sin(log x) x > 0 ,Find the integrals of the given function with respect to x.
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HELLO DEAR,
GIVEN:-
∫sin(logx).dx
put logx = t so that x = e^t and dx/x = dt or dx = e^t.dt
therefore, ∫sin(logx).dx = ∫e^t sint.dx------( 1 )
now, ∫e^t sint dt = e^t(-cost) - ∫e^t(-cost).dt [integrating by parts]
= -e^t cost + ∫e^t cost .dt
= -e^t cost + [e^t sint - ∫e^t sint.dt]
[integrating e^t cost by parts]
= -e^t cost + e^t sint - ∫e^t sint.dt
form-----( 1 )
therefore, 2∫e^t sint.dt = -e^t cost + e^t sint
=> ∫e^t sint.dt = 1/2(-e^t cost + e^tsint) + C.
putting this value in--(1), we get,
∫sin(logx).dx = ∫e^t sint.dt
= 1/2(-e^t cost + e^t sint) + C.
= 1/2[-x cos(logx) + x sin(logx)] + C.
= -1/2xcos(logx) + 1/2xsin(logx) + C.
I HOPE ITS HELP YOU DEAR,
THANKS
GIVEN:-
∫sin(logx).dx
put logx = t so that x = e^t and dx/x = dt or dx = e^t.dt
therefore, ∫sin(logx).dx = ∫e^t sint.dx------( 1 )
now, ∫e^t sint dt = e^t(-cost) - ∫e^t(-cost).dt [integrating by parts]
= -e^t cost + ∫e^t cost .dt
= -e^t cost + [e^t sint - ∫e^t sint.dt]
[integrating e^t cost by parts]
= -e^t cost + e^t sint - ∫e^t sint.dt
form-----( 1 )
therefore, 2∫e^t sint.dt = -e^t cost + e^t sint
=> ∫e^t sint.dt = 1/2(-e^t cost + e^tsint) + C.
putting this value in--(1), we get,
∫sin(logx).dx = ∫e^t sint.dt
= 1/2(-e^t cost + e^t sint) + C.
= 1/2[-x cos(logx) + x sin(logx)] + C.
= -1/2xcos(logx) + 1/2xsin(logx) + C.
I HOPE ITS HELP YOU DEAR,
THANKS
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