Its beautiful, if you know the secret

Set a=2k+1 (a becomes and odd number). And change the integral with the sumation: $\sum\limits_{k=1}^{\infty }{\int_{0}^{\pi }{\frac{\sin (ax)}{a}dx}}={{\sum\limits_{k=1}^{\infty }{{{\left[ -\frac{\cos (ax)}{{{a}^{2}}} \right]}_{0}}}}^{\pi }}=\sum\limits_{k=1}^{\infty }{\frac{2}{{{a}^{2}}}}=2\sum\limits_{k=1}^{\infty }{\frac{1}{{{(2k+1)}^{2}}}}$

Knowing that: $\sum\limits_{k=1}^{\infty }{\frac{1}{{{k}^{2}}}}=\frac{{{\pi }^{2}}}{6}$ The sum of the inverse square off even integers is easy to calculate: $\sum\limits_{k=1}^{\infty }{\frac{1}{{{(2k)}^{2}}}}=\frac{1}{4}\sum\limits_{k=1}^{\infty }{\frac{1}{{{k}^{2}}}}=\frac{1}{4}\frac{{{\pi }^{2}}}{6}=\frac{{{\pi }^{2}}}{24}$ and: $\sum\limits_{k=1}^{\infty }{\frac{1}{{{(2k+1)}^{2}}}}=\sum\limits_{k=1}^{\infty }{\frac{1}{{{k}^{2}}}}-\sum\limits_{k=1}^{\infty }{\frac{1}{{{(2k)}^{2}}}}=\frac{{{\pi }^{2}}}{6}-\frac{{{\pi }^{2}}}{24}=\frac{{{\pi }^{2}}}{8}$

Therefore the result is: $\frac{{{\pi }^{2}}}{4}$

and 10a+b = 24

u can do this without interchanging sum and integral