Question

If ${\int }_0^{\pi }xf({\cos }^{2}x+{\tan }^{4}x)dx=\frac{\pi k}{1997}{\int }_0^{\pi /2}f({\cos }^{2}x+{\tan }^{4}x)dx$ then the value of k is

Answer 1

$I={\int }_0^{\pi }xf({\cos }^{2}x+{\tan }^{4}x)dx$

Using ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$

$I={\int }_0^{\pi }(\pi -x)f\left({\cos }^2\right(\pi -x)+{\tan }^4(\pi -x\left)\right)dx$

$={\int }_0^{\pi }\pi f({\cos }^{2}x+{\tan }^{4}x)dx-{\int }_0^{\pi }xf({\cos }^{2}x+{\tan }^{4}x)dx$

$\Rightarrow I=\frac{\pi }{2}{\int }_0^{\pi }f({\cos }^{2}x+{\tan }^{4}x)dx$

$=\frac{\pi }{2}\cdot 2{\int }_0^{\pi /2}f({\cos }^{2}x+{\tan }^{4}x)dx$
Therefore $k=1997$