Evaluate the integral from 0 to π/2 of (x sin(x)) / (1 + cos^2(x)) dx.

Steps to Solve

1. Set up the integral We start with the integral we want to evaluate: $$ I = \int_0^{\frac{\pi}{2}} \frac{x \sin(x)}{1 + \cos^2(x)} , dx $$

2. Consider integration by parts To tackle this integral, we apply the technique of integration by parts. We let:

• ( u = x ) → then ( du = dx )
• ( dv = \frac{\sin(x)}{1 + \cos^2(x)} , dx )

3. Compute ( v ) Next, we need to find ( v ). To do this, we integrate ( dv ). We can express ( v ) as: $$ v = \int \frac{\sin(x)}{1 + \cos^2(x)} , dx $$

Let ( w = \cos(x) ), then ( dw = -\sin(x) , dx ). The integral transforms to: $$ v = -\int \frac{1}{1 + w^2} , dw = -\tan^{-1}(w) + C = -\tan^{-1}(\cos(x)) + C $$

4. Apply integration by parts formula Now using the integration by parts formula ( \int u , dv = uv - \int v , du ), we substitute: $$ I = \left[-x \tan^{-1}(\cos(x))\right]_0^{\frac{\pi}{2}} + \int_0^{\frac{\pi}{2}} \tan^{-1}(\cos(x)) , dx $$

5. Evaluate boundary terms Evaluate the boundary term:

• At ( x = \frac{\pi}{2} ): ( -\frac{\pi}{2} \tan^{-1}(\cos(\frac{\pi}{2})) = -\frac{\pi}{2} \cdot 0 = 0 )
• At ( x = 0 ): ( -0 \cdot \tan^{-1}(\cos(0)) = -0 \cdot \frac{\pi}{4} = 0 )

Thus, the boundary term is 0, and we have: $$ I = \int_0^{\frac{\pi}{2}} \tan^{-1}(\cos(x)) , dx $$

6. Final form of the integral Now, we've reduced our original integral to the evaluation of: $$ I = \int_0^{\frac{\pi}{2}} \tan^{-1}(\cos(x)) , dx $$

This integral can be evaluated using symmetry or known results.