Question

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

Answer 1

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

& 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{aligned}$$ Since we know the property $$\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{aligned} & 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{aligned}$$ Now on splitting the above integral using the identity that $${{\sin }^{2}}\left( \pi -x \right)={{\sin }^{2}}x$$we get, $$\begin{aligned} & 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{aligned}$$ Now since we have $$I=\int\limits_{0}^{\pi }{x{{\sin }^{2}}xdx}$$, we have $$\begin{aligned} & 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{aligned}$$ Taking $$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{aligned} & 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{aligned}$$ Now we will divide the equation by 2 both sides, we get $$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.