What is #int sin(lnx) dx#?

Question

Ezekiel Alexander · Accepted Answer

Use integration by parts twice and solve for the integral to find

#intsin(ln(x))dx = (x(sin(ln(x)) - cos(ln(x))))/2 + C# We will proceed through integration by parts. Let #I = intsin(ln(x))dx# Integration by Parts (1) Let #u = sin(ln(x))# and #dv = dx# Then #du = cos(ln(x))/xdx# and #v = x# Applying the integration by parts formula #intudv = uv - intvdu# #I = intsin(ln(x))dx# #= xsin(ln(x)) - intcos(ln(x))dx# Integration by Parts (2) Let #u = cos(ln(x))# and #dv = dx# Then #du = -sin(ln(x))/x# and #v = x# Applying the formula, we have #intcos(ln(x))dx = xcos(ln(x)) - int-sin(ln(x))dx# #= xcos(ln(x)) + intsin(ln(x))dx# #= xcos(ln(x)) + I# Substituting this into the result from the first integration by parts, we obtain #I = xsin(ln(x)) - xcos(ln(x)) - I# #=> 2I = x(sin(ln(x)) - cos(ln(x)))# #=> I = (x(sin(ln(x)) - cos(ln(x))))/2# But in the process of adding #I# to both sides, we lost the constant, and so for our final answer, we add it back in to get #I = intsin(ln(x))dx = (x(sin(ln(x)) - cos(ln(x))))/2 + C#

Charles Anctil · Answer

undefined The integral of $ \int \sin(\ln(x)) \, dx $ cannot be expressed in terms of elementary functions. It is an example of a non-elementary integral. However, it can be evaluated using techniques such as integration by parts or substitution, leading to expressions involving special functions such as the exponential integral.