Use king theorem definite integral

$\int _{a}^{b}f(x)dx=\int _{a}^{b}f(a+b-x)dx$

$y=\int _{0}^{\pi /2}\frac{{{\cos }^{\frac{1}{4}}}\left( \frac{\pi }{2}-x \right)}{{{\sin }^{\frac{1}{4}}}\left( \frac{\pi }{2}-x \right)+{{\cos }^{\frac{1}{4}}}\left( \frac{\pi }{2}-x \right)}dx$

$y=\int _{0}^{\pi /2}\frac{{{\sin }^{\frac{1}{4}}}x}{{{\sin }^{\frac{1}{4}}}x+{{\cos }^{\frac{1}{4}}}x}dx….(2)$

Adding equation (1) and (2)

$2y=\int _{0}^{\pi /2}\frac{{{\cos }^{\frac{1}{4}}}x}{{{\sin }^{\frac{1}{4}}}x+{{\cos }^{\frac{1}{4}}}x}dx+\int _{0}^{\pi /2}\frac{{{\sin }^{\frac{1}{4}}}x}{{{\sin }^{\frac{1}{4}}}x+{{\cos }^{\frac{1}{4}}}x}dx$

$2y=\int _{0}^{\pi /2}\frac{{{\cos }^{\frac{1}{4}}}x+{{\sin }^{\frac{1}{4}}}x}{{{\sin }^{\frac{1}{4}}}x+{{\cos }^{\frac{1}{4}}}x}dx$

$2y=\int _{0}^{\pi /2}1dx$

$2y=(x)_{0}^{\pi /2}$

$y=\frac{\pi }{4}$