Question

Evaluate \( \int_{0}^{\pi/4} \frac{\cos^2 x}{\cos^2 x + 4 \sin^2 x} \, dx \):

• A. \( \frac{\pi}{4} + \frac{2}{3} \tan^{-1} 2 \)
• B. \( -\frac{\pi}{3} + \frac{2}{3} \tan^{-1} 3 \)
• C. \( -\frac{\pi}{12} + \frac{2}{3} \tan^{-1} 2 \)
• D. \( \frac{\pi}{6} - \frac{2}{3} \tan^{-1} 4 \)

Hint:

Substituting \( \tan x = t \) in integrals involving trigonometric functions can help simplify the expression and convert it into a rational function.

Answer 1

The Correct Option is C

We are tasked with evaluating the integral: \[ I = \int_{0}^{\pi/4} \frac{\cos^2 x}{\cos^2 x + 4\sin^2 x} \, dx. \] Step 1: substituting \( \tan x = t \)
Let:
\[ \tan x = t, \quad \sec^2 x \, dx = dt. \] The limits of integration change as: \[ x = 0 \implies t = 0, \quad x = \frac{\pi}{4} \implies t = 1. \] Substituting into the integral: \[ I = \int_{0}^{1} \frac{\frac{1}{1+t^2}}{\frac{1}{1+t^2} + 4 \cdot \frac{t^2}{1+t^2}} \, dt. \] Simplifying: \[ I = \int_{0}^{1} \frac{1}{1 + t^2 + 4t^2} \, dt = \int_{0}^{1} \frac{1}{1 + 4t^2 + t^2} \, dt. \] Step 2: Simplify further Combining terms, we get: \[ I = \int_{0}^{1} \frac{1}{1 + 4t^2} \, dt. \] Step 3: Decompose and integrate Using partial fractions: \[ I = \frac{1}{3} \int_{0}^{1} \left( \frac{4}{1 + 4t^2} - \frac{1}{1 + t^2} \right) \, dt. \] Splitting the integral: \[ I = \frac{1}{3} \left[ \int_{0}^{1} \frac{4}{1 + 4t^2} \, dt - \int_{0}^{1} \frac{1}{1 + t^2} \, dt \right]. \] Step 4: Evaluate each term 1. For the first term: \[ \int \frac{4}{1 + 4t^2} \, dt = \int \frac{1}{\frac{1}{4} + t^2} \, dt = 2 \tan^{-1}(2t). \] 2. For the second term: \[ \int \frac{1}{1 + t^2} \, dt = \tan^{-1}(t). \] Substituting back: \[ I = \frac{1}{3} \left[ 2 \tan^{-1}(2) - \tan^{-1}(1) \right]. \] Step 5: Simplify the result Using \(\tan^{-1}(1) = \frac{\pi}{4}\), we get: \[ I = \frac{1}{3} \left[ 2 \tan^{-1}(2) - \frac{\pi}{4} \right]. \] Simplifying further: \[ I = -\frac{\pi}{12} + \frac{2}{3} \tan^{-1}(2). \] Final Answer: \[ \boxed{-\frac{\pi}{12} + \frac{2}{3} \tan^{-1} 2}. \]