Indefinite IntegrationHard
Question
∫sin2(lnx) dx is equal to -
Options
A.
(5 + 2sin (2lnx)) + c
(5 + 2sin (2lnx)) + cB.(5 + 2sin (2lnx)) - cos (2lnx)) + c
(5 + 2sin (2lnx)) - cos (2lnx)) + cC.(5 - 2sin(2lnx) - cos (2lnx)) + c
(5 - 2sin(2lnx) - cos (2lnx)) + cD.(5 - 2sin (2lnx) + cos (2lnx)) + c
(5 - 2sin (2lnx) + cos (2lnx)) + cSolution
I = ∫sin2(lnx) dx,
Let t = lnx ⇒ dt =
⇒ dx = et dt
⇒ I = ∫etsin2t dt
I =
∫et(1 - cos 2t)dt
⇒ 2I = et - ∫et cos 2t dt
Let I1 = ∫et cos 2t dt
=
[cos 2t + 2 sin 2t] + C
(∵ ∫eax cos bx dx =
[ a cos bx + b sin bx] )
⇒ I =
et (5 - 2sin2 t - cos2t) + C
=
(5 - 2sin(2lnx)) + C
Let t = lnx ⇒ dt =
⇒ dx = et dt⇒ I = ∫etsin2t dt
I =
∫et(1 - cos 2t)dt⇒ 2I = et - ∫et cos 2t dt
Let I1 = ∫et cos 2t dt
=
[cos 2t + 2 sin 2t] + C(∵ ∫eax cos bx dx =
[ a cos bx + b sin bx] )⇒ I =
et (5 - 2sin2 t - cos2t) + C=
(5 - 2sin(2lnx)) + C Create a free account to view solution
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