Checkout JEE MAINS 2022 Question Paper Analysis : Checkout JEE MAINS 2022 Question Paper Analysis :

# What is the ordered pair (λ, f(θ))

If

$$\begin{array}{l}\int \frac{d\theta }{\cos ^{2}\theta (\tan 2\theta +\sec 2\theta )}= \lambda \tan \theta +2\: log_{e}\left | f(\theta ) \right |+C\end{array}$$
where C is constant if integration, then the ordered pair (λ, f(θ)) is equal to:

a. (-1, 1-tan θ)

b. (-1, 1+tan θ)

c. (1, 1+tan θ)

d. (1, 1-tan θ)

Solution:

Let

$$\begin{array}{l}I=\int \frac{d \theta}{\cos ^{2} \theta(\sec 2 \theta+\tan 2 \theta)}\end{array}$$

$$\begin{array}{l}I=\int \frac{\sec ^{2} \theta d \theta}{\left(\frac{1+\tan ^{2} \theta}{1-\tan ^{2} \theta}\right)+\left(\frac{2 \tan \theta}{1-\tan ^{2} \theta}\right)}\end{array}$$

$$\begin{array}{l}I=\int \frac{\left(1-\tan ^{2} \theta\right)\left(\sec ^{2} \theta\right) d \theta}{(1+\tan \theta)^{2}}\end{array}$$

Let tan θ= k

sec2 θ = dk

$$\begin{array}{l}I=\int \frac{\left(1-k^{2}\right)}{(1+k)^{2}} d k=\int \frac{(1-k)}{(1+k)} d k\end{array}$$

I = (2/(1+k))-1)dk

I = 2 lnǀ1+kǀ -k+c

I = 2 lnǀ1+tan θǀ -tan θ +c

Given I = tan θ+ 2 log f(θ) + c

λ= -1, f(θ) = ǀ1+ tan θǀ