$\int\{\sin (\log x)+\cos (\log x)\} d x$
Answer & explanation
Correct answer: option 1
Put log x = t ⇒ dx = et dt
x = et
$I=\int e^t(\sin t+\cos t) d t=e^{t} \sin t=x \sin (\log x)+c$
Hence (1) is the correct answer.
$\int\{\sin (\log x)+\cos (\log x)\} d x$
Correct answer: option 1
Put log x = t ⇒ dx = et dt
x = et
$I=\int e^t(\sin t+\cos t) d t=e^{t} \sin t=x \sin (\log x)+c$
Hence (1) is the correct answer.