Question

How do you integrate $\int {\sin \left( {\log x} \right)} $ by integration by parts methods?

Answer 1

Hint: In this question they have given$\int {\sin \left( {\log x} \right)} $and asked us to solve it by using parts method. We will have to take any one term as \[u\] and the other as \[dv\] and have to find its derivative and the value of $v$. Then we need to substitute it in the formula \[I = uv - \smallint vdu\] and then simply and solve it to find the correct answer.

Formula used: Formula for integrating in parts method:
\[I = uv - \smallint vdu\]

Complete step-by-step solution:
In this question they have given$\int {\sin \left( {\log x} \right)} $and asked us to solve it by using parts method.
We know that the formula for using part method is \[I = uv - \smallint vdu\]
First we need to decide which one to choose as \[u\]and which one as\[dv\].
Let us take \[u = \sin (\log x)\;\] and \[dv = dx.\]
Now, we need to find the derivative of \[u\]and the value of $v$
\[ \Rightarrow u = \sin (\log x)\;\]
$ \Rightarrow \dfrac{{du}}{{dx}} = \dfrac{d}{{dx}}\sin (\log x)\;$
$ \Rightarrow \dfrac{{du}}{{dx}} = \dfrac{1}{x}\cos (\log x)$
With this we need to find $du$ , which is
$ \Rightarrow du = \dfrac{1}{x}\cos (\log x)dx$
Now we know that \[dv = dx.\]
$ \Rightarrow \dfrac{{dv}}{{dx}} = 1$
$ \Rightarrow v = \int {1dx} $
$ \Rightarrow v = x$
Now, substituting in the formula,\[I = uv - \smallint vdu\]
\[ \Rightarrow \int {\sin (\log x)dx} = x\sin (\log x) - \smallint \cos (\log x)dx\]
Now, we need to repeat the same process for $\int {\cos (\log x)dx} $
Let us take \[u = \cos (\log x)\;\] and \[dv = dx.\]
Now, we need to find the derivative of \[u\]and the value of $v$
\[\, \Rightarrow u = \cos (\log x)\;\]
$ \Rightarrow \dfrac{{du}}{{dx}} = \dfrac{d}{{dx}}\cos (\log x)\;$
$ \Rightarrow \dfrac{{du}}{{dx}} = - \dfrac{1}{x}\sin (\log x)$
And \[dv = dx.\]then $v = x$
Substituting in the integration
$ \Rightarrow I = x\sin (\log x) - x\cos (\log x) - \int {\sin (\log x)dx} $
We know that $\int {\sin \left( {\log x} \right)} $is the given question i.e. $I$
Therefore,
$ \Rightarrow I = x\sin (\log x) - x\cos (\log x) - I$
Transferring the $I$ to the other side,
$ \Rightarrow I + I = x\sin (\log x) - x\cos (\log x)$
$ \Rightarrow 2I = x\sin (\log x) - x\cos (\log x)$
Dividing $2$ in the right hand side,
$ \Rightarrow I = \dfrac{1}{2}x\sin (\log x) - x\cos (\log x)$

Therefore, $\int {\sin (\log x)dx = } \dfrac{1}{2}x\sin (\log x) - x\cos (\log x)$

Note: Integration by Parts formula is used for integrating the product of two functions. This method is used to find the integrals by reducing them into standard forms
Further note that, Integration by parts is not applicable for functions such as \[\smallint \;\surd x{\text{ }}sin{\text{ }}x{\text{ }}dx.\]
We do not add any constant while finding the integral of the second function.
Usually, if any function is a power of \[\;x\] or a polynomial in \[\;x\], then we take it as the first function. However, in cases where another function is an inverse trigonometric function or logarithmic function, then we take them as the first function.