Using trigonometric identities sin(π - x) = sin x and cos(π - x) = -cos x, we see that cos²(π - x) = (-cos x)² = cos² x. The denominator remains unchanged.

When x = 0, t = cos(0) = 1. When x = π, t = cos(π) = -1. Substituting these gives 2I = π ∫_1^(-1) (-dt)/(1 + t²) = π ∫_-1^(1) (dt)/(1 + t²).

The anti-derivative of (1)/(1 + t²) is tan^(-1)(t). Evaluating from -1 to 1 yields tan^(-1)(1) - tan^(-1)(-1) = (π)/(4) - (-(π)/(4)) = (π)/(2). Multiplying by π gives 2I = (π²)/(2), so I = (π²)/(4).

The anti-derivative of (1)/(1 + t²) is tan^(-1)(t). Evaluating from -1 to 1 yields tan^(-1)(1) - tan^(-1)(-1) = (π)/(4) - (-(π)/(4)) = (π)/(2). Multiplying by π gives 2I = (π²)/(2), so I = (π²)/(4).

Integrate: ((x)sinx)/(1+cos²x)

Answer: $\frac{\pi^2}{4}$

Step-by-step solution

Step 1: Set up the integral and apply definite integral property

We denote the given standard integral as $I$ . To eliminate the variable $x$ from the numerator, we apply the property $\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx$ , where $a = \pi$ .

Step 2: Substitute x with (π - x)

Using trigonometric identities $\sin(\pi - x) = \sin x$ and $\cos(\pi - x) = -\cos x$ , we see that $\cos^2(\pi - x) = (-\cos x)^2 = \cos^2 x$ . The denominator remains unchanged.

Step 3: Add both equations to eliminate x

Adding the two expressions for $I$ eliminates the $x \sin x$ term, leaving only $\pi \sin x$ in the numerator.

Step 4: Substitute $t = cos x$

When $x = 0$ , $t = \cos(0) = 1$ . When $x = \pi$ , $t = \cos(\pi) = -1$ . Substituting these gives $2I = \pi \int_{1}^{-1} \frac{-dt}{1 + t^2} = \pi \int_{-1}^{1} \frac{dt}{1 + t^2}$ .

Step 5: Evaluate the resulting integral

The anti-derivative of $\frac{1}{1 + t^2}$ is $\tan^{-1}(t)$ . Evaluating from $-1$ to $1$ yields $\tan^{-1}(1) - \tan^{-1}(-1) = \frac{\pi}{4} - \left(-\frac{\pi}{4}\right) = \frac{\pi}{2}$ . Multiplying by $\pi$ gives $2I = \frac{\pi^2}{2}$ , so $I = \frac{\pi^2}{4}$ .

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