The integral
Question:

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

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

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

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

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

  4. (4) $x\left[\cos \left(\log _{e} x\right)-\sin \left(\log _{e} x\right)\right]+\mathrm{C}$


Correct Option: , 3

Solution:

Let the integral, $I=\int \cos (\ln x) d x$

$\Rightarrow \quad I=\cos (\ln x) \cdot x-\int \frac{-\sin (\ln x)}{x} x d x$

$=x \cos (\ln x)+\int \sin (\ln x) d x$

$=x \cos (\ln x)+\sin (\ln x) x-\int \frac{\cos (\ln x)}{x} x d x$

$=x \cos (\ln x)+\sin (\ln x) \cdot x-I$

$\Rightarrow \quad 2 I=x(\cos (\ln x)+\sin (\ln x))+C$

$\Rightarrow \quad I=\frac{x}{2}[\cos (\ln x)+\sin (\ln x)]+C$

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