Indefinite IntegrationHard
Question
The value of 
dx is equal to:
dx is equal to:Options
A.cot-1 (cot2 x) + C
B.- cot-1 (tan2 x) + C
C.tan-1 (tan2 x) + C
D.- tan-1 (cos 2 x) + C
Solution
I = 
dx =
put tan2x = t ⇒ 2 tan x sec2x dx = dt
=
= tan-1(t) + C
= tan-1(tan2x) + C =
- cot-1(tan2x) + C
= - cot-1 (tan2 x) + C1 = - cot-1
+ C1
= cot-1 (cot2x) + C1
also cos2x =
⇒ 
= tan2x, using these values in given integral
put cos 2x = t ⇒ - 2 sin2x dx = dt
⇒ I =
= - tan-1 t + C2 = - tan-1 (cos 2 x) + C2
dx = put tan2x = t ⇒ 2 tan x sec2x dx = dt=
= tan-1(t) + C= tan-1(tan2x) + C =
- cot-1(tan2x) + C= - cot-1 (tan2 x) + C1 = - cot-1
+ C1= cot-1 (cot2x) + C1
also cos2x =
⇒ = tan2x, using these values in given integral put cos 2x = t ⇒ - 2 sin2x dx = dt⇒ I =
= - tan-1 t + C2 = - tan-1 (cos 2 x) + C2Create a free account to view solution
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