JEE Main On 16 April 2018 Question 30
Question: If $ \int _{{}}^{{}}{\frac{\tan x}{1+\tan x+{{\tan }^{2}}x}}dx=x-\frac{K}{\sqrt{A}}{{\tan }^{-1}}( \frac{K\tan x+1}{\sqrt{A}} )+C, $ (C is a constant of integration), then the ordered pair (K, A) is equal to. [JEE Main 16-4-2018]
Options:
A) (2, 3)
B) (2, 1)
C) $ (-2,1) $
D) $ (-2,3) $
Show Answer
Answer:
Correct Answer: A
Solution:
Let $ \tan x=t\Rightarrow dt={{\sec }^{2}}xdx $ $ =(1+{{\tan }^{2}}x)dx $ $ =(1+t^{2})dx $
$ \Rightarrow $ $ dx=\frac{dt}{1+t^{2}} $
$ \therefore $ $ \int _{{}}^{{}}{\frac{\tan x}{{{\tan }^{2}}x+\tan x+1}dx} $ $ =\int _{{}}^{{}}{\frac{t}{(1+t^{2})(1+t+t^{2})}dt} $ $ ={{\tan }^{-1}}t-\frac{2}{\sqrt{3}}{{\tan }^{-1}}( \frac{2t+1}{\sqrt{3}} ) $
$ =x-\frac{2}{\sqrt{3}}{{\tan }^{-1}}( \frac{2\tan x+1}{\sqrt{3}} ) $ Clearly then $ (K,A)=(2,3). $
so option A is the correct answer. $ \int _{{}}^{{}}{\frac{\tan x+1+{{\tan }^{2}}x}{\tan x+1+{{\tan }^{2}}x}}dx-\int _{{}}^{{}}{\frac{(1+{{\tan }^{2}}x)}{1+\tan +{{\tan }^{2}}x}}dx $
$ =x-\int _{{}}^{{}}{\frac{{{\sec }^{2}}dx}{1+\tan x+{{\tan }^{2}}x}} $ $ =x-\int _{{}}^{{}}{\frac{dt}{t^{2}+t+\frac{1}{4}+1-\frac{1}{4}}} $
$ =x-\int _{{}}^{{}}{\frac{dt}{{{( t+\frac{1}{2} )}^{2}}+{{( \frac{3}{2} )}^{2}}}} $ $ =x-\frac{2}{c _3}{{\tan }^{-}}( \frac{t+11 _2}{c _3/2} )+c $ $ x-\frac{2}{c _3}{{\tan }^{-}}( \frac{2\tan x+1}{\sqrt{3}} )+c $
$ A=3 $ $ K=2. $ Thus the answer is (2, 3)
