What is the value of the integral \[\int\limits_0^\pi {\dfrac{{x\tan x}}{{\sec x + \cos x}} dx} \]?
A. \[\dfrac{{{\pi ^2}}}{4}\]
B. \[\dfrac{{{\pi ^2}}}{2}\]
C. \[\dfrac{{3{\pi ^2}}}{2}\]
D. \[\dfrac{{{\pi ^2}}}{3}\]

Hint: Here, a definite integral is given. First, apply the property of the definite integral \[\int\limits_0^a {f\left( x \right) dx} = \int\limits_0^a {f\left( {a - x} \right) dx} \] and rewrite the given integral. Then, add both integrals and simplify it. After that, solve the right-hand side by using the basic trigonometric ratios. Then, substitute \[\cos x = u\] in the given integral and calculate the new values of the upper and lower limits. Simplify the integral by applying the integral property \[\int\limits_b^a {f\left( x \right) dx} = - \int\limits_a^b {f\left( x \right) dx} \]. In the end, solve apply the upper and lower limit of the integration and solve it to get the required answer.

Formula Used: Properties of definite integral:
\[\int\limits_b^a {f\left( x \right) dx} = - \int\limits_a^b {f\left( x \right) dx} \]
\[\int\limits_0^a {f\left( x \right) dx} = \int\limits_0^a {f\left( {a - x} \right) dx} \]
Trigonometric ratios:
\[\tan x = \dfrac{{\sin x}}{{\cos x}}\]
\[\sec x = \dfrac{1}{{\cos x}}\]
Integration Formula: \[\int {\dfrac{{dx}}{{{a^2} + {x^2}}} = \dfrac{1}{a}} {\tan ^{ - 1}}\left( {\dfrac{x}{a}} \right)\] , where \[a\] is a constant.

Complete step by step solution: The given definite integral is \[\int\limits_0^\pi {\dfrac{{x\tan x}}{{\sec x + \cos x}} dx} \].

Let consider,
\[I = \int\limits_0^\pi {\dfrac{{x\tan x}}{{\sec x + \cos x}} dx} \]
Apply the property of the definite integral \[\int\limits_0^a {f\left( x \right) dx} = \int\limits_0^a {f\left( {a - x} \right) dx} \] on the right-hand side.
We get,
\[I = \int\limits_0^\pi {\dfrac{{\left( {\pi - x} \right)\tan \left( {\pi - x} \right)}}{{\sec \left( {\pi - x} \right) + \cos \left( {\pi - x} \right)}} dx} \]
\[ \Rightarrow I = \int\limits_0^\pi {\dfrac{{\left( {\pi - x} \right)\tan x}}{{\sec x + \cos x}} dx} \]
Now add above equation with the given equation.
\[2I = \int\limits_0^\pi {\dfrac{{x\tan x}}{{\sec x + \cos x}} dx} + \int\limits_0^\pi {\dfrac{{\left( {\pi - x} \right)\tan x}}{{\sec x + \cos x}} dx} \]
\[ \Rightarrow 2I = \int\limits_0^\pi {\left[ {\dfrac{{x\tan x}}{{\sec x + \cos x}} + \dfrac{{\left( {\pi - x} \right)\tan x}}{{\sec x + \cos x}}} \right] dx} \]
\[ \Rightarrow 2I = \int\limits_0^\pi {\left[ {\dfrac{{\left( {x + \pi - x} \right)\tan x}}{{\sec x + \cos x}}} \right] dx} \]
\[ \Rightarrow 2I = \pi \int\limits_0^\pi {\dfrac{{\tan x}}{{\sec x + \cos x}} dx} \]
\[ \Rightarrow I = \dfrac{\pi }{2}\int\limits_0^\pi {\dfrac{{\tan x}}{{\sec x + \cos x}} dx} \]
Rewrite the right-hand side by using the basic trigonometric ratios.
 \[I = \dfrac{\pi }{2}\int\limits_0^\pi {\dfrac{{\dfrac{{\sin x}}{{\cos x}}}}{{\dfrac{1}{{\cos x}} + \cos x}} dx} \]
\[ \Rightarrow I = \dfrac{\pi }{2}\int\limits_0^\pi {\dfrac{{\dfrac{{\sin x}}{{\cos x}}}}{{\dfrac{{1 + {{\cos }^2}x}}{{\cos x}}}} dx} \]
\[ \Rightarrow I = \dfrac{\pi }{2}\int\limits_0^\pi {\dfrac{{\sin x}}{{1 + {{\cos }^2}x}} dx} \] \[.....\left( 1 \right)\]

Now substitute \[\cos x = u\] in the above equation.
Differentiate the substituting equation, we get
\[ - \sin xdx = du\]
\[ \Rightarrow \sin xdx = - du\]
And limits changes as follows:
\[x = 0 \Rightarrow u = 1\] and \[x = \pi \Rightarrow u = - 1\]

We get the equation \[\left( 1 \right)\] as follows:
\[I = \dfrac{\pi }{2}\int\limits_1^{ - 1} {\dfrac{{ - du}}{{1 + {u^2}}}} \]
Apply the property of definite integral \[\int\limits_b^a {f\left( x \right) dx} = - \int\limits_a^b {f\left( x \right) dx} \] on the right-hand side.
\[ \Rightarrow I = \dfrac{\pi }{2}\int\limits_{ - 1}^1 {\dfrac{{du}}{{1 + {u^2}}}} \]
Solve the integral by using the standard integral formula \[\int {\dfrac{{dx}}{{{a^2} + {x^2}}} = \dfrac{1}{a}} {\tan ^{ - 1}}\left( {\dfrac{x}{a}} \right)\].
\[ \Rightarrow I = \dfrac{\pi }{2}\left[ {{{\tan }^{ - 1}}u} \right]_{ - 1}^1\]
\[ \Rightarrow I = \dfrac{\pi }{2}\left[ {{{\tan }^{ - 1}}\left( 1 \right) - {{\tan }^{ - 1}}\left( { - 1} \right)} \right]\]
\[ \Rightarrow I = \dfrac{\pi }{2}\left[ {\dfrac{\pi }{4} - \left( { - \dfrac{\pi }{4}} \right)} \right]\]
\[ \Rightarrow I = \dfrac{\pi }{2}\left[ {\dfrac{\pi }{4} + \dfrac{\pi }{4}} \right]\]
\[ \Rightarrow I = \dfrac{\pi }{2}\left[ {\dfrac{\pi }{2}} \right]\]
\[ \Rightarrow I = \dfrac{{{\pi ^2}}}{4}\]
Therefore,
\[\int\limits_0^\pi {\dfrac{{x\tan x}}{{\sec x + \cos x}} dx} = \dfrac{{{\pi ^2}}}{4}\]


Option ‘A’ is correct

Note: Students often get confused about the formula of the definite integral of the function. They used \[\int\limits_a^b {f\left( x \right)} dx = F\left( b \right) + F\left( a \right)\] , which is incorrect. The correct formula is \[\int\limits_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right)\].
Sometimes they also add integration constant \[c\] in the definite integral. But definite integral is calculated for a certain interval. So, there is no need to write the integration constant.