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

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(D) Let $I = \int [\sin (\log x) + \cos (\log x)] \;dx$. We can split the integral as $I = \int \sin (\log x) \;dx + \int \cos (\log x) \;dx$. Conside

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

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(D) Let $I = \int [\sin (\log x) + \cos (\log x)] \;dx$. We can split the integral as $I = \int \sin (\log x) \;dx + \int \cos (\log x) \;dx$. Consider the first integral $\int \sin (\log x) \;dx$. Let $u = \log x$,then $x = e^u$ and $dx = e^u \;du$. So,$\int \sin (\log x) \;dx = \int e^u \sin u \;du$. Using the formula $\int e^u \sin u \;du = \frac{e^u}{2} (\sin u - \cos u) + c_1$,we get $\int \sin (\log x) \;dx = \frac{x}{2} [\sin (\log x) - \cos (\log x)] + c_1$. Similarly,$\int \cos (\log x) \;dx = \int e^u \cos u \;du = \frac{e^u}{2} (\sin u + \cos u) + c_2 = \frac{x}{2} [\sin (\log x) + \cos (\log x)] + c_2$. Adding these,$I = \frac{x}{2} [\sin (\log x) - \cos (\log x)] + \frac{x}{2} [\sin (\log x) + \cos (\log x)] + c$. $I = \frac{x}{2} [2 \sin (\log x)] + c = x \sin (\log x) + c$.

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

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