The value of int cos (log x) , dx is | vedclass

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Question

(B) Let $I = \int \cos (\log x) \cdot 1 \, dx \dots (i)$ Using integration by parts,$\int u \, dv = uv - \int v \, du$. Let $u = \cos (\log x)$ and $d

Vedclass · Accepted Answer

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

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(B) Let $I = \int \cos (\log x) \cdot 1 \, dx \dots (i)$ Using integration by parts,$\int u \, dv = uv - \int v \, du$. Let $u = \cos (\log x)$ and $dv = 1 \, dx$. Then $du = -\sin (\log x) \cdot \frac{1}{x} \, dx$ and $v = x$. $I = x \cos (\log x) - \int x \cdot (-\sin (\log x)) \cdot \frac{1}{x} \, dx$ $I = x \cos (\log x) + \int \sin (\log x) \, dx$ Now,apply integration by parts again to $\int \sin (\log x) \cdot 1 \, dx$: $I = x \cos (\log x) + [x \sin (\log x) - \int x \cdot \cos (\log x) \cdot \frac{1}{x} \, dx] + C$ $I = x \cos (\log x) + x \sin (\log x) - \int \cos (\log x) \, dx + C$ $I = x [\cos (\log x) + \sin (\log x)] - I + C$ $2I = x [\sin (\log x) + \cos (\log x)] + C$ $I = \frac{x}{2} [\sin (\log x) + \cos (\log x)] + C$

The value of $\int \cos (\log x) \, dx$ is

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