Question 127 of 246MediumMCQ
The value of \[\int\limits_{ - \pi /2}^{\pi /2} {{{{{\sin }^2}x} \over {1 + {2^x}}}} dx\] is
Solutions · 1
Official solution
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, \[\,\,\,\] \[I = \int\limits_0^{{\pi \over 2}} {{{\sin }^2}\left( {{b \over 2} - x} \right)} \,\,dx\,\,\,\,\]
\[\,\,\,I = \int\limits_0^{{\pi \over 2}} {{{\cos }^2}x} \,\,dx\,\,\,\,\]
\[\therefore\,\,\,\,\] \[\,\,\,\,2I = \int\limits_0^{{\pi \over 2}} {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)} \,\,dx\,\,\,\,\]
\[ \Rightarrow \,\,\,\,2I = \int\limits_0^{{\pi \over 2}} \, \,\,dx\,\,\,\,\]
\[ \Rightarrow \,\,\,\,2I = {\left[ x \right]_0}^{{\pi \over 2}}\]
\[ \Rightarrow \,\,\,\,2I = {\pi \over 2}\]
\[ \Rightarrow \,\,\,\,I = {\pi \over 4}\]
Answer the question to unlock