1. ⇒ (MHT CET 2023 9th May Evening Shift )

Let $α \in (0, \frac{π}{2})$ be fixed. If the integral $\int \frac{\tan x + \tan α}{\tan x - \tan α} d x = A (x) \cos 2 α + B (x) \sin 2 α + c,$ (where $c$ is a constant of integration), then functions $A (x)$ and $B (x)$ are respectively

A. $x + α$ and $\log | \sin (x + α) |$ .

B. $x - α$ and $\log | \sin (x - α) |$ .

C. $x - α$ and $\log | \cos (x - α) |$ .

D. $x + α$ and $\log | \sin (x - α) |$ .

Correct Option is (B)

$\begin{aligned} Let I = \int \frac{\tan x + \tan α}{\tan x - \tan α} d x \\ = \int \frac{\frac{\sin x}{\cos x} + \frac{\sin α}{\cos α}}{\frac{\sin x}{\cos x} - \frac{\sin α}{\cos α}} d x \\ = \int \frac{\sin x \cos α + \sin α \cos x}{\sin x \cos α - \sin α \cos x} d x \\ = \int \frac{\sin (x + α)}{\sin (x - α)} d x \\ Let x - α = t \\ ∴ I = \frac{\sin (t + 2 α)}{\sin t} \\ = \int \frac{\sin (t) \cos 2 α + \cos (t) \sin 2 α}{\sin (t)} d x \\ = \cos 2 α \int 1 d t + \sin 2 α \int \cot (t) dt \\ = \cos 2 α \cdot t + \sin 2 α \cdot \log | \sin (t) | + c \\ ∴ I = (x - α) \cos 2 α + \log | \sin (x - α) | \sin 2 α + c \\ But \int \frac{\tan x + \tan α}{\tan x - \tan α} d x = A (x) \cos 2 α + B (x) \sin 2 α + c .... [Given] \\ \Rightarrow A (x) = x - α, B (x) = \log | \sin (x - α) | + c \end{aligned}$

2. ⇒ (MHT CET 2023 9th May Morning Shift )

$\int \frac{1}{\sin (x - a) \sin x} d x =$

A. $\sin a (\log (\sin (x - a) \cdot cosec x)) + c$ , where $c$ is a constant of integration.

B. $cosec a (\log (\sin (x - a) \cdot cosec x)) + c$ , where $c$ is a constant of integration.

C. $- \sin a (\log (\sin (x - a) \cdot \sin x)) + c$ , where c is a constant of integration.

D. $- cosec a (\log (\sin (x - a) \cdot \sin x)) + c$ , where $c$ is a constant of integration.

Correct Option is (B)

$\begin{aligned} Let I = \int \frac{1}{\sin (x - a) \sin x} d x \\ Put x - a = t \to x = a + t \\ d x = dt \end{aligned}$

$\begin{aligned} ∴ I & = \int \frac{1}{\sin t \cdot \sin (a + t)} d t \\ = \frac{1}{\sin a} \int \frac{\sin a}{\sin t \cdot \sin (a + t)} d t \\ = \frac{1}{\sin a} \int \frac{\sin ((a + t) - t)}{\sin (a + t) \cdot \sin t} d t \\ = \frac{1}{\sin a} [\int \frac{\sin (a + t) \cos t}{\sin (a + t) \sin t} d t - \int \frac{\sin t \cos (a + t)}{\sin (a + t) \sin t} d t] \\ = \frac{1}{\sin a} [\int \cot t d t - \int \cot (a + t) d t] \\ = cosec a [\log | \sin t | - \log | \sin (a + t) |] + c \\ = cosec a [\log ∣ \frac{\sin t}{\sin (a + t) ∣}] + c \\ = cosec a [\log (\sin (x - a) \cdot cosec x)] + c \end{aligned}$

3. ⇒ (MHT CET 2021 20th September Evening Shift )

$\int cosec (x - a) cosec x d x =$

A. $cosec a \cdot \log [\sin (x - a) cosec x] + c$

B. $cosec a \log [\sin (x - a) \sin x] + c$

C. $\sin a \cdot \log [\sin (x - a) \sin x] + c$

D. $cosec a \cdot \log [cosec (x - a) \sin x] + c$

Correct Option is (A)

$\begin{aligned} Let I = \int cosec (x - a) cosec x d x \\ = \int \frac{d x}{\sin (x - a) \sin x} \\ = \int \frac{\sin a}{\sin a \sin (x - a) \sin x} d x \\ = \frac{1}{\sin a} \int \frac{\sin (a + x - x)}{\sin (x - a) \sin x} d x \\ = \frac{1}{\sin a} \int \frac{\sin - [x - a - x]}{\sin (x - a) \sin x} d x \\ = \frac{1}{\sin a} \int \frac{\sin - [(x - a) - (x)]}{\sin (x - a) \sin x} d x = \frac{- 1}{\sin a} \int \frac{\sin [(x - a) - x]}{\sin (x - a) \sin x} d x \\ = \frac{- 1}{\sin a} \int \frac{\sin (x - a) \cos x - \cos (x - a) \sin x}{\sin (x - a) \sin x} d x \\ = \frac{- 1}{\sin a} \int [\cot x - \cot (x - a)] d x \\ = \frac{- 1}{\sin a} [\log | \sin x | - \log | \sin (x - a) |] + c \\ = \frac{- 1}{\sin a} [\int \cot x d x - \int \cot (x - a) d x] \\ = \frac{1}{\sin a} [\log | \sin (x - a) | - \log | \sin x |] + c \\ = (cosec a) [\log | \frac{\sin (x - a)}{\sin x} |] + c \\ = cosec a \cdot \log | \sin (x - a) \cdot cosec x | + c \end{aligned}$

⇒Indefinite Integration