Question

The value of ${\int }_{-1/2}^{1/2}\left|xcos\left(\frac{\pi x}{2}\right)\right|dx$, is

• A. $\frac{\pi \sqrt{2}+4\sqrt{2}-8}{{\pi }^2}$
• B. $\frac{\sqrt{2}+4\sqrt{2}-8}{{\pi }^2}$
• C. $\frac{\pi \sqrt{2}+4\sqrt{2}+8}{{\pi }^2}$
• D. none of these

Answer 1

The correct option is A $\frac{\pi \sqrt{2}+4\sqrt{2}-8}{{\pi }^2}$

${\int }_{-1/2}^{1/2}\left|x\cos \left(\frac{\pi x}{2}\right)\right|dx=-{\int }_{-1/2}^{0}x\cos \left(\frac{\pi x}{2}\right)+{\int }_0^{1/2}x\cos \left(\frac{\pi x}{2}\right)dx$
$=-{\left[\frac{2x\sin \left(\frac{\pi x}{2}\right)}{\pi }\right]}_{-1/2}^0+\frac{4}{{\pi }^2}{\int }_{-1/2}^0\frac{\sin \pi x}{2}dx$
$+{\left[\frac{2x\sin \left(\frac{\pi x}{2}\right)}{\pi }\right]}_0^{1/2}-\frac{4}{{\pi }^2}{\int }_0^{1/2}\sin \left(\frac{\pi x}{2}\right)dx$
$=-{\left[\frac{2x\sin \left(\frac{\pi x}{2}\right)}{\pi }\right]}_{-1/2}^0-{\left[\frac{2x\sin \left(\frac{\pi x}{2}\right)}{\pi }\right]}_{-1/2}$
$+{\left[\frac{2x\sin \left(\frac{\pi x}{2}\right)}{\pi }\right]}_0^{1/2}+{\left[\frac{2x\sin \left(\frac{\pi x}{2}\right)}{\pi }\right]}_0^{1/2}$
$=\frac{1}{{\pi }^2}(\pi \sqrt{2}+4\sqrt{2}-8)+c$