If ∫ (dθ)/(cos ²θ (tan 2θ + sec 2θ)) = λ tan θ + 2 log e | f(θ) | + C where C is a constant of integration, then the ordered pair (λ, ƒ(θ)) is equal to:

Integral Calculus: JEE Main 2020 Mathematics question

JEE Main 2020 · 9 Jan, Evening shift · Q4021

JEE Main 20209 Jan, Evening shiftMathematicsHardIntegral Calculus369× since 2002

\int {{{d\theta } \over {{{\cos }^2}\theta \left( {\tan 2\theta + \sec 2\theta } \right)}}} = \lambda \tan \theta + 2{\log _e}\left| {f\left( \theta \right)} \right| + C

where C is a constant of integration, then the ordered pair (

\lambda

, ƒ(

\theta

)) is equal to :

JEE Main 2020 Integral Calculus Question with Solution (9 Jan Evening shift, Q4021) | JEE Arc

JEE Main 2020 · 9 Jan, Evening shift · Q4021

JEE Main 20209 Jan, Evening shiftMathematicsHardIntegral Calculus369× since 2002

\int {{{d\theta } \over {{{\cos }^2}\theta \left( {\tan 2\theta + \sec 2\theta } \right)}}} = \lambda \tan \theta + 2{\log _e}\left| {f\left( \theta \right)} \right| + C

where C is a constant of integration, then the ordered pair (

\lambda

, ƒ(

\theta

)) is equal to :