Question

The value of $\int\limits_{ - \pi }^\pi {\dfrac{{{{\cos }^2}x}}{{1 + {a^x}}}dx,a > 0} $ is
(a) $2\pi $
(b) $\dfrac{\pi }{a}$
(c) $\dfrac{\pi }{2}$
(d) $a\pi $

Answer 1

Hint: Here we have to use definite integral properties and trigonometry identities to solve the given integral.

Complete step-by-step answer:
Let \[I = \int\limits_{ - \pi }^\pi {\dfrac{{{{\cos }^2}x}}{{1 + {a^x}}}dx} \] ...... (1)
Now we know from the properties of definite integral that $\int\limits_{ - a}^a {f\left( x \right)} dx = \int\limits_{ - a}^a {f\left( { - a + a - x} \right)} dx$
Applying this, we get
\[I = \int\limits_{ - \pi }^\pi {\dfrac{{{{\cos }^2}\left( { - \pi + \pi - x} \right)}}{{1 + {a^{ - \pi + \pi - x}}}}dx} \]
\[I = \int\limits_{ - \pi }^\pi {\dfrac{{{{\cos }^2}\left( { - x} \right)}}{{1 + {a^{ - x}}}}dx} \]
As you know $\cos \left( { - \theta } \right) = \cos \theta $
\[ \Rightarrow I = \int\limits_{ - \pi }^\pi {\dfrac{{{{\cos }^2}\left( x \right)}}{{1 + {a^{ - x}}}}dx} \]
\[ \Rightarrow I = \int\limits_{ - \pi }^\pi {\dfrac{{{{\cos }^2}\left( x \right)}}{{1 + \dfrac{1}{{{a^x}}}}}dx} \]
\[ \Rightarrow I = \int\limits_{ - \pi }^\pi {\dfrac{{{a^x}{{\cos }^2}\left( x \right)}}{{1 + {a^x}}}dx} \] …… (2)
Now add equation (1) and (2)
\[ \Rightarrow 2I = \int\limits_{ - \pi }^\pi {\dfrac{{{{\cos }^2}x}}{{1 + {a^x}}}dx} + \int\limits_{ - \pi }^\pi {\dfrac{{{a^x}{{\cos }^2}\left( x \right)}}{{1 + {a^x}}}dx} \]
\[ \Rightarrow 2I = \int\limits_{ - \pi }^\pi {\dfrac{{\left( {1 + {a^x}} \right){{\cos }^2}\left( x \right)}}{{1 + {a^x}}}dx} = \int\limits_{ - \pi }^\pi {{{\cos }^2}\left( x \right)dx} \]
Now you know ${\cos ^2}x$ is written as $\left( {\dfrac{{1 + \cos 2x}}{2}} \right)$
$2I = \int\limits_{ - \pi }^\pi {\left( {\dfrac{{1 + \cos 2x}}{2}} \right)dx} $
Now for even functions
$ \Rightarrow I = \int\limits_{ - a}^a {f\left( x \right)dx = 2\int\limits_0^a {f\left( x \right)dx} } $
Now you know that cos is even function so, this integral is written as
$2I = 2\int\limits_0^\pi {\left( {\dfrac{{1 + \cos 2x}}{2}} \right)dx} $
\[I = \dfrac{1}{2}\int\limits_0^\pi {\left( {1 + \cos 2x} \right)dx} \]
Now apply the integral
\[ \Rightarrow I = \dfrac{1}{2}\left[ {x + \dfrac{{\sin 2x}}{2}} \right]_0^\pi \]
\[ \Rightarrow I = \dfrac{1}{2}\left[ {\pi + \dfrac{{\sin 2\pi }}{2} - 0 - \dfrac{{\sin 0}}{2}} \right]\]
\[ \Rightarrow I = \dfrac{1}{2}\left[ {\pi + \dfrac{0}{2} - 0 - \dfrac{0}{2}} \right]\]
\[ \Rightarrow I = \dfrac{\pi }{2}\]
So option C is correct.

Note: Always remember integral properties. It will help you a lot in solving the integration problems. Also, be careful with signs. Converting higher power trigonometric functions to lower power is always advisable to ease the solving process.