\( J = \int _ { 0 } ^ { \frac { \pi } { 2 } } ( 1 + \sin x ) ^ { 2 } \cdot \cos x d x \)

To solve the integral \( J = \int _ { 0 } ^ { \frac { \pi } { 2 } } ( 1 + \sin x ) ^ { 2 } \cdot \cos x \, dx \), we can start by expanding the integrand: \[ (1 + \sin x)^2 = 1 + 2\sin x + \sin^2 x \] Thus, we can rewrite the integral as: \[ J = \int _ { 0 } ^ { \frac { \pi } { 2 } } (1 + 2\sin x + \sin^2 x) \cos x \, dx \] Now, split this into three separate integrals: \[ J = \int _ { 0 } ^ { \frac { \pi } { 2 } } \cos x \, dx + 2 \int _ { 0 } ^ { \frac { \pi } { 2 } } \sin x \cos x \, dx + \int _ { 0 } ^ { \frac { \pi } { 2 } } \sin^2 x \cos x \, dx \] 1. The first integral: \[ \int _ { 0 } ^ { \frac { \pi } { 2 } } \cos x \, dx = \left[ \sin x \right] _ { 0 } ^ { \frac { \pi } { 2 } } = 1 - 0 = 1 \] 2. The second integral can be simplified using the substitution \( u = \sin x \). Then, \( du = \cos x \, dx \): \[ \int _ { 0 } ^ { \frac { \pi } { 2 } } \sin x \cos x \, dx = \int _ { 0 } ^ { 1 } u \, du = \left[ \frac{u^2}{2} \right] _{0}^{1} = \frac{1}{2} \] Thus, \[ 2 \int _ { 0 } ^ { \frac { \pi } { 2 } } \sin x \cos x \, dx = 2 \cdot \frac{1}{2} = 1 \] 3. For the third integral, again using \( u = \sin x \): \[ \int _ { 0 } ^ { \frac { \pi } { 2 } } \sin^2 x \cos x \, dx = \int _ { 0 } ^ { 1 } u^2 \, du = \left[ \frac{u^3}{3} \right] _{0}^{1} = \frac{1}{3} \] Now, we combine all parts: \[ J = 1 + 1 + \frac{1}{3} = 2 + \frac{1}{3} = \frac{6}{3} + \frac{1}{3} = \frac{7}{3} \] Thus, the final answer is: \[ \boxed{\frac{7}{3}} \]