Question

Evaluate the integral: \[ \int_0^{\frac{\pi}{4}} \frac{\cos^2 x}{\cos^2 x + 4 \sin^2 x} \, dx \]

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

Hint:

For integrals involving trigonometric functions, use substitution or trigonometric identities to simplify the expression before integrating.

Answer 1

The Correct Option is C

We are given the integral \( \int_0^{\frac{\pi}{4}} \frac{\cos^2 x}{\cos^2 x + 4 \sin^2 x} \, dx \). Step 1: Rewrite the integral using trigonometric identities. The denominator can be rewritten as a sum involving the tangent function. Step 2: Apply substitution to simplify the expression and evaluate the integral. After solving, the result is: \[ -\frac{\pi}{12} + \frac{2}{3} \tan^{-1} 2 \] % Final Answer The value of the integral is \( -\frac{\pi}{12} + \frac{2}{3} \tan^{-1} 2 \).

Answer 2

Step 1: Simplify the integrand
Given integral:
\[ \int_0^{\frac{\pi}{4}} \frac{\cos^2 x}{\cos^2 x + 4 \sin^2 x} \, dx \]
Rewrite denominator:
\[ \cos^2 x + 4 \sin^2 x = \cos^2 x + 4 (1 - \cos^2 x) = 4 - 3 \cos^2 x \]

Step 2: Express integral using cosine only
\[ \int_0^{\frac{\pi}{4}} \frac{\cos^2 x}{4 - 3 \cos^2 x} \, dx \]

Step 3: Use substitution for easier integration
Let \( t = \tan x \), then \( dx = \frac{dt}{1 + t^2} \), and
\[ \cos^2 x = \frac{1}{1 + t^2} \] Substitute in the integral:
\[ \int_0^1 \frac{\frac{1}{1 + t^2}}{4 - 3 \frac{1}{1 + t^2}} \cdot \frac{dt}{1 + t^2} = \int_0^1 \frac{1}{(1 + t^2)(4 - \frac{3}{1 + t^2})} \, dt \]
Simplify denominator inside integral:
\[ 4 - \frac{3}{1 + t^2} = \frac{4(1 + t^2) - 3}{1 + t^2} = \frac{4 + 4t^2 - 3}{1 + t^2} = \frac{1 + 4 t^2}{1 + t^2} \]
So integrand becomes:
\[ \frac{1}{(1 + t^2)} \cdot \frac{1 + t^2}{1 + 4 t^2} = \frac{1}{1 + 4 t^2} \]

Step 4: Integral reduces to
\[ \int_0^1 \frac{1}{1 + 4 t^2} dt = \int_0^1 \frac{1}{1 + (2t)^2} dt \]

Step 5: Evaluate the integral
\[ \int \frac{1}{1 + (2t)^2} dt = \frac{1}{2} \tan^{-1} (2 t) + C \]
Apply limits:
\[ = \frac{1}{2} \left[ \tan^{-1} (2 \times 1) - \tan^{-1} (0) \right] = \frac{1}{2} \tan^{-1} 2 \]

Step 6: Reconsider original integral expression
Wait, step 3 shows an inconsistency in substitution since the integral simplified too much.
Let's solve integral using a different approach:

Alternate Step 1: Express integrand in terms of \(\tan x\)
Rewrite denominator:
\[ \cos^2 x + 4 \sin^2 x = \cos^2 x + 4 \sin^2 x = \cos^2 x (1 + 4 \tan^2 x) \] So integrand becomes:
\[ \frac{\cos^2 x}{\cos^2 x (1 + 4 \tan^2 x)} = \frac{1}{1 + 4 \tan^2 x} \]
Since \( dx = \frac{dt}{1 + t^2} \) for \( t = \tan x \), integral becomes:
\[ \int_0^1 \frac{1}{1 + 4 t^2} \cdot \frac{dt}{1 + t^2} \]

Step 7: Evaluate
This integral is
\[ I = \int_0^1 \frac{dt}{(1 + t^2)(1 + 4 t^2)} \] Use partial fractions:
\[ \frac{1}{(1 + t^2)(1 + 4 t^2)} = \frac{A}{1 + t^2} + \frac{B}{1 + 4 t^2} \] Multiply both sides:
\[ 1 = A(1 + 4 t^2) + B(1 + t^2) = (A + B) + (4A + B) t^2 \] Equate coefficients:
\[ A + B = 1, \quad 4A + B = 0 \] Solving:
\[ B = 1 - A \] \[ 4A + 1 - A = 0 \implies 3A = -1 \implies A = -\frac{1}{3}, \quad B = 1 - \left(-\frac{1}{3}\right) = \frac{4}{3} \]

Step 8: Write integral as sum
\[ I = \int_0^1 \frac{-\frac{1}{3}}{1 + t^2} dt + \int_0^1 \frac{\frac{4}{3}}{1 + 4 t^2} dt = -\frac{1}{3} \int_0^1 \frac{dt}{1 + t^2} + \frac{4}{3} \int_0^1 \frac{dt}{1 + 4 t^2} \]

Step 9: Integrate
\[ \int_0^1 \frac{dt}{1 + t^2} = \tan^{-1} 1 - \tan^{-1} 0 = \frac{\pi}{4} \] \[ \int_0^1 \frac{dt}{1 + 4 t^2} = \frac{1}{2} \tan^{-1} (2 t) \bigg|_0^1 = \frac{1}{2} \tan^{-1} 2 \]

Step 10: Combine results
\[ I = -\frac{1}{3} \cdot \frac{\pi}{4} + \frac{4}{3} \cdot \frac{1}{2} \tan^{-1} 2 = -\frac{\pi}{12} + \frac{2}{3} \tan^{-1} 2 \]

Final answer:
\[ \boxed{ -\frac{\pi}{12} + \frac{2}{3} \tan^{-1} 2 } \]