can any body show me the steps to solve this integral problem
The answer is: e-1
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I = sec²x.e^tanx.dx where Limit [ 0, π/4]
Let ,
tanx = z
differentiate ,
sec²x.dx = dz
put this in integration ,
I = e^z { tanx.dx}
I = e^z.dz
I =[ e^z ]
now put z = tanx
I = [ e^tanx ] now take Limit
I = { e^tanπ/4 - e^0 }
I = {e^1 - 1}
I = e -1
Let ,
tanx = z
differentiate ,
sec²x.dx = dz
put this in integration ,
I = e^z { tanx.dx}
I = e^z.dz
I =[ e^z ]
now put z = tanx
I = [ e^tanx ] now take Limit
I = { e^tanπ/4 - e^0 }
I = {e^1 - 1}
I = e -1
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