int [sin (log x) + cos (log x)] , dx is equal | vedclass

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Question

(C) Let $I = \int [\sin (\log x) + \cos (\log x)] \, dx$. We know that $\frac{d}{dx} [x \sin (\log x)] = \sin (\log x) \cdot \frac{d}{dx}(x) + x \cdot

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let $I = \int [\sin (\log x) + \cos (\log x)] \, dx$. We know that $\frac{d}{dx} [x \sin (\log x)] = \sin (\log x) \cdot \frac{d}{dx}(x) + x \cdot \cos (\log x) \cdot \frac{d}{dx}(\log x)$. $= \sin (\log x) \cdot 1 + x \cdot \cos (\log x) \cdot \frac{1}{x}$. $= \sin (\log x) + \cos (\log x)$. Therefore,$\int [\sin (\log x) + \cos (\log x)] \, dx = \int \frac{d}{dx} [x \sin (\log x)] \, dx$. $= x \sin (\log x) + c$.

$\int [\sin (\log x) + \cos (\log x)] \, dx$ is equal to

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