If ∫2dxcosxsin2x=f(x)+c,then f(x)=

∫2 dxcos xsin2x=f(x)+c =∫2 dxcos x2 sin x cos x =∫dxcos2x tan x =∫ dtt, Put tan x t sec2x dx=dt =2t+c =2tan x+c

If ∫2dxcosxsin2x=f(x)+c,then f(x)=

∫2 dxcos xsin2x=f(x)+c =∫2 dxcos x2 sin x cos x =∫dxcos2x tan x =∫ dtt, Put tan x t sec2x dx=dt =2t+c =2tan x+c

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If $\int \frac{\sqrt{2} d x}{\cos x \sqrt{\sin 2 x}} = f (x) + c, t h e n f (x) =$

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$2 \sqrt{s e c x}$

$\sqrt{\tan x}$

$2 \sqrt{\tan x}$

$2 \sqrt{2} \sqrt{\tan x}$

answer is C.

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Detailed Solution

$\int \frac{\sqrt{2} dx}{\cos x \sqrt{\sin 2 x}} = f (x) + c = \int \frac{\sqrt{2} dx}{\cos x \sqrt{2 s in x \cos x}} = \int \frac{dx}{\cos^{2} x \sqrt{\tan x}} = \int \frac{dt}{\sqrt{t}}, Put \tan x t \sec^{2} x dx = dt = 2 \sqrt{t} + c = 2 \sqrt{\tan x} + c$