Question

∫e²log cot x dx =

1) cot x-x+c 2) cot x+x+c 3) tan x-x+c 4)-cot x-x+c​

Answer 1

Given,

$\displaystyle\longrightarrow I=\int e^{2\log\cot x}\ dx$

Since $e^{b\log a}=a^b,$

$\displaystyle\longrightarrow I=\int \cot^2x\ dx$

Since $\csc^2x-\cot^2x=1,$

$\displaystyle\longrightarrow I=\int(\csc^2x-1)\ dx$

$\displaystyle\longrightarrow I=\int\csc^2x\ dx-\int dx$

$\displaystyle\longrightarrow\underline{\underline{I=-\cot x-x+c}}$

Hence (4) is the answer.

Some Integral Results:-

$\displaystyle\int\cos x\ dx=\sin x+c$

$\displaystyle\int\sin x\ dx=-\cos x+c$

$\displaystyle\int\sec^2x\ dx=\tan x+c$

$\displaystyle\int\csc^2x\ dx=-\cot x+c$

$\displaystyle\int\sec x\tan x\ dx=\sec x+c$

$\displaystyle\int\csc x\cot x\ dx=-\csc x+c$

$\displaystyle\int\tan x\ dx=\log|\sec x|+c$

$\displaystyle\int\cot x\ dx=-\log|\csc x|+c$

$\displaystyle\int\sec x\ dx=\log|\sec x+\tan x|+c$

$\displaystyle\int\csc x\ dx=-\log|\csc x+\cot x|+c$