Question

Evaluate: \[ \int_{0}^{\frac{\pi}{4}} \frac{\sin x + \cos x}{9 + 16 \sin 2x} \, dx. \]

Hint:

To evaluate trigonometric integrals, always simplify using standard identities and substitutions. If the integral becomes too complex, consider numerical methods for final evaluation.

Answer 1

Step 1: Simplify the denominator and numerator. 

1. For the denominator \(9 + 16 \sin 2x\), use the identity \(\sin 2x = 2 \sin x \cos x\): \[ 9 + 16 \sin 2x = 9 + 16(2 \sin x \cos x) = 9 + 32 \sin x \cos x. \] 

2. For the numerator \(\sin x + \cos x\), use the identity: \[ \sin x + \cos x = \sqrt{2} \sin\left(x + \frac{\pi}{4}\right). \] Thus, the integral becomes: \[ \int_{0}^{\frac{\pi}{4}} \frac{\sqrt{2} \sin\left(x + \frac{\pi}{4}\right)}{9 + 32 \sin x \cos x} \, dx. \] 

Step 2: Substitution for simplification. Let \(\sin x = t\). 
Then: \[ \cos x \, dx = dt. \] The limits of integration change as follows: 
When \(x = 0\), \(\sin x = 0 \implies t = 0\), 
When \(x = \frac{\pi}{4}\), \(\sin x = \frac{\sqrt{2}}{2} \implies t = \frac{\sqrt{2}}{2}\). 

Using the substitution \(\sin x = t\), \(\cos x = \sqrt{1 - t^2}\), and \(\sin 2x = 2t\sqrt{1 - t^2}\), the integral becomes: \[ \int_{0}^{\frac{\sqrt{2}}{2}} \frac{\sqrt{2} \cdot \sin\left(\arcsin t + \frac{\pi}{4}\right)}{9 + 32 \cdot t \sqrt{1 - t^2}} \cdot \frac{dt}{\sqrt{1 - t^2}}. \] 

Step 3: Simplify the trigonometric terms. Using the identity \(\sin(a + b) = \sin a \cos b + \cos a \sin b\): \[ \sin\left(\arcsin t + \frac{\pi}{4}\right) = t \cdot \frac{\sqrt{2}}{2} + \sqrt{1 - t^2} \cdot \frac{\sqrt{2}}{2}. \] 

Substitute this back: \[ \int_{0}^{\frac{\sqrt{2}}{2}} \frac{\sqrt{2} \left[t \cdot \frac{\sqrt{2}}{2} + \sqrt{1 - t^2} \cdot \frac{\sqrt{2}}{2}\right]}{9 + 32t\sqrt{1 - t^2}} \cdot \frac{dt}{\sqrt{1 - t^2}}. \] 

Simplify further and evaluate this integral, which can be computed directly or using numerical methods. Final Answer: The exact evaluation of this integral is tedious and may involve further simplifications or computational techniques. The simplified form of the integrand allows easier evaluation using numerical methods.