Question

$\int \cos \left(\log x\right)dx=F\left(x\right)+c$, where $c$ is an arbitrary constant. Here $F\left(x\right)=$

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

Answer 1

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

Given : $\int \cos \left(\log x\right)dx=F\left(x\right)+c$,

Let $I=\int \cos \left(\log x\right)dx$,

Take $\log x=t,x=e^t⟹dx=e^{t}dt$

$\therefore I=\int e^t\cos tdt$

$=e^t\cos t+e^t\sin t-\int e^t\cos tdt$

$=e^t\cos t+e^t\sin t-I$
$\therefore I=\frac{e^t\cos t+e^t\sin t}{2}=\frac{x}{2}\left[\cos \right(\log x)+\sin (\log x\left)\right]$

Hence, $F\left(x\right)=\frac{x}{2}\left[\cos \right(\log x)+\sin (\log x\left)\right]$