If ∫dθcos⁡2θ(tan⁡2θ+sec⁡2θ)=λtan⁡θ+2log⁡e∣f(θ)∣+C\int \frac{d\theta}{\cos^2 \theta (\tan 2\theta + \sec 2\theta)} = \lambda \tan \theta + 2 \log_e |f(\theta)| + C…

Mathematics

Problem

If ∫dθcos⁡2θ(tan⁡2θ+sec⁡2θ)=λtan⁡θ+2log⁡e∣f(θ)∣+C\int \frac{d\theta}{\cos^2 \theta (\tan 2\theta + \sec 2\theta)} = \lambda \tan \theta + 2 \log_e |f(\theta)| + C, where CC is a constant of integration, then the ordered pair (λ,f(θ))(\lambda, f(\theta)) is equal to (a) (−1,1−tan⁡θ)(-1, 1 - \tan \theta) (b) (−1,1+tan⁡θ)(-1, 1 + \tan \theta) (c) (1,1+tan⁡θ)(1, 1 + \tan \theta) (d) (1,1−tan⁡θ)(1, 1 - \tan \theta)

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