JEE MAIN - Mathematics (2020 - 9th January Morning Slot - No. 2)
The value of
$$\int\limits_0^{2\pi } {{{x{{\sin }^8}x} \over {{{\sin }^8}x + {{\cos }^8}x}}} dx$$ is equal to :
$$\int\limits_0^{2\pi } {{{x{{\sin }^8}x} \over {{{\sin }^8}x + {{\cos }^8}x}}} dx$$ is equal to :
4$$\pi $$
2$$\pi $$
$$\pi $$2
2$$\pi $$2
Explanation
I = $$\int\limits_0^{2\pi } {{{x{{\sin }^8}x} \over {{{\sin }^8}x + {{\cos }^8}x}}} dx$$ .....(1)
I = $$\int\limits_0^{2\pi } {{{\left( {2\pi - x} \right){{\sin }^8}x} \over {{{\sin }^8}x + {{\cos }^8}x}}} dx$$ ......(2)
Adding (1) and (2)
2I = $$\int\limits_0^{2\pi } {{{2\pi {{\sin }^8}x} \over {{{\sin }^8}x + {{\cos }^8}x}}} dx$$
$$ \Rightarrow $$ I = $$2\pi \int\limits_0^\pi {{{{{\sin }^8}x} \over {{{\sin }^8}x + {{\cos }^8}x}}} dx$$
$$ \Rightarrow $$ I = $$2\pi \left[ {\int\limits_0^{\pi /2} {{{{{\sin }^8}x} \over {{{\sin }^8}x + {{\cos }^8}x}}} dx + \int\limits_0^{\pi /2} {{{{{\cos }^8}x} \over {{{\sin }^8}x + {{\cos }^8}x}}} dx} \right]$$
= $${2\pi \int\limits_0^{\pi /2} 1 dx}$$
= $$2\pi .{\pi \over 2}$$ = $${\pi ^2}$$
I = $$\int\limits_0^{2\pi } {{{\left( {2\pi - x} \right){{\sin }^8}x} \over {{{\sin }^8}x + {{\cos }^8}x}}} dx$$ ......(2)
Adding (1) and (2)
2I = $$\int\limits_0^{2\pi } {{{2\pi {{\sin }^8}x} \over {{{\sin }^8}x + {{\cos }^8}x}}} dx$$
$$ \Rightarrow $$ I = $$2\pi \int\limits_0^\pi {{{{{\sin }^8}x} \over {{{\sin }^8}x + {{\cos }^8}x}}} dx$$
$$ \Rightarrow $$ I = $$2\pi \left[ {\int\limits_0^{\pi /2} {{{{{\sin }^8}x} \over {{{\sin }^8}x + {{\cos }^8}x}}} dx + \int\limits_0^{\pi /2} {{{{{\cos }^8}x} \over {{{\sin }^8}x + {{\cos }^8}x}}} dx} \right]$$
= $${2\pi \int\limits_0^{\pi /2} 1 dx}$$
= $$2\pi .{\pi \over 2}$$ = $${\pi ^2}$$
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