Question

# $$\displaystyle \int \cot^{-1} (\sec x-{\it \tan x} )dx =\ldots\ldots$$

A
πx4x24+c
B
πx4+x24+c
C
x24+c
D
x22+c

Solution

## The correct option is A $$\displaystyle \frac{\pi x}{4}+\frac{x^{2}}{4}+c$$$$\displaystyle I= \int \cot^{-1} (\sec x-{\tan x} )dx$$$$\displaystyle =\int \cot^{-1} \left(\dfrac {1-\sin x}{\cos x}\right)$$$$\displaystyle =\int \cot ^{ -1 } \left(\dfrac { \cos{ \dfrac { x }{ 2 } }-\sin { \dfrac { x }{ 2 } } }{ \cos{ \dfrac { x }{ 2 } }+\sin { \dfrac { x }{ 2 } } } \right)dx$$$$\displaystyle =\int \cot ^{ -1 } \left(\dfrac { 1-\tan { \dfrac { x }{ 2 } } }{ 1+\tan { \dfrac { x }{ 2 } } } \right)dx$$$$\displaystyle =\int \cot ^{ -1 } \left(\tan { \left(\dfrac { \pi }{ 4 } -\dfrac { x }{ 2 } \right)}\right)dx$$$$\displaystyle =\int \cot^{ -1 } \left(\cot\left(\dfrac { \pi }{ 2 } -\dfrac { \pi }{ 4 } +\dfrac { x }{ 2 } \right)\right)dx$$$$\Rightarrow \displaystyle I= \int \left(\dfrac { \pi }{ 4 } +\dfrac { x }{ 2 }\right) dx$$$$\Rightarrow \displaystyle I= \dfrac{\pi x}{4}+\dfrac{x^{2}}{4}+c$$Mathematics

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