Question

Evaluate the integral: \[ \int \frac{4\tan^4 x + 3 \tan^2 x - 1}{\tan^2 x + 4} \, dx. \]

• A. \( 4 \tan x - \frac{17}{4} \tan ^{-1} \frac{tan x}{4} \)
• B. \( 4 \tan x - \frac{17}{4} \tan ^{-1} \frac{tan x}{2} \)
• C. \( 4 \tan x - \frac{17}{2} \tan ^{-1} \frac{tan x}{2} \)
• D. \( 2 \tan x - \frac{17}{2} \tan ^{-1} \frac{tan x}{2} \) \bigskip

Hint:

Use substitution for trigonometric integrals and reduce the powers of tangent to simplify the expression.

Answer 1

The Correct Option is C

We need to evaluate the integral: \[ I = \int \frac{4\tan^4 x + 3\tan^2 x - 1}{\tan^2 x + 4} \, dx. \] --- Step 1: Substituting \( t = \tan x \) Let: \[ t = \tan x \Rightarrow dt = \sec^2 x \, dx. \] Rewriting the integral in terms of \( t \): \[ I = \int \frac{4t^4 + 3t^2 - 1}{t^2 + 4} \, dx. \] --- Step 2: Polynomial Division We perform polynomial division of: \[ 4t^4 + 3t^2 - 1 \quad \text{by} \quad t^2 + 4. \] Dividing the leading term: \[ \frac{4t^4}{t^2} = 4t^2. \] Multiplying: \[ (4t^2) (t^2 + 4) = 4t^4 + 16t^2. \] Subtracting: \[ (4t^4 + 3t^2 - 1) - (4t^4 + 16t^2) = -13t^2 - 1. \] Now divide \( -13t^2 \) by \( t^2 + 4 \): \[ \frac{-13t^2}{t^2 + 4} = -1 + \frac{17}{t^2 + 4}. \] Thus, the division gives: \[ \frac{4t^4 + 3t^2 - 1}{t^2 + 4} = 4t^2 - 1 + \frac{17}{t^2 + 4}. \] Rewriting the integral: \[ I = \int (4t^2 - 1) dx + \int \frac{17}{t^2 + 4} dx. \] --- Step 3: Evaluate Integrals # First Integral: \[ \int (4t^2 - 1) dx = 4 \int \tan^2 x \, dx - \int dx. \] Using: \[ \int \tan^2 x \, dx = \tan x - x, \] we obtain: \[ 4 \int \tan^2 x \, dx = 4 (\tan x - x). \] Thus, \[ \int (4t^2 - 1) dx = 4 \tan x - x. \] # Second Integral: \[ \int \frac{17}{t^2 + 4} dx. \] Using the standard result: \[ \int \frac{1}{x^2 + a^2} dx = \frac{1}{a} \tan^{-1} \frac{x}{a}, \] we get: \[ \int \frac{17}{t^2 + 4} dx = \frac{17}{2} \tan^{-1} \frac{t}{2}. \] --- Step 4: Final Expression \[ I = 4 \tan x - x + \frac{17}{2} \tan^{-1} \frac{\tan x}{2}. \] Since \( x = \tan^{-1} \tan x \), we rewrite: \[ I = 4 \tan x - \frac{17}{2} \tan^{-1} \frac{\tan x}{2}. \] --- Final Answer: \[ \boxed{4 \tan x - \frac{17}{2} \tan^{-1} \frac{\tan x}{2}} \] which matches option (3). \bigskip