QJEEM20Mae709: JEE Main 2020 Mathematics - Class 12 JEEM20Mae709 Solution

Question

Verified Solution • 2025–26 Session moderate

The value of

\int\limits_0^{2\pi } {{{x{{\sin }^8}x} \over {{{\sin }^8}x + {{\cos }^8}x}}} dx

is equal to :

Accepted Answer

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}

Step-by-Step Verified Solution