\( \int _{0}^{1}\int _{x}^{1}\cos (y)^{2} d y d x \)

To solve the double integral \( \int _{0}^{1}\int _{x}^{1}\cos (y)^{2} \, d y \, d x \), we first change the order of integration. The region of integration is defined by \( 0 \leq x \leq 1 \) and \( x \leq y \leq 1 \). Reversing the order gives us the bounds: \( 0 \leq y \leq 1 \) and \( 0 \leq x \leq y \). Thus, we rewrite the integral as: \[ \int_{0}^{1} \int_{0}^{y} \cos^2(y) \, d x \, d y \] Now, we can integrate with respect to \( x \): \[ \int_{0}^{1} \left( \int_{0}^{y} \cos^2(y) \, d x \right) d y = \int_{0}^{1} (y \cos^2(y)) \, d y \] Here, the inner integral \( \int_{0}^{y} \cos^2(y) \, d x \) just gives \( y \cos^2(y) \) since we are integrating \( \cos^2(y) \) with respect to \( x \). Next, we need to evaluate \( \int_{0}^{1} y \cos^2(y) \, d y \). To do this, we'll use the identity: \[ \cos^2(y) = \frac{1 + \cos(2y)}{2} \] Substituting this into our integral yields: \[ \int_{0}^{1} y \cos^2(y) \, d y = \int_{0}^{1} y \cdot \frac{1 + \cos(2y)}{2} \, d y = \frac{1}{2} \int_{0}^{1} y \, d y + \frac{1}{2} \int_{0}^{1} y \cos(2y) \, d y \] Calculating the first integral: \[ \int_{0}^{1} y \, d y = \left[\frac{y^2}{2}\right]_{0}^{1} = \frac{1}{2} \] Now, we evaluate the second integral \( \int_{0}^{1} y \cos(2y) \, d y \) using integration by parts. Letting \( u = y \) and \( dv = \cos(2y) \, dy \), we have \( du = dy \) and \( v = \frac{1}{2} \sin(2y) \): \[ \int_{0}^{1} y \cos(2y) \, d y = \left[ \frac{y}{2} \sin(2y) \right]_{0}^{1} - \int_{0}^{1} \frac{1}{2} \sin(2y) \, dy \] Evaluating the boundary term: \[ \frac{1}{2} \sin(2) - 0 = \frac{1}{2} \sin(2) \] Now, we calculate \( \int_{0}^{1} \sin(2y) \, dy \): \[ \int_{0}^{1} \sin(2y) \, dy = -\frac{1}{2} \cos(2y) \bigg|_{0}^{1} = -\frac{1}{2} (\cos(2) - \cos(0)) = -\frac{1}{2} (\cos(2) - 1) \] Thus, \[ = -\frac{1}{2} \cos(2) + \frac{1}{2} = \frac{1 - \cos(2)}{2} \] Putting everything together, we have: \[ \int_{0}^{1} y \cos(2y) \, dy = \frac{1}{2} \sin(2) - \frac{1}{4} \left( 1 - \cos(2) \right) \] Finally, combining all terms in the calculation: \[ \int_{0}^{1} y \cos^2(y) \, d y = \frac{1}{2} \left( \frac{1}{2} + \left( \frac{1}{2} \sin(2) - \frac{1}{4} (1 - \cos(2)) \right) \right) \] Now let's finalize it: