int [sin(log x) + cos(log x)] dx = | vedclass

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(D) Let $I = \int [\sin(\log x) + \cos(\log x)] dx$. Substitute $\log x = t$,which implies $x = e^t$ and $dx = e^t dt$. Substituting these into the in

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$x \sin(\log x) + c$

Vedclass · Answer

(D) Let $I = \int [\sin(\log x) + \cos(\log x)] dx$. Substitute $\log x = t$,which implies $x = e^t$ and $dx = e^t dt$. Substituting these into the integral,we get: $I = \int (\sin t + \cos t) e^t dt$. Using the standard integral formula $\int e^t [f(t) + f'(t)] dt = e^t f(t) + c$,where $f(t) = \sin t$ and $f'(t) = \cos t$. Therefore,$I = e^t \sin t + c$. Substituting back $t = \log x$,we get $I = x \sin(\log x) + c$.

$\int [\sin(\log x) + \cos(\log x)] dx = $

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