Question

Evaluate \( \int_0^{\pi/4} \frac{x}{1+\cos 2x + \sin 2x} \, dx \).

Hint:

For definite integrals with symmetric limits, apply symmetry properties to simplify.

Answer 1

Step 1: Apply the symmetry property of definite integrals.
Let \( I = \int_0^{\pi/4} \frac{x}{1+\cos 2x + \sin 2x} \, dx \). Using the property \( \int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx \), we obtain \( I = \int_0^{\pi/4} \frac{\pi/4 - x}{1+\cos 2x + \sin 2x} \, dx \).
Step 2: Combine the integrals.
Adding the two expressions for \( I \) yields \( 2I = \int_0^{\pi/4} \frac{\pi/4}{1+\cos 2x + \sin 2x} \, dx \).
Step 3: Simplify the integrand.
Rewrite \( 1 + \cos 2x + \sin 2x \) as \( \cos 2x + \sin 2x = \sqrt{2} \sin(2x + \pi/4) \) and simplify the integral to \( I = \frac{\pi}{16} \int_0^{\pi/4} \frac{1}{\cos^2 x + \sin x \cos x} \, dx \).
Step 4: Integrate and simplify.
The integral evaluates to \( I = \frac{\pi}{16} (\log |1 + \tan x|)_0^{\pi/4} \). Substituting the limits gives \( I = \frac{\pi}{16} \log 2 \).
Conclusion: The integral evaluates to \( \frac{\pi}{16} \log 2 \).