Question 170 of 246MediumMCQ
The value of \[\int\limits_{ - \pi /2}^{\pi /2} {{{{{\cos }^2}x} \over {1 + {3^x}}}} dx\] is :
Solutions · 1
Official solution
Let \[I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{{{\cos }^2}x} \over {1 + {3^x}}}} dx\] .... (1)
Replace x with -x,
\[ \therefore \] \[I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{{{\cos }^2}x} \over {1 + {1 \over {{3^x}}}}}} \] .... (2)
Adding (1) and (2), we get,
\[2I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{{{\cos }^2}x + {3^x}{{\cos }^2}x} \over {1 + {3^x}}}} dx\]
\[ = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{{{\cos }^2}x(1 + {3^x})} \over {1 + {3^x}}}dx} \]
\[ = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\cos }^2}x} \,dx\]
[\[{{{\cos }^2}x}\] is a even function as f(x)=f(−x) for \[{\cos ^2}x \]]
\[= 2\int\limits_0^{{\pi \over 2}} {{{\cos }^2}x} \,dx\]
\[ = 2\int\limits_0^{{\pi \over 2}} {\left( {{{1 + \cos 2x} \over 2}} \right)} \,dx\]
\[ \Rightarrow I = {1 \over 2}\int\limits_0^{{\pi \over 2}} {(1 + \cos 2x)} \,dx\]
\[ = {1 \over 2}\left[ {x + \sin 2x} \right]_0^{{\pi \over 2}}\]
\[ = {1 \over 2}\left[ {{\pi \over 2} - 0} \right]\]
\[ = {\pi \over 4}\]
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