Question

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

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

Answer 1

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

$I=\int \cos \left({\log }_{e}x\right)dx$
Put ${\log }_{e}x=t$
$\Rightarrow \frac{1}{x}dx=dt$
$\Rightarrow dx=e^{t}dt$
$\Rightarrow I=\int \cos t.e^{t}dt$
$=\cos t\int e^{t}dt-\int \left(\frac{d}{dt}\right(\cos t\left)\int e^{t}dt\right)dt$
$=\cos t.e^t+\int \sin t{e}^{t}dt$
$\Rightarrow I=e^t\cos t+e^t\sin t-\int \cos t.e^{t}dt$
$\Rightarrow 2I=e^t\cos t+e^t\sin t$

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