Question

The integral $\int \cos \left({\log }_{e}x\right)dx$ is equal to :
(where $C$ is a constant of integration)

• A. $\frac{x}{2}[\sin \left({\log }_{e}x\right)-\cos \left({\log }_{e}x\right)]+C$
• B. $\frac{x}{2}[\cos \left({\log }_{e}x\right)+\sin \left({\log }_{e}x\right)]+C$
• C. $x[\cos \left({\log }_{e}x\right)+\sin \left({\log }_{e}x\right)]+C$
• D. $x[\cos \left({\log }_{e}x\right)-\sin \left({\log }_{c}x\right)]+C$

Answer 1

Answer: (B)

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

$I=\int \cos \left(\ell nx\right)dx$
$I=\cos \left(\ln x\right)\cdot x+\int \sin \left(\ell nx\right)dx$
$\cos \left(\ell nx\right)x+\left[\sin \right(\ell nx)\cdot x-\int \cos (\ell nx\left)dx\right]$
$I=\frac{x}{2}\left[\sin \right(\ell nx)+\cos (\ell nx\left)\right]+C$