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

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

Explanation

Solution

In this question they have given∫sin⁡(log⁡x)\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 uu and the other as dvdv and have to find its derivative and the value of vv. Then we need to substitute it in the formula I=uv−∫vduI = uv - \smallint vdu and then simply and solve it to find the correct answer.

Formula used: Formula for integrating in parts method:
I=uv−∫vduI = uv - \smallint vdu

Complete step-by-step solution:
In this question they have given∫sin⁡(log⁡x)\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−∫vduI = uv - \smallint vdu
First we need to decide which one to choose as uuand which one asdvdv.
Let us take u=sin⁡(log⁡x)  u = \sin (\log x)\; and dv=dx.dv = dx.
Now, we need to find the derivative of uuand the value of vv
⇒u=sin⁡(log⁡x)  \Rightarrow u = \sin (\log x)\;
⇒dudx=ddxsin⁡(log⁡x)  \Rightarrow \dfrac{{du}}{{dx}} = \dfrac{d}{{dx}}\sin (\log x)\;
⇒dudx=1xcos⁡(log⁡x)\Rightarrow \dfrac{{du}}{{dx}} = \dfrac{1}{x}\cos (\log x)
With this we need to find dudu , which is
⇒du=1xcos⁡(log⁡x)dx\Rightarrow du = \dfrac{1}{x}\cos (\log x)dx
Now we know that dv=dx.dv = dx.
⇒dvdx=1\Rightarrow \dfrac{{dv}}{{dx}} = 1
⇒v=∫1dx\Rightarrow v = \int {1dx}
⇒v=x\Rightarrow v = x
Now, substituting in the formula,I=uv−∫vduI = uv - \smallint vdu
⇒∫sin⁡(log⁡x)dx=xsin⁡(log⁡x)−∫cos⁡(log⁡x)dx\Rightarrow \int {\sin (\log x)dx} = x\sin (\log x) - \smallint \cos (\log x)dx
Now, we need to repeat the same process for ∫cos⁡(log⁡x)dx\int {\cos (\log x)dx}
Let us take u=cos⁡(log⁡x)  u = \cos (\log x)\; and dv=dx.dv = dx.
Now, we need to find the derivative of uuand the value of vv
 ⇒u=cos⁡(log⁡x)  \, \Rightarrow u = \cos (\log x)\;
⇒dudx=ddxcos⁡(log⁡x)  \Rightarrow \dfrac{{du}}{{dx}} = \dfrac{d}{{dx}}\cos (\log x)\;
⇒dudx=−1xsin⁡(log⁡x)\Rightarrow \dfrac{{du}}{{dx}} = - \dfrac{1}{x}\sin (\log x)
And dv=dx.dv = dx.then v=xv = x
Substituting in the integration
⇒I=xsin⁡(log⁡x)−xcos⁡(log⁡x)−∫sin⁡(log⁡x)dx\Rightarrow I = x\sin (\log x) - x\cos (\log x) - \int {\sin (\log x)dx}
We know that ∫sin⁡(log⁡x)\int {\sin \left( {\log x} \right)} is the given question i.e. II
Therefore,
⇒I=xsin⁡(log⁡x)−xcos⁡(log⁡x)−I\Rightarrow I = x\sin (\log x) - x\cos (\log x) - I
Transferring the II to the other side,
⇒I+I=xsin⁡(log⁡x)−xcos⁡(log⁡x)\Rightarrow I + I = x\sin (\log x) - x\cos (\log x)
⇒2I=xsin⁡(log⁡x)−xcos⁡(log⁡x)\Rightarrow 2I = x\sin (\log x) - x\cos (\log x)
Dividing 22 in the right hand side,
⇒I=12xsin⁡(log⁡x)−xcos⁡(log⁡x)\Rightarrow I = \dfrac{1}{2}x\sin (\log x) - x\cos (\log x)

Therefore, ∫sin⁡(log⁡x)dx=12xsin⁡(log⁡x)−xcos⁡(log⁡x)\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 ∫  √x sin x dx.\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\;x or a polynomial in   x\;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.