the value of ∫etanθ(secθ−sinθ)dθis

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etanθsecθ+c

etanθsinθ+c

etanθ(secθ+sinθ)+c

etanθcosθ+c

answer is D.

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Detailed Solution

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

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