Answer
$$\dfrac{4}{\pi^2} $$
Work Step by Step
Our aim is to integrate the triple integral as follows:
$$ V=\int^1_0 \int^{1-x}_0 \int^{cos(\pi x/2)}_0 dz dy dx \\= \int^1_0 \int^{1-x}_0 \cos(\dfrac{\pi x}{2}) \space dy \space dx \\=\int^1_0 (\ cos\dfrac{\pi x}{2})(1-x) \space dx \\=\int^1_0 \cos (\dfrac{\pi x}{2})dx-\int^1_0 x \cos (\dfrac{\pi x}{2}) dx \\=[\dfrac{2}{\pi} sin \dfrac{\pi x}{2}]^1_0 -\dfrac{4}{\pi^2} \int^{\pi/2}_0 u \cos u du \\=\dfrac{2}{\pi}-\dfrac{4}{\pi^2}[\cos u+u\sin u]^{\pi/2}_0 \\ = \dfrac{2}{\pi}-\dfrac{4}{\pi^2}(\dfrac{\pi}{2}-1) \\ =\dfrac{4}{\pi^2} $$
