If int [ cos(x) cdot frac{d}{dx}(csc(x)) ] dx | vedclass

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(A) We are given the integral $I = \int [ \cos(x) \cdot \frac{d}{dx}(\csc(x)) ] dx$. First,we know that $\frac{d}{dx}(\csc(x)) = -\csc(x) \cot(x)$. Su

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$x \cot(x)$

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(A) We are given the integral $I = \int [ \cos(x) \cdot \frac{d}{dx}(\csc(x)) ] dx$. First,we know that $\frac{d}{dx}(\csc(x)) = -\csc(x) \cot(x)$. Substituting this into the integral,we get: $I = \int [ \cos(x) \cdot (-\csc(x) \cot(x)) ] dx$ $I = - \int [ \cos(x) \cdot \frac{1}{\sin(x)} \cdot \frac{\cos(x)}{\sin(x)} ] dx$ $I = - \int [ \frac{\cos^2(x)}{\sin^2(x)} ] dx$ $I = - \int \cot^2(x) dx$ Using the identity $\cot^2(x) = \csc^2(x) - 1$,we have: $I = - \int (\csc^2(x) - 1) dx$ $I = - \int \csc^2(x) dx + \int 1 dx$ Since $\int \csc^2(x) dx = -\cot(x)$,we get: $I = -(-\cot(x)) + x + c$ $I = \cot(x) + x + c$ Comparing this with $f(x) + g(x) + c$,we have $f(x) = \cot(x)$ and $g(x) = x$ (or vice versa). Therefore,$f(x) \cdot g(x) = x \cot(x)$.

If $\int [ \cos(x) \cdot \frac{d}{dx}(\csc(x)) ] dx = f(x) + g(x) + c$,then $f(x) \cdot g(x) =$

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