$\int\limits_0^{\pi /2} {\cfrac{{{{\cos }^5}xdx}}{{{{\sin }^5}x + {{\cos }^5}x}}}$
$\int\limits_0^{\pi /2} {\cfrac{{{{\cos }^5}xdx}}{{{{\sin }^5}x + {{\cos }^5}x}}}$
Official Solution
Let$I = \int\limits_0^{\pi /2} {\cfrac{{{{\cos }^5}x}}{{{{\sin }^5}x + {{\cos }^5}x}}} dx$
…(i)
Also, $I = \int\limits_0^{\pi /2} {\cfrac{{{{\cos }^5}\left( {\cfrac{\pi }{2} - x} \right)}}{{{{\sin }^5}\left( {\cfrac{\pi }{2} - x} \right) + {{\cos }^5}\left( {\cfrac{\pi }{2} - x} \right)}}} dx$
$= \int\limits_0^{\pi /2} {\cfrac{{{{\sin }^5}x}}{{{{\cos }^5}x + {{\sin }^5}x}}dx}$
….(ii)
Adding (i) and (ii),
we have
$2I = \int\limits_0^{\pi /2} {\cfrac{{{{\cos }^2}x}}{{{{\cos }^5}x + {{\sin }^5}x}}dx + \int\limits_0^{\pi /2} {\cfrac{{{{\sin }^5}x}}{{{{\cos }^5}x + {{\sin }^5}x}}dx} }$
$= \int\limits_0^{\pi /2} {\cfrac{{{{\cos }^5}x + {{\sin }^5}x}}{{{{\cos }^5}x + {{\sin }^5}x}}dx} = \int\limits_1^{\pi /2} {1\,dx} = \left[ x \right]_0^{\pi /2} = \cfrac{\pi }{2}$
$\therefore$ $I = \cfrac{\pi }{4}$
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