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If $\int\frac{f(x)}{\log(\sin x)}dx=\log[\log \sin x]+c$ , then f(x)  is equal to

Answer:

Explanation:

Given,   $\int\frac{f(x)}{\log(\sin x)}dx=\log[\log \sin x]+c$

On differentiating both sides , we get

$\frac{f(x)}{\log (\sin x)}=\frac{1}{\log \sin x}\frac{d}{dx}(\log \sin x)+0$

 $\Rightarrow$    $\frac{f(x)}{\log (\sin x)}=\frac{1}{\log \sin x}\times\frac{1}{\sin x}\times \cos x$

 $\Rightarrow$    f(x)  = cot x