The integral

Question:
The integral $\int \cos \left(\log _{\mathrm{e}} \mathrm{x}\right) \mathrm{dx}$ is equal to :

(where $\mathrm{C}$ is a constant of integration)

$\frac{x}{2}\left[\sin \left(\log _{e} x\right)-\cos \left(\log _{e} x\right)\right]+C$
$\frac{x}{2}\left[\cos \left(\log _{e} x\right)+\sin \left(\log _{e} x\right)\right]+C$
$x\left[\cos \left(\log _{e} x\right)+\sin \left(\log _{e} x\right)\right]+C$
$x\left[\cos \left(\log _{e} x\right)-\sin \left(\log _{e} x\right)\right]+C$

Correct Option: , 2

Solution:

$I=\int \cos (\ell n x) d x$

$I=\cos (\ln x) \cdot x+\int \sin (\ell n x) d x$

$\cos (\ell n x) x+\left[\sin (\ell n x) \cdot x-\int \cos (\ell n x) d x\right]$

$\mathrm{I}=\frac{\mathrm{x}}{2}[\sin (\ell \mathrm{nx})+\cos (\ell \mathrm{n} x)]+\mathrm{C}$

The integral

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