Question

$\int \sin \left(\log x\right)dx=$

• A. $\frac{x}{2}\left\{\sin \right(\log x)+\cos (\log x\left)\right\}+c$
• B. $\frac{x}{2}\left\{\cos \right(\log x)-\sin (\log x\left)\right\}+c$
• C. $\frac{x}{2}\left\{\sin \right(\log x)-\cos (\log x\left)\right\}+c$
• D. None of these

Answer 1

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

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