Question

Evaluate the value of $\int\limits_0^\pi {\dfrac{{x\sin x}}{{1 + {{\cos }^2}x}}} dx$ .
A) $\dfrac{{{\pi ^2}}}{4}$
B) $\dfrac{{{\pi ^2}}}{2}$
C) $\dfrac{{{\pi ^2}}}{{2\sqrt 2 }}$
D) $\dfrac{{{\pi ^2}}}{{4\sqrt 2 }}$

Answer 1

Hint:
Let $I = \int\limits_0^\pi {\dfrac{{x\sin x}}{{1 + {{\cos }^2}x}}} dx$
Then, applying the property $\int\limits_a^b {xdx = \int\limits_a^b {\left( {a + b - x} \right)dx}} $.
Now add both the equations before property and after property and solve further.
Finally, find I.

Complete step by step solution:
Let $I = \int\limits_0^\pi {\dfrac{{x\sin x}}{{1 + {{\cos }^2}x}}} dx$ … (1)
Now, applying the property $\int\limits_a^b {xdx = \int\limits_a^b {\left( {a + b - x} \right)dx} } $ in equation (1)
 $\therefore I = \int\limits_0^\pi {\dfrac{{\left( {\pi - x} \right)\sin x}}{{1 + {{\cos }^2}\left( {\pi - x} \right)}}} dx$
$ = \int\limits_0^\pi {\dfrac{{\left( {\pi - x} \right)\sin x}}{{1 + {{\cos }^2}x}}} dx$ … (2)
On adding equation (1) and equation (2), we get
 $
  I + I = \int\limits_0^\pi {\dfrac{{x\sin x}}{{1 + {{\cos }^2}x}}} dx + \int\limits_0^\pi {\dfrac{{\left( {\pi - x} \right)\sin x}}{{1 + {{\cos }^2}x}}} dx \\
\Rightarrow 2I = \int\limits_0^\pi {\dfrac{{\left( {\pi - x + x} \right)\sin x}}{{1 + {{\cos }^2}x}}} dx \\
 \Rightarrow 2I = \int\limits_0^\pi {\dfrac{{\pi \sin x}}{{1 + {{\cos }^2}x}}} dx \\
 \Rightarrow 2I = \pi \int\limits_0^\pi {\dfrac{{\sin x}}{{1 + {{\cos }^2}x}}} dx \\
 $
Let $\cos x = t$.
 $\therefore - \sin xdx = dt$
Also, at $x = 0,t = \cos \left( 0 \right) = 1$ and at $x = \pi ,t = \cos \left( \pi \right) = - 1$ .
So, \[2I = - \pi \int\limits_1^{ - 1} {\dfrac{{dt}}{{1 + {t^2}}}} \]
Applying the property, $\int\limits_{ - a}^a {xdx} = - \int\limits_a^{ - a} {xdx} $ , we get
 \[2I = \pi \int\limits_{ - 1}^1 {\dfrac{{dt}}{{1 + {t^2}}}} \]
 \[
  \therefore 2I = \pi \left[ {{{\tan }^{ - 1}}t} \right]_{ - 1}^1 \\
  \therefore 2I = \pi \left[ {{{\tan }^{ - 1}}\left( 1 \right) - {{\tan }^{ - 1}}\left( { - 1} \right)} \right] \\
  \therefore 2I = \pi \left[ {{{\tan }^{ - 1}}\left( 1 \right) + {{\tan }^{ - 1}}\left( 1 \right)} \right] \\
  \therefore 2I = \pi \times 2 \times {\tan ^{ - 1}}\left( 1 \right) \\
  \therefore \dfrac{{2I}}{2} = \pi \times \dfrac{\pi }{4} \\
  \therefore I = \dfrac{{{\pi ^2}}}{4} \\
 \]

Thus, option (A) is correct.

Note:
Some properties of definite integration:
$\int\limits_a^b {f\left( x \right)dx = \int\limits_a^b {\left( {a + b - f\left( x \right)} \right)dx} } $
 $\int\limits_{ - a}^a {f\left( x \right)dx} = - \int\limits_a^{ - a} {f\left( x \right)dx} $
 $\int\limits_a^b {kxdx = } k\int\limits_a^b {xdx} $
 $\int\limits_a^a {f\left( x \right)dx = } 0$
 $\int\limits_a^b {f\left( x \right)dx} = - \int\limits_b^a {f\left( x \right)dx}$