Question

If \[ \int \frac{5 \tan x}{\tan x - 2} dx = x + a \log | \sin x - 2 \cos x | + c \], then \( a = \) (Where \( c \) is constant of integration)

• A. 1
• B. \(-2\)
• C. \(-1\)
• D. 2

Hint:

When an integral is given in the form \( \int f(x) dx = g(x) + c \), differentiate \(g(x)\) to recover \(f(x)\) and solve for unknown constants.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are given an integral result and need to find the constant \(a\).

Step 2: Key Formula or Approach:
Differentiate the right-hand side and equate it to the integrand.

Step 3: Detailed Explanation:
Let \( I = x + a \log |\sin x - 2 \cos x| + c \). Differentiate w.r.t. \(x\): \[ \frac{dI}{dx} = 1 + a \cdot \frac{\cos x + 2 \sin x}{\sin x - 2 \cos x}. \] This must equal the integrand \(\frac{5 \tan x}{\tan x - 2}\). Write \(\tan x = \frac{\sin x}{\cos x}\): \[ \frac{5 \tan x}{\tan x - 2} = \frac{5 \sin x / \cos x}{(\sin x / \cos x) - 2} = \frac{5 \sin x}{\sin x - 2 \cos x}. \] Set equal: \[ 1 + a \cdot \frac{\cos x + 2 \sin x}{\sin x - 2 \cos x} = \frac{5 \sin x}{\sin x - 2 \cos x}. \] Multiply both sides by \((\sin x - 2 \cos x)\): \[ (\sin x - 2 \cos x) + a (\cos x + 2 \sin x) = 5 \sin x. \] Collect coefficients of \(\sin x\) and \(\cos x\): \[ \sin x - 2 \cos x + 2a \sin x + a \cos x = 5 \sin x. \] \[ (1 + 2a) \sin x + (-2 + a) \cos x = 5 \sin x + 0 \cdot \cos x. \] Comparing coefficients: \(1 + 2a = 5 \implies 2a = 4 \implies a = 2\). And \(-2 + a = 0 \implies a = 2\). Both consistent.
Thus \(a = 2\).

Step 4: Final Answer:
Option (D) is correct.