Question

Solve this integral​

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:-

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

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

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

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

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

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

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

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

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

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

Answer 2

Answer:

$\large\mathbb{\underbrace\purple{ANSWER}}$

Given,

$\displaystyle\longrightarrow I=\int e^{2\log\cot x}$ $dx⟶I=∫e2logcotx dx$

$Since e^{b\log a}=a^b,ebloga=ab,$

$\displaystyle\longrightarrow I=\int \cot^2x\ dx⟶I=∫cot2x dx$

$Since \csc^2x-\cot^2x=1,csc2x−cot2x=1,$

$\displaystyle\longrightarrow I=\int(\csc^2x-1)\ dx⟶I=∫(csc2x−1) dx$

$\displaystyle\longrightarrow I=\int\csc^2x\ dx-\int dx⟶I=∫csc2x dx−∫dx$

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

Hence (4) is the answer.

Some Integral Results:-

$\displaystyle\int\cos x\ dx=\sin x+c∫cosx dx=sinx+c$

$\displaystyle\int\sin x\ dx=-\cos x+c∫sinx dx=−cosx+c$

$\displaystyle\int\sec^2x\ dx=\tan x+c∫sec2x dx=tanx+c$

$\displaystyle\int\csc^2x\ dx=-\cot x+c∫csc2x dx=−cotx+c$

$\displaystyle\int\sec x\tan x\ dx=\sec x+c∫secxtanx dx=secx+c$

$\displaystyle\int\csc x\cot x\ dx=-\csc x+c∫cscxcotx dx=−cscx+c</p><p>\displaystyle\int\tan x\ dx=\log|\sec x|+c∫tanx dx=log∣secx∣+c$

$\displaystyle\int\cot x\ dx=-\log|\csc x|+c∫cotx dx=−log∣cscx∣+c</p><p>\displaystyle\int\sec x\ dx=\log|\sec x+\tan x|+c∫secx dx=log∣secx+tanx∣+c</p><p>\displaystyle\int\csc x\ dx=-\log|\csc x+\cot x|+c∫cscx dx=−log∣cscx+cotx∣+c$