The value of [_ - ^ ^2 1 + a^xdx,,,a 0, ] is

Question

The value of $$\int\limits_{ - \pi }^\pi  {{{{{\cos }^2}} \over {1 + {a^x}}}dx,\,\,a > 0,} $$  is

Accepted Answer

Let $$I = \int\limits_{ - \pi }^\pi  {{{{{\cos }^2}x} \over {1 + {a^x}}}} dx\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

$$ = \int\limits_{ - \pi }^\pi  {{{{{\cos }^2}\left( { - x} \right)} \over {1 + {a^{ - x}}}}} dx$$

$$\left[ \, \right.$$ Using $$\int\limits_a^b {f\left( x \right)} dx = \int\limits_a^b {f\left( {a + b - x} \right)dx} $$ $$\left. \, \right]$$

$$ = \int\limits_{ - \pi }^\pi  {{{{{\cos }^2}x} \over {1 + {\alpha ^x}}}} dx\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)$$

Adding equations $(1)$ and $(2)$ we get

$$2I = \int\limits_{ - \pi }^\pi  {{{\cos }^2}} x\left( {{{1 + {a^x}} \over {1 + {a^x}}}} \right)dx$$

$$ = \int\limits_{ - \pi }^\pi  {{{\cos }^2}} x\,dx$$

$$ = 2\int\limits_0^\pi  {{{\cos }^2}} x\,dx$$

$$ = 2 \times 2\int\limits_0^{{\pi  \over 2}} {{{\cos }^2}} x\,dx$$

$$ = 4\int\limits_0^{{\pi  \over 2}} {{{\sin }^2}} x\,dx$$

$$ \Rightarrow I = 2\int\limits_0^{{\pi  \over 2}} {{{\sin }^2}} \,x\,dx$$

$$ = 2\int\limits_0^{{\pi  \over 2}} {\left( {1 - {{\cos }^2}x\,dx} \right)} $$

$$ \Rightarrow I\,\, = 2\int\limits_0^{{\pi  \over 2}} {dx - 2\int\limits_0^{{\pi  \over 2}} {{{\cos }^2}} } \,x\,dx$$

$$ \Rightarrow I + I = 2\left( {{\pi  \over 2}} \right) = \pi $$

$$ \Rightarrow I = {\pi  \over 2}$$

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