If ∫ dθ/cos²θ(tan 2θ+sec 2θ)=λ tan θ+2logₑ|f(θ)|+C
where C is a constant of integration, then the ordered pair (λ, ƒ(θ)) is equal to :

Solution

Maths sir · 4 steps

Turn tan 2θ + sec 2θ into a fraction in tan θ, and sec² θ dθ becomes dt.

1. Simplify the bracket. tan 2θ + sec 2θ = (1 + sin 2θ)/(cos 2θ). The top is (cos θ + sin θ)² and the bottom is (cos θ - sin θ)(cos θ + sin θ). Cancel one factor, then divide top and bottom by cos θ to get (1 + tan θ)/(1 - tan θ).
2. Substitute. The integrand is now sec² θ · (1 - tan θ)/(1 + tan θ). Put t = tan θ, so dt = sec² θ dθ.

   ∫ (1 - t)/(1 + t) dt = ∫ (-1 + 2/(1 + t))dt = -t + 2 ln|1 + t| + C

3. Split the fraction. The split works because 1 - t = -(1 + t) + 2. Putting t = tan θ back gives -tan θ + 2 ln|1 + tan θ| + C.
4. So. Comparing with λ tan θ + 2 ln|f(θ)| gives λ = -1 and f(θ) = 1 + tan θ, which is option (c).

The idea to keep. When you see sec² θ beside double-angle terms, rewrite everything in tan θ. Then t = tan θ turns it into an ordinary algebraic fraction.

Answer from the NTA final answer key. Solution by PrepWiser.