If ∫ dθ cos^2θ ( tan2θ + sec2θ )= λ tanθ +2 log_e|f(θ )|+C where C is a constant of integration, then the ordered pair (λ ,f(θ )) is equal to:

$$\int \frac{{\mathrm{sec}}^{2}\theta }{\frac{1+{\mathrm{tan}}^{2}\theta }{1-{\mathrm{tan}}^{2}\theta }+\frac{2\mathrm{tan}\theta }{1-{\mathrm{tan}}^{2}\theta }}d\theta$$∫1−tan2θ1+tan2θ​+1−tan2θ2tanθ​sec2θ​dθ

$$=\int \frac{{\mathrm{sec}}^{2}\theta (1-{\mathrm{tan}}^{2}\theta )}{{(1+tan\theta )}^{2}}d\theta$$=∫(1+tanθ)2sec2θ(1−tan2θ)​dθ

$$=\int \frac{{\mathrm{sec}}^{2}\theta (1-\mathrm{tan}\theta )}{1+\mathrm{tan}\theta }d\theta$$=∫1+tanθsec2θ(1−tanθ)​dθ

Let $\mathrm{tan}\theta =t\Rightarrow {\mathrm{sec}}^{2}\theta d\theta =dt$tanθ=t⇒sec2θdθ=dt

$$=\int (\frac{1-t}{1+t})dt=\int (-1+\frac{2}{1+t})dt$$=∫(1+t1−t​)dt=∫(−1+1+t2​)dt

$$=-t+2\mathrm{ln}(1+t)+C$$=−t+2ln(1+t)+C

$$=-\mathrm{tan}\theta +2\mathrm{ln}(1+\mathrm{tan}\theta )+C$$=−tanθ+2ln(1+tanθ)+C

$$\Rightarrow \lambda =-1andf(\theta )=1+\mathrm{tan}\theta$$⇒λ=−1andf(θ)=1+tanθ