The integral int {cos left( {{{log }_e}x} rig | vedclass

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Question

(C) Let $I = \int \cos(\log_e x) dx$. Using integration by parts,let $u = \cos(\log_e x)$ and $dv = dx$. Then $du = -\sin(\log_e x) \cdot \frac{1}{x}

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let $I = \int \cos(\log_e x) dx$. Using integration by parts,let $u = \cos(\log_e x)$ and $dv = dx$. Then $du = -\sin(\log_e x) \cdot \frac{1}{x} dx$ and $v = x$. $I = x \cos(\log_e x) - \int x \left( -\sin(\log_e x) \cdot \frac{1}{x} \right) dx$ $I = x \cos(\log_e x) + \int \sin(\log_e x) dx$. Now,apply integration by parts to $\int \sin(\log_e x) dx$: Let $u = \sin(\log_e x)$ and $dv = dx$. Then $du = \cos(\log_e x) \cdot \frac{1}{x} dx$ and $v = x$. $\int \sin(\log_e x) dx = x \sin(\log_e x) - \int x \left( \cos(\log_e x) \cdot \frac{1}{x} \right) dx = x \sin(\log_e x) - I$. Substituting this back into the equation for $I$: $I = x \cos(\log_e x) + x \sin(\log_e x) - I$ $2I = x(\cos(\log_e x) + \sin(\log_e x))$ $I = \frac{x}{2}(\cos(\log_e x) + \sin(\log_e x)) + C$.

The integral $\int {\cos \left( {{{\log }_e}x} \right)dx} $ is equal to: (where $C$ is a constant of integration)

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