The ∫(1−cos⁡x)cosec⁡e2x dx=f(x)+c⇒f(x)=

∫2sin2⁡x/2(2sin⁡x/2cos⁡x/2)2dx =∫2sin2⁡x/24sin2⁡x/2cos2⁡x/2dx =12∫sec2⁡x/2dx=12tan⁡x2x2+c=tan⁡x2+c

The ∫(1−cos⁡x)cosec⁡e2x dx=f(x)+c⇒f(x)=

∫2sin2⁡x/2(2sin⁡x/2cos⁡x/2)2dx =∫2sin2⁡x/24sin2⁡x/2cos2⁡x/2dx =12∫sec2⁡x/2dx=12tan⁡x2x2+c=tan⁡x2+c

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The $\int (1 - \cos x) cosec e^{2} x$ $dx = f (x) + c \Rightarrow f (x) =$

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tan x/2

2tan x/2

cot x/2

$\frac{1}{2} \tan x / 2$

answer is A.

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Detailed Solution

$\begin{array}{l} \int \frac{2 \sin^{2} x / 2}{(2 \sin x / 2 \cos x / 2)^{2}} dx \\ = \int \frac{2 \sin^{2} x / 2}{4 \sin^{2} x / 2 \cos^{2} x / 2} dx \\ = \frac{1}{2} \int \sec^{2} x / 2 dx \\ = \frac{1}{2} \tan \frac{x}{2} x 2 + c \\ = \tan \frac{x}{2} + c \end{array}$