Indefinite integral sinlnxSolution sin ln x dx Let uln x the

Solution

?sin (ln x) dx Let u=ln x, then e^u = x and dx = e^u du. ?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 substitute u = ln x back in, (x sin (ln x) - x cos (ln x))/2 + C