Question

If the value of $I = \int\limits_{ - \pi }^\pi {\dfrac{{{{\sin }^2}x}}{{1 + {a^x}}}dx} $, a>0, then $I$ equals?

Answer 1

Hint: If we assume the value of $I$to be $I = \int\limits_a^b {f\left( x \right)dx = \int\limits_a^b {f\left( {a + b - x} \right)} } dx$, using the property we will put the value of $x$ in the equation given in the question as $\left( {a + b - x} \right)$ which in this case is $\left( { - \pi + \pi - x} \right)$and we get this equation: $I = \int\limits_{ - \pi }^\pi {\dfrac{{{{\sin }^2}\left( { - x} \right)}}{{1 + {a^{ - x}}}}dx} $.

Complete step-by-step answer:
Now, in order to solve this question further, we will convert the negative value in the denominator i.e. ${a^{ - x}}$into positive, making it $\dfrac{1}{{{a^x}}}$and replace it in the above equation and ${\sin ^2}\left( { - x} \right)$can also be written as ${\sin ^2}x$.
We have $I = \int\limits_{ - \pi }^\pi {\dfrac{{{{\sin }^2}\left( { - x} \right)}}{{1 + \dfrac{1}{{{a^x}}}}}dx} $, taking the LCM of the denominator $\left( {1 + \dfrac{1}{{{a^x}}}} \right)$ we have $\left( {\dfrac{{{a^x} + 1}}{{{a^x}}}} \right)$
Replacing the denominator in the above equation,
$I = \int\limits_{ - \pi }^\pi {\dfrac{{{{\sin }^2}\left( { - x} \right)}}{{\dfrac{{{a^x} + 1}}{{{a^x}}}}}dx} $ = $I = \int\limits_{ - \pi }^\pi {\dfrac{{{a^x}{{\sin }^2}\left( x \right)}}{{{a^x} + 1}}dx} $
Now, taking the equation $I = \int\limits_{ - \pi }^\pi {\dfrac{{{a^x}{{\sin }^2}\left( x \right)}}{{{a^x} + 1}}dx} $ as equation 1
$I = \int\limits_{ - \pi }^\pi {\dfrac{{{a^x}{{\sin }^2}\left( x \right)}}{{{a^x} + 1}}dx} $ $ \to $ equation 1
And taking the equation given in the question as equation 2
$I = \int\limits_{ - \pi }^\pi {\dfrac{{{{\sin }^2}x}}{{1 + {a^x}}}dx} $ $ \to $ equation 2
Adding equation 1 and 2 we get,
$2I = I = \int\limits_{ - \pi }^\pi {\dfrac{{{a^x}{{\sin }^2}x}}{{1 + {a^x}}}dx} + \int\limits_{ - \pi }^\pi {\dfrac{{{{\sin }^2}x}}{{1 + {a^x}}}} dx$
Solving further,
$2I = \int\limits_{ - \pi }^\pi {{{\sin }^2}x\left( {\dfrac{{1 + {a^x}}}{{1 + {a^x}}}} \right)} dx$
taking the 2 from the left side to the right side of the equation we have,
$I = \dfrac{1}{2}\int\limits_{ - \pi }^\pi {{{\sin }^2}xdx} $
Using $
  \cos 2x = 1 - 2{\sin ^2}x \\
    \\
  {\sin ^2}x = \dfrac{{1 - \cos 2x}}{2} \\
$ we have,
$I = \dfrac{1}{4}\int\limits_{ - \pi }^\pi {\left( {1 - \cos 2x} \right)} dx$
Using the property $
  \int\limits_{ - a}^a {f\left( x \right)} dx, \\
  f\left( x \right) = f\left( { - x} \right), \\
  2\int\limits_0^a {f\left( x \right)} dx \\
 $
$I = \dfrac{1}{4} \times 2\int\limits_0^\pi {\left( {1 - \cos 2x} \right)} dx$
$I = \dfrac{1}{2}\left[ {x - \dfrac{{\sin 2x}}{2}} \right]_0^\pi $
Putting the value of $x$as $\pi $and 0,
$I = \dfrac{1}{2}\left[ {\pi - \dfrac{{\sin 2\pi }}{2} - 0} \right]_0^\pi $
$I = \dfrac{\pi }{2}$

Note: Using properties such as $
  \int\limits_{ - a}^a {f\left( x \right)} dx, \\
  f\left( x \right) = f\left( { - x} \right), \\
  2\int\limits_0^a {f\left( x \right)} dx \\
 $ makes the solution easy and solving the questions by using equation 1 and equation 2 is better.