The Integral of Sin ln x by Parts

In this tutorial we shall derive the integral of sin(lnx) and solve this problem with the help of the integration by parts methods as well as with the help of the substitution method.

The integral of sin(lnx) is of the form

I = \int {\sin \left( {\ln x} \right)dx}

Suppose that u = \ln x \Rightarrow x = {e^x}, then by differentiation dx = {e^u}du, we have

I = \int {{e^u}\sin u} du\,\,\,\,{\text{ - - - }}\left( {\text{i}} \right)

Here the first function is {e^u} and the second function is \sin u

Using the formula for integration by parts, we have

\int {\left[ {f\left( x \right)g\left( x \right)} \right]dx = f\left( x \right)\int {g\left( x \right)dx - \int {\left[ {\frac{d}{{dx}}f\left( x \right)\int {g\left( x \right)dx} } \right]dx} } }

Using the formula above, equation (i) becomes

\begin{gathered} I = {e^u}\int {\sin udu - \int {\left[ {\frac{d}{{du}}{e^u}\left( {\int {\sin udu} } \right)} \right]} du} \\ \Rightarrow I = {e^u}\left( { - \cos u} \right) - \int {\left[ {{e^u}\left( { - \cos u} \right)} \right]} du \\ \Rightarrow I = - {e^u}\cos u + \int {{e^u}\cos u} du \\ \end{gathered}

Again using integration by parts, we have

\begin{gathered} I = - {e^u}\cos u + \left[ {{e^u}\int {\cos udu - \int {\left[ {\frac{d}{{du}}{e^u}\left( {\int {\cos udu} } \right)} \right]du} } } \right] \\ \Rightarrow I = - {e^u}\cos u + {e^u}\sin u - \int {{e^u}\sin udu} \\ \end{gathered}

Using the original integral form as I = \int {{e^u}\sin u} du, we have

\begin{gathered} I = - {e^u}\cos u + {e^u}\sin u - I \\ \Rightarrow I + I = - {e^u}\cos u + {e^u}\sin u + c \\ \Rightarrow 2I = - {e^u}\cos u + {e^u}\sin u + c \\ \Rightarrow 2I = - {e^u}\cos u + {e^u}\sin u + c \\ \Rightarrow I = \frac{{\left( {\sin u - \cos u} \right){e^u}}}{2} + c \\ \Rightarrow \int {\sin \left( {\ln x} \right)dx} = \frac{{x\left[ {\sin \left( {\ln x} \right) - \cos \left( {\ln x} \right)} \right]}}{2} + c \\ \end{gathered}