If $\int \frac{d \theta}{\cos ^{2} \theta(\tan 2 \theta+\sec 2 \theta)}=\lambda \tan \theta+2 \log _{e}|f(\theta)|+C$ where $C$ is a constant of integration, then the ordered pair $(\lambda, f(\theta))$ is equal to:

JEE Main · 2020 · 09 Jan 2020 · Shift 2 · Indefinite Integrals

\int \frac{d θ}{c o s ^{2} θ ( t a n 2 θ + s e c 2 θ )} = λ tan θ + 2 lo g_{e} ∣ f (θ) ∣ + C

where

C

is a constant of integration, then the ordered pair

(λ, f (θ))

is equal to:

Answer: (A)

Convert the double angle:

tan 2 θ + sec 2 θ = \frac{2 tan θ + 1 + tan ^{2} θ}{1 - tan ^{2} θ} = \frac{( 1 + tan θ ) ^{2}}{( 1 - tan θ ) ( 1 + tan θ )} = \frac{1 + tan θ}{1 - tan θ}

. The integral becomes

\int \frac{sec ^{2} θ ( 1 - tan θ )}{1 + tan θ} d θ

, and with

t = tan θ

d t = sec^{2} θ d θ

, this is

\int \frac{1 - t}{1 + t} d t = \int (- 1 + \frac{2}{1 + t}) d t = - t + 2 ln ∣1 + t ∣ + C

. Restoring

t = tan θ

gives

- tan θ + 2 ln ∣1 + tan θ ∣ + C

, so

λ = - 1

and

f (θ) = 1 + tan θ

JEE Main · 2020 · 09 Jan 2020 · Shift 2 · Single correct option