Nắng
Anh sẽ vì em làm cha thằng bé
$\sin ^4x+\cos ^4x = \left(\sin ^2x+\cos ^2x\right)^2-2 \ \text{s}in^2x.\cos ^2x = 1-\dfrac{\left(2 \ \text{s}inx\cos x\right)^2}{2} = 1-\dfrac{\sin ^2 2x}{2}<br \dfrac{ }{}>\int_{0}^{\dfrac{\pi }{2}}{\cos _2x\left(\sin ^4x+\cos ^4x\right)dx} = \int_{0}^{\dfrac{\pi }{2}}{\cos _2x\left(1-\dfrac{\sin ^2 2x}{2}\right)dx} = \int_{0}^{\dfrac{\pi }{2}}{\cos _2xdx}-\dfrac{1}{4}\int_{0}^{\dfrac{\pi }{2}}{\sin ^2 2xd\left(\sin 2x\right)}<br \dfrac{ }{}>$