Indefinite IntegrationHard
Question
The value of ∫[1 + tan x. tan(x + α)] dx is equal to
Options
A.cos α . ln 
+ C
+ CB.tan α . ln + C
+ CC.cot α . ln 
+ C
+ CD.cot α . ln 
+ C
+ CSolution
∫(1 + tan x tan(x + α))dx
= cot α
= cot α
= cot α [∫tan(x + α)dx - ∫tan x dx]
= cot α [ln|sec|(x + α)|-ln|sec x]
= cot α ln
+ C
= cot α ln
+ C
= cot α
= cot α
= cot α [∫tan(x + α)dx - ∫tan x dx]
= cot α [ln|sec|(x + α)|-ln|sec x]
= cot α ln
+ C= cot α ln
+ CCreate a free account to view solution
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