du: ?sin u e^u du = sin u e^u - ?cos u e^u du Integrate by parts again: Let v=cos u, dw = e^u du, w=e^u, dv = -sin u du: ?sin u e^u du = sin u e^u - (cos u e^u + ?sin u e^u du) ?sin u e^u du = sin u e^u - cos u e^u - ?sin u e^u du 2 ?sin u e^u du= sin u e^u - cos u e^u ?sin u e^u du = (sin u e^u - cos u e^u)/2 + C Note that since we didn\'t have a \"natural\" opportunity to add the constant of integration earlier, we had to tack it on at the end. Anyway, now that we have the integral, just substitute u = ln x back in, and we have: (x sin (ln x) - x cos (ln x))/2 + C Check of result: d((x sin (ln x) - x cos (ln x))/2 + C)/dx = 1/2 (d(x sin (ln x))/dx - d(x cos (ln x))/dx) = 1/2 (dx/dx * sin (ln x) + x*d(sin (ln x))/dx - dx/dx * cos (ln x) - x*d(cos (ln x))/dx) = 1/2 (sin (ln x) + x cos (ln x) * d(ln x)/dx - cos (ln x) + x sin (ln x) * d(ln x)/dx) = 1/2 (sin (ln x) + cos (ln x) - cos (ln x) + sin (ln x)) = 1/2 (2 sin (ln x)) = sin (ln x) ?