The value of int cos (log _e x) dx is equal t | vedclass

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(C) Let $I = \int \cos (\log _e x) dx$. Substitute $\log _e x = t$,which implies $x = e^t$ and $dx = e^t dt$. Then,$I = \int e^t \cos t dt$. Using the

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$\frac{x}{2}[\sin (\log x)+\cos (\log x)]+C$

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(C) Let $I = \int \cos (\log _e x) dx$. Substitute $\log _e x = t$,which implies $x = e^t$ and $dx = e^t dt$. Then,$I = \int e^t \cos t dt$. Using the integration by parts formula $\int e^t f(t) dt = e^t f(t) - \int e^t f'(t) dt$,we have: $I = e^t \cos t - \int e^t (-\sin t) dt = e^t \cos t + \int e^t \sin t dt$. Applying integration by parts again to $\int e^t \sin t dt$: $I = e^t \cos t + [e^t \sin t - \int e^t \cos t dt]$. $I = e^t \cos t + e^t \sin t - I$. $2I = e^t (\cos t + \sin t)$. $I = \frac{e^t}{2} (\cos t + \sin t) + C$. Substituting $e^t = x$ and $t = \log _e x$: $I = \frac{x}{2} [\cos (\log _e x) + \sin (\log _e x)] + C$.

The value of $\int \cos (\log _e x) dx$ is equal to (where $C$ is a constant of integration.)

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