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 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 get: \[ I = \int_{0}^{\pi/4} \frac{\pi/4 - x}{1+\cos 2x + \sin 2x} \, dx. \] Step 2: {Combine integrals}
Adding the two forms of \( I \): \[ 2I = \int_{0}^{\pi/4} \frac{\pi/4}{1+\cos 2x + \sin 2x} \, dx. \] Step 3: {Simplify the integrand}
Rewrite \( 1 + \cos 2x + \sin 2x \): \[ \cos 2x + \sin 2x = \sqrt{2} \sin(2x + \pi/4), \] and simplify: \[ I = \frac{\pi}{16} \int_{0}^{\pi/4} \frac{1}{\cos^2 x + \sin x \cos x} \, dx. \] Step 4: {Integrate and simplify}
The integral evaluates to: \[ I = \frac{\pi}{16} (\log |1 + \tan x|)_{0}^{\pi/4}. \] Substitute limits: \[ I = \frac{\pi}{16} \log 2. \] Conclusion: The integral evaluates to \( \frac{\pi}{16} \log 2 \).