Let $\alpha \in(0, \pi / 2)$ be fixed. If the integral $\int \frac{\tan x+\tan \alpha}{\tan x-\tan \alpha} dx= A(x) \cos 2 \alpha+B(x) \sin 2 \alpha+C$, where $C$ is a constant of integration, then the functions $A(x)$ and $B(x)$ are respectively:

(C) $x-\alpha$ and $\log_{e}|\sin (x-\alpha)|$

(A) $x+\alpha$ and $\log_{e}|\sin (x-\alpha)|$

(B) $x-\alpha$ and $\log_{e}|\cos (x-\alpha)|$

(D) $x+\alpha$ and $\log_{e}|\sin (x+\alpha)|$

JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Indefinite Integrals

Let

α \in (0, π /2)

be fixed. If the integral

\int \frac{t a n x + t a n α}{t a n x - t a n α} d x = A (x) cos 2 α + B (x) sin 2 α + C

, where

C

is a constant of integration, then the functions

A (x)

and

B (x)

are respectively:

Answer: (C)

Combine the tangents over a common denominator:

\frac{t a n x + t a n α}{t a n x - t a n α} = \frac{s i n ( x + α )}{s i n ( x - α )}

. Put

u = x - α

, so

x + α = u + 2 α

and

sin (u + 2 α) = sin u cos 2 α + cos u sin 2 α

. The integrand becomes

cos 2 α + sin 2 α cot u

, and integrating gives

(x - α) cos 2 α + sin 2 α ln ∣ sin (x - α) ∣ + C

. So

A (x) = x - α

and

B (x) = ln ∣ sin (x - α) ∣

JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Single correct option