Question

Evaluate the integral \[\int\limits_{0}^{\pi }{\dfrac{x\tan x}{\sec x\cdot \text{cosec}x}dx}\].

Answer 1

Hint: In this question, in order to evaluate the definite integral \[\int\limits_{0}^{\pi }{\dfrac{x\tan x}{\sec x\cdot \text{cosec}x}dx}\], we will first substitute the values \[\tan x=\dfrac{\sin x}{\cos x}\], \[\sec x=\dfrac{1}{\cos x}\] and \[\text{cosec}x=\dfrac{1}{\sin x}\] in the integrant of the given integral to simplify it into \[\int\limits_{0}^{\pi }{x{{\sin }^{2}}xdx}\]. Then we will use the property of the integral that \[\int\limits_{0}^{a}{f\left( x \right)dx}=\int\limits_{0}^{a}{f\left( a-x \right)dx}\] in the integral \[\int\limits_{0}^{\pi }{x{{\sin }^{2}}xdx}\] and then will substituting the value \[2{{\sin }^{2}}x=1-\cos 2x\] in the simplified form of the integral. We will then evaluate the same in order to get the desired answer.

Complete step by step answer:
Let \[I\] denote the integral \[\int\limits_{0}^{\pi }{\dfrac{x\tan x}{\sec x\cdot \text{cosec}x}dx}\].
That is, let \[I=\int\limits_{0}^{\pi }{\dfrac{x\tan x}{\sec x\cdot \text{cosec}x}dx}\].
Now on substituting the values \[\tan x=\dfrac{\sin x}{\cos x}\], \[\sec x=\dfrac{1}{\cos x}\] and \[\text{cosec}x=\dfrac{1}{\sin x}\] in the integrant of the above integral we get
\[\begin{align}
  & I=\int\limits_{0}^{\pi }{\dfrac{x\dfrac{\sin x}{\cos x}}{\dfrac{1}{\cos x}\cdot \dfrac{1}{\sin x}}dx} \\
 & =\int\limits_{0}^{\pi }{x{{\sin }^{2}}xdx}
\end{align}\]
Since we know the property \[\int\limits_{0}^{a}{f\left( x \right)dx}=\int\limits_{0}^{a}{f\left( a-x \right)dx}\] of definite integral.
On comparing the integral \[\int\limits_{0}^{\pi }{x{{\sin }^{2}}xdx}\] with the general form \[\int\limits_{0}^{a}{f\left( x \right)dx}\] of definite integral, we will get
\[a=\pi \] and \[f\left( x \right)=x{{\sin }^{2}}x\].
Now using the above mention property, we will get
\[\begin{align}
  & I=\int\limits_{0}^{\pi }{x{{\sin }^{2}}xdx} \\
 & =\int\limits_{0}^{\pi }{\left( \pi -x \right){{\sin }^{2}}\left( \pi -x \right)dx}
\end{align}\]
Now on splitting the above integral using the identity that \[{{\sin }^{2}}\left( \pi -x \right)={{\sin }^{2}}x\]we get,
\[\begin{align}
  & I=\int\limits_{0}^{\pi }{\left( \pi -x \right){{\sin }^{2}}\left( \pi -x \right)dx} \\
 & =\int\limits_{0}^{\pi }{\left( \pi -x \right){{\sin }^{2}}xdx} \\
 & =\int\limits_{0}^{\pi }{\pi {{\sin }^{2}}xdx}-\int\limits_{0}^{\pi }{x{{\sin }^{2}}xdx}
\end{align}\]

Now since we have \[I=\int\limits_{0}^{\pi }{x{{\sin }^{2}}xdx}\], we have
\[\begin{align}
  & I=\int\limits_{0}^{\pi }{\pi {{\sin }^{2}}xdx}-\int\limits_{0}^{\pi }{x{{\sin }^{2}}xdx} \\
 & =\int\limits_{0}^{\pi }{\pi {{\sin }^{2}}xdx}-I
\end{align}\]
Taking \[I\] common in one side, we side
\[2I=\int\limits_{0}^{\pi }{\pi {{\sin }^{2}}xdx}\]
We will further simplify the above integral into
\[2I=\dfrac{\pi }{2}\int\limits_{0}^{\pi }{2{{\sin }^{2}}xdx}\]
Since we have \[2{{\sin }^{2}}x=1-\cos 2x\], we get
\[2I=\dfrac{\pi }{2}\int\limits_{0}^{\pi }{2\left( 1-\cos 2x \right)dx}\]
Now we will spit the above integral.
On solving using the values \[\sin n\pi =0\,\,\,\,\,\forall n\in Z\], we will have
\[\begin{align}
  & 2I=\dfrac{\pi }{2}\int\limits_{0}^{\pi }{2dx}-\dfrac{\pi }{2}\int\limits_{0}^{\pi }{\cos 2xdx} \\
 & =\dfrac{\pi }{2}\left[ x \right]_{0}^{\pi }-\dfrac{\pi }{2}\left[ \dfrac{\sin 2x}{2} \right]_{0}^{\pi } \\
 & =\dfrac{\pi }{2}\left( \pi -0 \right)-\dfrac{\pi }{4}\left( \sin 2\pi -\sin 0 \right) \\
 & =\dfrac{{{\pi }^{2}}}{2}-0 \\
 & =\dfrac{{{\pi }^{2}}}{2}
\end{align}\]
Now we will divide the equation by 2 both sides, we get
\[I=\dfrac{{{\pi }^{2}}}{4}\]
Therefore have that the integral \[\int\limits_{0}^{\pi }{\dfrac{x\tan x}{\sec x\cdot \text{cosec}x}dx}=\dfrac{{{\pi }^{2}}}{4}\].

Note:
In this problem, we can also evaluate the integral \[I=\int\limits_{0}^{\pi }{x{{\sin }^{2}}xdx}\] by first substituting the value \[2{{\sin }^{2}}x=1-\cos 2x\] in the integrant to get \[I=\int\limits_{0}^{\pi }{x\left( \dfrac{1-\cos 2x}{2} \right)dx}\]and then we can split the integral into two parts . Further we can evaluate the same integral by using integration by parts to get the desired result.