JEE PYQ: Indefinite Integration Question 26
Question 26 - 2019 (12 Apr Shift 1)
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:
(1) $x + \alpha$ and $\log_e|\sin(x+\alpha)|$
(2) $x - \alpha$ and $\log_e|\sin(x-\alpha)|$
(3) $x - \alpha$ and $\log_e|\cos(x-\alpha)|$
(4) $x + \alpha$ and $\log_e|\sin(x-\alpha)|$
Show Answer
Answer: (2)
Solution
$\frac{\tan x + \tan\alpha}{\tan x - \tan\alpha} = \frac{\sin(x+\alpha)}{\sin(x-\alpha)}$. Let $x - \alpha = t$: $\int \frac{\sin(t+2\alpha)}{\sin t},dt = \int (\cos 2\alpha + \sin 2\alpha\cdot\cot t),dt = t\cos 2\alpha + \sin 2\alpha\cdot\ln|\sin t| + C = (x-\alpha)\cos 2\alpha + \sin 2\alpha\cdot\log_e|\sin(x-\alpha)| + C$. So $A(x) = x - \alpha$, $B(x) = \log_e|\sin(x-\alpha)|$.
BETA
