Let I=∫−1+11+3xcosαxdx ...(I) I=∫−1+11+3−xcosαxdx (using∫abf(x)dx=∫abf(a+b−x)dx) ...(II) Add (1) and (II) $\begin{aligned}
& 2 I=\int_{-1}^{+1} \cos (\alpha x) d x=2 \int_0^1 \cos (\alpha x) d x \
& I=\frac{\sin \alpha}{\alpha}=\frac{2}{\pi}(\text { given }) \
& \therefore \alpha=\frac{\pi}{2}
\end{aligned}$