Calculus

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Evaluate the indefinite integral

integral sec(t/2) dt=

a)ln |sec t +tan t| +C
b)ln |sec (t/2) +tan (t/2)| +C
c)2tan^2 (t/2)+C
d)2ln cos(t/2) +C
e)2ln |sec (t/2)+tan (t/2)| +C

  • Calculus -

    integral sec(t/2) dt =
    2*integral sec(t/2) d(t/2)=
    2*ln[sec (t/2) + tan (t/2)] +C

  • Calculus -

    Thanks

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