Integration

next put cosφ = t^2

-sinφ dφ = 2t .dt

1+ cosφ = 1+t^2

1-cosφ = 1-t^2

and proceed

2nd method

put

cosφ = t

-sinφdφ = dt

multiply numerator and denominator by t-1

then proceed!!!!

this is my working.am not getting the answer.plz correct me.
I=₀∫¹ e^2x-[2x] dx [as [x]=0 from 0 to 1]
=₀∫^1/2 e^2xdx + _1/2∫¹ e^2x-1dx [as [2x]=0 from 0 to 1/2 and 1 from 1/2 to 1]
=e/2 + (e²-e)/2e
=e-1/2
But ans is given as e-1.
plz tell me my mistake.

yes,dint even c that earlier[3].thanks uttara and ∫.