Integral of Sin ln x by Parts

In this tutorial we shall drive integral of sin(lnx), and solve this problem with the help of integration by parts methods as well as with the help of 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 first function is {e^u} and second function will be \sin u
Using 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} } }


Equation (i) becomes using above formula, we have

\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}

Comments

comments