The integral ∫sec3/2θ−sec1/2θ2+tan2θtanθdθ is equal to

The given integral is I=∫(secθ−1)secθ1+sec2θtanθdθ                 =∫(secθ−1)secθtanθ(1+sec2θ)secθdθ Put secθ=t⇒12secθsecθtanθdθ=dt ∴I=∫(t2−1)1+t4.2dt =2∫1−1t2t2+1t2dt=2∫(1−1t2)dt(1+1t)2−2 Put t+1t=u⇒(1−1t2)dt=du ∴I=2∫duu2−2=222loge|u−2u+2|+C =12loge|t+1t−2t+1t+2|+C =12loge|t2−2t+1t2+2t+1|+C =12loge|secθ−2secθ+1secθ+2secθ+1|+C