How do you find the integral of #cos(log_e(x)) dx#?

Question

Emilia Aguilar · Accepted Answer

First, try integration-by-parts with #u=cos(ln(x))#, #du=-sin(ln(x))*1/x#, #dv=dx#, and #v=x# to get: #\int cos(ln(x))\ dx=uv-\int v\ du=x cos(ln(x))+\int sin(ln(x))\ dx#. Now use integration-by-parts again on this second integral with #u=sin(ln(x))#, #du=cos(ln(x))*1/x#, #dv=dx#, and #v=x# to get  #\int cos(ln(x))\ dx=x cos(ln(x))+(xsin(ln(x))-\int cos(ln(x))\ dx)#. Adding #\int cos(ln(x))\ dx# to both sides of this last equation, dividing both sides by 2, and including the #+C# at the very end gives the answer: #\int cos(ln(x))\ dx=1/2xcos(ln(x))+1/2xsin(ln(x))+C#.

Christopher Agresta · Answer

undefined To find the integral of cos(log_e(x)) dx, you can use the substitution method. Let u = log_e(x). Then, du/dx = 1/x, and dx = e^u du. Substituting these into the integral, you get ∫cos(u) * e^u du. Now, use integration by parts, letting dv = e^u du, which gives v = e^u. Thus, the integral becomes e^u * cos(u) - ∫e^u * (-sin(u)) du. Simplify this to get e^u * cos(u) + ∫e^u * sin(u) du. Finally, integrate by parts again for the remaining integral, and substitute back u = log_e(x) to get the final expression in terms of x.