Integral Calculus Question 300

Question: $ \int_{{}}^{{}}{\frac{\sin x+cosec,x}{\tan x}dx=} $

Options:

A) $ \sin x-cosec,x+c $

B) $ cosec,x-\sin x+c $

C) $ \log \tan x+c $

D) $ \log \cot x+c $

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Answer:

Correct Answer: A

Solution:

$ \int_{{}}^{{}}{\frac{\sin x+cosec,x}{\tan x}},dx=\int_{{}}^{{}}{(\cos x+cosec,x\cot x),dx} $ $ =\sin x-cosec,x+c $