If the primitive of cos(log x) is f(x){cos(g( | vedclass

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(B) The primitive means the indefinite integral. Let $I = \int \cos(\log x) dx$. Using integration by parts,let $u = \cos(\log x)$ and $dv = dx$. Then

Vedclass · Accepted Answer

$f^{\prime}(x) = \frac{1}{2}$

Vedclass · Answer

(B) The primitive means the indefinite integral. Let $I = \int \cos(\log x) dx$. Using integration by parts,let $u = \cos(\log x)$ and $dv = dx$. Then $du = -\sin(\log x) \cdot \frac{1}{x} dx$ and $v = x$. $I = x \cos(\log x) - \int x \cdot (-\sin(\log x)) \cdot \frac{1}{x} dx = x \cos(\log x) + \int \sin(\log x) dx$. Now,integrate $\int \sin(\log x) dx$ by parts: $u = \sin(\log x)$,$dv = dx$. $I = x \cos(\log x) + [x \sin(\log x) - \int x \cdot \cos(\log x) \cdot \frac{1}{x} dx]$. $I = x \cos(\log x) + x \sin(\log x) - I$. $2I = x [\cos(\log x) + \sin(\log x)] + C$. $I = \frac{x}{2} [\cos(\log x) + \sin(\log x)] + C$. Comparing this with $f(x)\{\cos(g(x)) + \sin(h(x))\}$,we get $f(x) = \frac{x}{2}$,$g(x) = \log x$,and $h(x) = \log x$. Thus,$f^{\prime}(x) = \frac{d}{dx}(\frac{x}{2}) = \frac{1}{2}$. Therefore,option $B$ is correct.

If the primitive of $\cos(\log x)$ is $f(x)\{\cos(g(x)) + \sin(h(x))\}$,then which among the following is true?

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