∫1−tan2θ1+tan2θ+1−tan2θ2tanθsec2θdθ
=∫(1+tanθ)2sec2θ(1−tan2θ)dθ
=∫1+tanθsec2θ(1−tanθ)dθ
Let tanθ=t⇒sec2θdθ=dt
=∫(1+t1−t)dt=∫(−1+1+t2)dt
=−t+2ln(1+t)+C
=−tanθ+2ln(1+tanθ)+C
⇒λ=−1andf(θ)=1+tanθ
If ∫cos2θ(tan2θ+sec2θ)dθ= λtanθ+2loge∣f(θ)∣+C where C is a constant of integration, then the ordered pair (λ,f(θ)) is equal to:
Held on 9 Jan 2020 · Verified 6 Jul 2026.
(1,1−tanθ)
(−1,1−tanθ)
(−1,1+tanθ)
(1,1+tanθ)
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