JEE MAIN - Mathematics (2018 (Offline) - No. 14)
The value of $$\int\limits_{ - \pi /2}^{\pi /2} {{{{{\sin }^2}x} \over {1 + {2^x}}}} dx$$ is
$${\pi \over 4}$$
$${\pi \over 8}$$
$${\pi \over 2}$$
$${4\pi }$$
Explanation
As we know,
$$\int\limits_a^b { + \left( x \right)dx} = \int\limits_a^b {f\left( {a + b - x} \right)} dx$$
Let $$I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {\,\,{{{{\sin }^2}x} \over {1 + {2^x}}}} \,dx$$
$$ \Rightarrow \,\,\,\,I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {\,\,{{{{\sin }^2}\left( { - {\pi \over 2} + {\pi \over 2} - x} \right)} \over {1 + {2^{ - {\pi \over 2} + {\pi \over 2} - x}}}}} $$
$$ \Rightarrow\,\,\,\, I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{{{\sin }^2}x} \over {1 + {2^{ - x}}}}} dx$$
$$ \Rightarrow \,\,\,\,I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{{2^x}{{\sin }^2}x} \over {1 + {2^x}}}} \,\,dx$$
$$\therefore\,\,\,$$ $$2I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {\left( {{{{{\sin }^2}x} \over {1 + {2^x}}} + {{{2^x}{{\sin }^2}x} \over {1 + {2^x}}}} \right)} \,\,dx$$
$$ \Rightarrow \,\,\,\,2I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\sin }^2}x\left( {{{1 + {2^x}} \over {1 + {2^x}}}} \right)} \,\,dx$$
$$ \Rightarrow \,\,\,\,2I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\sin }^2}x} \,\,dx$$
$$ \Rightarrow \,\,\,\,2I = 2\int\limits_0^{{\pi \over 2}} {{{\sin }^2}x} \,\,dx\,\,\,$$ [ as sin2 x is an even function ]
$$ \Rightarrow \,\,\,\,I = \int\limits_0^{{\pi \over 2}} {{{\sin }^2}x} \,\,dx\,\,\,\,$$
[ as $$\,\,\,\,$$ $$\int\limits_0^a {f\left( a \right)\,\,dx = \int\limits_0^a {f\left( {a - x} \right)\,dx]} } $$
So, $$\,\,\,$$ $$I = \int\limits_0^{{\pi \over 2}} {{{\sin }^2}\left( {{b \over 2} - x} \right)} \,\,dx\,\,\,\,$$
$$\,\,\,I = \int\limits_0^{{\pi \over 2}} {{{\cos }^2}x} \,\,dx\,\,\,\,$$
$$\therefore\,\,\,\,$$ $$\,\,\,\,2I = \int\limits_0^{{\pi \over 2}} {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)} \,\,dx\,\,\,\,$$
$$ \Rightarrow \,\,\,\,2I = \int\limits_0^{{\pi \over 2}} \, \,\,dx\,\,\,\,$$
$$ \Rightarrow \,\,\,\,2I = {\left[ x \right]_0}^{{\pi \over 2}}$$
$$ \Rightarrow \,\,\,\,2I = {\pi \over 2}$$
$$ \Rightarrow \,\,\,\,I = {\pi \over 4}$$
$$\int\limits_a^b { + \left( x \right)dx} = \int\limits_a^b {f\left( {a + b - x} \right)} dx$$
Let $$I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {\,\,{{{{\sin }^2}x} \over {1 + {2^x}}}} \,dx$$
$$ \Rightarrow \,\,\,\,I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {\,\,{{{{\sin }^2}\left( { - {\pi \over 2} + {\pi \over 2} - x} \right)} \over {1 + {2^{ - {\pi \over 2} + {\pi \over 2} - x}}}}} $$
$$ \Rightarrow\,\,\,\, I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{{{\sin }^2}x} \over {1 + {2^{ - x}}}}} dx$$
$$ \Rightarrow \,\,\,\,I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{{2^x}{{\sin }^2}x} \over {1 + {2^x}}}} \,\,dx$$
$$\therefore\,\,\,$$ $$2I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {\left( {{{{{\sin }^2}x} \over {1 + {2^x}}} + {{{2^x}{{\sin }^2}x} \over {1 + {2^x}}}} \right)} \,\,dx$$
$$ \Rightarrow \,\,\,\,2I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\sin }^2}x\left( {{{1 + {2^x}} \over {1 + {2^x}}}} \right)} \,\,dx$$
$$ \Rightarrow \,\,\,\,2I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\sin }^2}x} \,\,dx$$
$$ \Rightarrow \,\,\,\,2I = 2\int\limits_0^{{\pi \over 2}} {{{\sin }^2}x} \,\,dx\,\,\,$$ [ as sin2 x is an even function ]
$$ \Rightarrow \,\,\,\,I = \int\limits_0^{{\pi \over 2}} {{{\sin }^2}x} \,\,dx\,\,\,\,$$
[ as $$\,\,\,\,$$ $$\int\limits_0^a {f\left( a \right)\,\,dx = \int\limits_0^a {f\left( {a - x} \right)\,dx]} } $$
So, $$\,\,\,$$ $$I = \int\limits_0^{{\pi \over 2}} {{{\sin }^2}\left( {{b \over 2} - x} \right)} \,\,dx\,\,\,\,$$
$$\,\,\,I = \int\limits_0^{{\pi \over 2}} {{{\cos }^2}x} \,\,dx\,\,\,\,$$
$$\therefore\,\,\,\,$$ $$\,\,\,\,2I = \int\limits_0^{{\pi \over 2}} {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)} \,\,dx\,\,\,\,$$
$$ \Rightarrow \,\,\,\,2I = \int\limits_0^{{\pi \over 2}} \, \,\,dx\,\,\,\,$$
$$ \Rightarrow \,\,\,\,2I = {\left[ x \right]_0}^{{\pi \over 2}}$$
$$ \Rightarrow \,\,\,\,2I = {\pi \over 2}$$
$$ \Rightarrow \,\,\,\,I = {\pi \over 4}$$
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