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If ∫ d θ cos 2 θ ( tan 2 θ + sec 2 θ ) = λ tan θ + 2 log 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 ∫ c o s 2 θ ( t a n 2 θ + s e c 2 θ ) d θ = λ tan θ + 2 log e ∣ f ( θ ) ∣ + C …
Mathematics
Problem If ∫ d θ cos 2 θ ( tan 2 θ + sec 2 θ ) = λ tan θ + 2 log 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 ∫ c o s 2 θ ( t a n 2 θ + s e c 2 θ ) d θ = λ tan θ + 2 log e ∣ f ( θ ) ∣ + C , where C C C is a constant of integration, then the ordered pair ( λ , f ( θ ) ) (\lambda, f(\theta)) ( λ , f ( θ )) is equal to
(a) ( − 1 , 1 − tan θ ) (-1, 1 - \tan \theta) ( − 1 , 1 − tan θ )
(b) ( − 1 , 1 + tan θ ) (-1, 1 + \tan \theta) ( − 1 , 1 + tan θ )
(c) ( 1 , 1 + tan θ ) (1, 1 + \tan \theta) ( 1 , 1 + tan θ )
(d) ( 1 , 1 − tan θ ) (1, 1 - \tan \theta) ( 1 , 1 − tan θ )
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