int {cos (log_e x)} , dx is equal to | vedclass

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(A) 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

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$\frac{1}{2}x\{ \cos (\log_e x) + \sin (\log_e x)\} + C$

Vedclass · Answer

(A) 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 \cdot (-\sin (\log_e x)) \cdot \frac{1}{x} \, dx$$I = x \cos (\log_e x) + \int \sin (\log_e x) \, dx$Now,apply integration by parts again to $\int \sin (\log_e x) \, dx$ with $u = \sin (\log_e x)$ and $dv = dx$.$I = x \cos (\log_e x) + [x \sin (\log_e x) - \int x \cdot \cos (\log_e x) \cdot \frac{1}{x} \, dx]$$I = x \cos (\log_e x) + x \sin (\log_e x) - I$$2I = x [\cos (\log_e x) + \sin (\log_e x)]$$I = \frac{1}{2} x [\cos (\log_e x) + \sin (\log_e x)] + C$.

$\int {\cos (\log_e x)} \, dx$ is equal to

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