If I = int sin(log x) , dx,then I is given by | vedclass

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(B) Let $\log x = t$,then $x = e^t$. Differentiating with respect to $t$,we get $dx = e^t \, dt$. Substituting these into the integral,we have: $I = \

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$\frac{x}{2}(\sin(\log x) - \cos(\log x)) + c$,where $c$ is a constant of integration.

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(B) Let $\log x = t$,then $x = e^t$. Differentiating with respect to $t$,we get $dx = e^t \, dt$. Substituting these into the integral,we have: $I = \int \sin(t) e^t \, dt$. Using integration by parts,$\int u \, dv = uv - \int v \, du$,let $u = \sin t$ and $dv = e^t \, dt$. Then $du = \cos t \, dt$ and $v = e^t$. $I = e^t \sin t - \int e^t \cos t \, dt$. Applying integration by parts again for $\int e^t \cos t \, dt$: $I = e^t \sin t - [e^t \cos t - \int e^t (-\sin t) \, dt]$. $I = e^t \sin t - e^t \cos t - \int e^t \sin t \, dt$. $I = e^t \sin t - e^t \cos t - I$. $2I = e^t(\sin t - \cos t) + c$. Substituting $t = \log x$ and $e^t = x$: $2I = x(\sin(\log x) - \cos(\log x)) + c$. $I = \frac{x}{2}(\sin(\log x) - \cos(\log x)) + c$.

If $I = \int \sin(\log x) \, dx$,then $I$ is given by

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