Let α ∈ ( 0, (π)/(2) ) , be constant.If the integral ∫ ( x + tanα)/( x - tanα)dx = A(x) 2α + B(x) 2α + C , where C is a constant of integration, then the functions A ( x ) and B ( x ) are respectively

Correct answer: A. x - α and lo g _ e | (x - α)| I = ∫ ( x + tanα)/( x - tanα)dx = ∫ (( x)/( x) + (sinα)/(cosα))/(( x)/( x) - (sinα)/(cosα))dx = ∫ ( x α + x α)/( x α - x α)dx = ∫ ( (x + α))/( (x - α))dx = ∫ ( ((x - α) + 2α))/( ( x - α ))dx = ∫ ( (x - α) 2α + (x - α) 2α)/( ( x - α ))dx = ∫ [ 2α + 2α cot(x - α)]dx = xcos2α + 2α _ e | (x - α)| + C = ( x - α ) 2α + 2α _ e | (x - α)| + C Hence, A(x) = x - α and B(x) = lo g _ e | (x - α)|

JEE Main 201912 AprilEvening Shift

Exam: JEE Main
Year: 2019
Date: 12 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Indefinite Integration

JEE Main 2019 · 12 April · Evening Shift · Mathematics · Indefinite Integration

Let α ∈ ( 0, (π)/(2) ) , be constant.If the integral ∫ ( x +...

Let

\alpha \in ( 0, \dfrac{\pi}{2} )

, be constant.If the integral

∫ \dfrac{\tan x + tan\alpha}{\tan x - tan\alpha}dx = A\left(x\right)\cos 2\alpha + B\left(x\right)\sin 2\alpha + C

, where C is a constant of integration, then the functions

A ( x )

and

B ( x )

are respectively

Options

A $x - \alpha$ and $lo{g}_{e}$ \left|\sin $\left(x - \alpha\right)\right|$
B $x + \alpha$ and $\log_{e}$ \left|\cos $\left(x - \alpha\right)\right|$
C $x + \alpha$ and $\log_{e}$ \left|\sin $\left(x + \alpha\right)\right|$
D $x - \alpha$ and $\log_{e}$ \left|\cos $\left(x - \alpha\right)\right|$

Correct answer

A. $x - \alpha$ and $lo{g}_{e}$ \left|\sin $\left(x - \alpha\right)\right|$

Step-by-step solution

I = ∫ \dfrac{\tan x + tan\alpha}{\tan x - tan\alpha}dx = ∫ \dfrac{\dfrac{\sin x}{\cos x} + \dfrac{sin\alpha}{cos\alpha}}{\dfrac{\sin x}{\cos x} - \dfrac{sin\alpha}{cos\alpha}}dx

= ∫ \dfrac{\sin x \cos\alpha + \cos x \sin\alpha}{\sin x \cos\alpha - \cos x \sin\alpha}dx = ∫ \dfrac{\sin \left(x + \alpha\right)}{\sin \left(x - \alpha\right)}dx

= ∫ \dfrac{\sin \left(\left(x - \alpha\right) + 2\alpha\right)}{\sin ( x - \alpha )}dx = ∫ \dfrac{\sin \left(x - \alpha\right)\cos 2\alpha + \cos \left(x - \alpha\right)\sin 2\alpha}{\sin ( x - \alpha )}dx

= ∫ \left[\cos 2\alpha + \sin 2\alpha cot\left(x - \alpha\right)\right]dx

= xcos2\alpha + \sin 2\alpha{\log}_{e}\left|\sin\left(x - \alpha\right)\right| + C

= ( x - \alpha ) \cos 2\alpha + \sin 2\alpha{\log}_{e}\left|\sin\left(x - \alpha\right)\right| + C

Hence,

A\left(x\right) = x - \alpha

and

B\left(x\right) = lo{g}_{e}\left|\sin\left(x - \alpha\right)\right|

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