Question

${\int }_0^{\pi /2}\frac{{\cos }^{9}x}{{\cos }^{3}x+{\sin }^{3}x}$

Answer 1

Let ${\int }_0^{\frac{\pi }{2}}\frac{{\cos }^{9}xdx}{{\cos }^{3}x+{\sin }^{3}x}=I$

${\int }_0^{\frac{\pi }{2}}\frac{{\sin }^{9}xdx}{{\cos }^{3}x+{\sin }^{3}x}=I$

adding both

$2I={\int }_0^{\frac{\pi }{2}}\frac{{\cos }^{9}x+{\sin }^9}{{\cos }^{3}x+{\sin }^{3}x}dx$

$I=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}({\cos }^{6}x+{\sin }^6-{\sin }^{3}x{\cos }^{3}x)dx$

. $=\frac{1}{2}[\frac{\mathrm{1.3.5}}{\mathrm{2.4.6}}.\frac{\pi }{2}+\frac{\mathrm{1.3.5}}{\mathrm{2.4.6}}\frac{\pi }{2}-\frac{2\times 2}{6\times 4\times 2}]$ (using walli's formula)

$I=\frac{15\pi }{96}-\frac{1}{24}$