If ∫tanx/(1+tanx+tan^2x)dx=x-K√Atan^(-1)Ktanx+1√A+C, (C is a constant of integration), then the ordered pair K,A is equal to

Correct answer: (B) 2,3. Let I=∫tanx/(1+tanx+tan^2x)dx ⇒I=∫1-(sec^2x)/(1+tanx+tan^2x)dx ⇒I=x-∫dt/(1+t+t^2) Where, tanx=t⇒sec^2xdx=dt ∴ I=x-1/(√32)tan^(-1)(t+12)/(√32)+C, where C is the constant of integration. =x-2√3tan^(-1)2tanx+1√3+C ∴ K=2 & A=3.

Q55 / 90 · JEE Main 2018 (16 Apr Online) · Mathematics

If $\int \frac{\tan x}{1 + \tan x + \tan^{2} x} d x = x - \frac{K}{\sqrt{A}} \tan^{- 1} (\frac{K \tan x + 1}{\sqrt{A}}) + C$ , ( $C$ is a constant of integration), then the ordered pair $(K, A)$ is equal to

A
$(2, 1)$
Correct answer
B
$(2, 3)$
C
$(- 2, 1)$
D
$(- 2, 3)$

Solution

Let $I = \int \frac{\tan x}{1 + \tan x + \tan^{2} x} d x$

$\Rightarrow I = \int (1 - \frac{\sec^{2} x}{1 + \tan x + \tan^{2} x}) d x$

$\Rightarrow I = x - \int \frac{d t}{1 + t + t^{2}}$

Where, $\tan x = t \Rightarrow \sec^{2} x d x = d t$

$∴ I = x - \frac{1}{\frac{\sqrt{3}}{2}} \tan^{- 1} (\frac{t + \frac{1}{2}}{\frac{\sqrt{3}}{2}}) + C$ , where $C$ is the constant of integration.

$= x - \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{2 \tan x + 1}{\sqrt{3}}) + C$

$∴ K = 2 & A = 3$ .

Hence, the correct answer is option: 'B'.

If ∫tanx1+tanx+tan2xdx=x-KAtan-1Ktanx+1A+C, (C is a constant of integration), then the ordered pair K,A is equal to

Solution

Relevant Questions

If $\int \frac{\tan x}{1 + \tan x + \tan^{2} x} d x = x - \frac{K}{\sqrt{A}} \tan^{- 1} (\frac{K \tan x + 1}{\sqrt{A}}) + C$ , ( $C$ is a constant of integration), then the ordered pair $(K, A)$ is equal to