the value of ∫etanθ(secθ−sinθ)dθis

Let I=∫etanθ(secθ−sinθ)dθPut tanθ=t⇒sec2θdθ=dt⇒dθ=dt1+t2⇒I=∫et(1+t2−t1+t2)dt1+t2=∫et(11+t2−1(1+t2)3/2)dt=∫et(f+f')dt=et.f(t)=et1+t2+c     =etanθcosθ+c