Evaluate the definite integral $\int\limits_0^\pi {\dfrac{{x\tan x}}{{\sec x + \tan x}}} dx$.

Hint: Before attempting this question, one should have prior knowledge about the concept of definite integrals and also remember to use $\int\limits_0^a {f\left( x \right)dx = \int\limits_0^a {f\left( {a - x} \right)} } dx$, \[\int\limits_b^a {f\left( x \right)} = f\left( a \right) - f\left( b \right)\]in the given integral, using this will help you to approach the solution of the question.

Complete step by step answer:
According to the question, we have to evaluate the following equation i.e.$I = \int\limits_0^\pi {\dfrac{{x\tan x}}{{\sec x + \tan x}}} dx$
We know that by the formula of integration i.e. $\int\limits_0^a {f\left( x \right)dx = \int\limits_0^a {f\left( {a - x} \right)} } dx$applying the formula in the given equation we get
$I = \int\limits_0^\pi {\dfrac{{\left( {\pi - x} \right) + \tan \left( {\pi - x} \right)}}{{\sec \left( {\pi - x} \right) + \tan \left( {\pi - x} \right)}}} $
We know that by the trigonometric identities i.e. \[\tan \left( {\pi - x} \right) = - \tan x\] and \[\sec \left( {\pi - x} \right) = - \sec x\]
Therefore, $\int\limits_0^\pi {\dfrac{{\left( {\pi - x} \right)\left( { - \tan x} \right)}}{{ - \sec x - \tan x}}} dx$
$ \Rightarrow $$\int\limits_0^\pi {\dfrac{{\left( {\pi - x} \right)\tan x}}{{\sec x + \tan x}}} dx$
$I = \int\limits_0^\pi {\dfrac{{\pi \tan xdx}}{{\sec x + \tan x}}} - \int\limits_0^\pi {\dfrac{{x\tan xdx}}{{\sec x + \tan x}}} $
Since it is given that$I = \int\limits_0^\pi {\dfrac{{x\tan xdx}}{{\sec x + \tan x}}} $
Now if you observe, you will find out that right equation is same as the value of I, so we can also write it as,
$I = \int\limits_0^\pi {\dfrac{{{\pi ^{\tan x}}dx}}{{\sec x - \tan x}}} - I$
If we move I to the left side of the equation, we have
$ \Rightarrow $$2I = \pi \int\limits_0^\pi {\dfrac{{\tan x}}{{\sec x + \tan x}}} dx$
Now, if we multiply both the numerator and denominator by$\left( {\sec x + \tan x} \right)$, we get
$I = \dfrac{\pi }{2}\int\limits_0^\pi {\dfrac{{\tan x\left( {\sec x - \tan x} \right)}}{{{{\sec }^2}x - {{\tan }^2}x}}} dx$
We know that by the trigonometric identities i.e. ${\sec ^2}x - {\tan ^2}x = 1$
Therefore, $I = \dfrac{\pi }{2}\int\limits_0^\pi {\sec x\tan x - {{\tan }^2}xdx} $
Since we know that by the trigonometric identities ${\sec ^2}x - {\tan ^2}x = 1$ or $ - {\tan ^2}x = 1 - {\sec ^2}x$
Now substituting the value in the above equation, we get
$I = \dfrac{\pi }{2}\int\limits_0^\pi {\sec x\tan x - \left( {se{c^2}x - 1} \right)} dx$
$ \Rightarrow $$I = \dfrac{\pi }{2}\int\limits_0^\pi {\sec x\tan x - {{\sec }^2}x + 1dx} $
We know that by the integration formulas i.e. $\int {\sec x\left( {\tan x} \right)} = \sec x$, $\int {{{\sec }^2}x} = \tan x$and $\int {1dx} = x$
Therefore, $I = \dfrac{\pi }{2}\left[ {secx - \tan x + x} \right]_0^\pi $
Since we know that in definite integrals the formula of integral is \[\int\limits_b^a {f\left( x \right)} = f\left( a \right) - f\left( b \right)\]
Therefore, by substituting the values of limit in the above equation we get$I = \dfrac{\pi }{2}\left[ {\sec \pi - \tan \pi + \pi - \sec 0 + \tan 0 - 0} \right]$
As we all know that $\sec x + \tan x = 1$
Therefore, $I = \dfrac{\pi }{2}\left( { - 1 + \pi - 1} \right)$
$ \Rightarrow $$I = \dfrac{\pi }{2}\left( {\pi - 2} \right)$
$ \Rightarrow $$I = \dfrac{{{\pi ^2}}}{2} - \pi $
So, the value of definite integral is $I = \dfrac{{{\pi ^2}}}{2} - \pi $

Note:
In the above solution we used the concepts of definite and indefinite integrals which are the two types of integration where definite integrals consist of two limits in integrals both of the limits are named as upper limit and lower limit which defines the lowest and height value of the given function whereas indefinite integrals doesn’t consists of upper and lower limits also the difference in definite and indefinite integrals is the definite and indefinite values given by the definite and indefinite integrals.