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# Evaluate the integral $\int\limits_{0}^{\pi }{\dfrac{x\sin x}{1+\sin x}dx}$.

Last updated date: 06th Aug 2024
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Hint: Solve the integral by replacing x by $\left( \pi -x \right)$as per $\int\limits_{0}^{a}{f\left( x \right)dx=}\int\limits_{0}^{a}{f\left( a-x \right)dx}$. Then simplify it using trigonometric identities. Finally, after integration substitute $\left( \pi ,0 \right)$in the place of x.

Complete step-by-step solution -
Given the integral, $\int\limits_{0}^{\pi }{\dfrac{x\sin x}{1+\sin x}dx}$.
Let’s put, $I=\int\limits_{0}^{\pi }{\dfrac{x\sin x}{1+\sin x}dx}$.
We know that, $\int\limits_{0}^{a}{f\left( x \right)dx=}\int\limits_{0}^{a}{f\left( a-x \right)dx}$.
Thus, x becomes $\left( \pi -x \right)$.
$\therefore I=\int\limits_{0}^{\pi }{\dfrac{\left( \pi -x \right)\sin x}{1+\sin \left( \pi -x \right)}dx}$
We know, $\sin \left( 180-\theta \right)=\sin \theta$
$\sin \left( \pi -x \right)=\sin x$
\begin{align} & I=\int\limits_{0}^{\pi }{\dfrac{\left( \pi -x \right)\sin x}{1+\sin x}}=\int\limits_{0}^{\pi }{\left( \dfrac{\pi \sin x-x\sin x}{1+\sin x} \right)dx} \\ & I=\int\limits_{0}^{\pi }{\left( \dfrac{\pi \sin x}{1+\sin x}-\dfrac{x\sin x}{1+\sin x} \right)dx} \\ \end{align}
$I=\int\limits_{0}^{\pi }{\dfrac{\pi \sin x}{1+\sin x}dx}-\int\limits_{0}^{\pi }{\dfrac{x\sin x}{1+\sin x}dx}$
$I=\int\limits_{0}^{\pi }{\dfrac{\pi \sin x}{1+\sin x}dx}-I$
\begin{align} & \Rightarrow I+I=\int\limits_{0}^{\pi }{\dfrac{\pi \sin x}{1+\sin x}dx} \\ & 2I=\int\limits_{0}^{\pi }{\dfrac{\pi \sin x}{1+\sin x}dx} \\ & \therefore I=\dfrac{\pi }{2}\int\limits_{0}^{\pi }{\dfrac{\sin x}{1+\sin x}dx} \\ \end{align}
Multiply numerator and denominator with $\left( 1-\sin x \right)$.
\begin{align} & I=\dfrac{\pi }{2}\int\limits_{0}^{\pi }{\dfrac{\sin x\left( 1-\sin x \right)}{\left( 1+\sin x \right)\left( 1-\sin x \right)}dx} \\ & =\dfrac{\pi }{2}\int\limits_{0}^{\pi }{\dfrac{\sin x\left( 1-\sin x \right)}{1-{{\sin }^{2}}x}dx} \\ & =\dfrac{\pi }{2}\int\limits_{0}^{\pi }{\dfrac{\sin x-{{\sin }^{2}}x}{1-{{\sin }^{2}}x}dx} \\ & =\dfrac{\pi }{2}\int\limits_{0}^{\pi }{\dfrac{\sin x}{{{\cos }^{2}}x}dx}-\dfrac{\pi }{2}\int\limits_{0}^{\pi }{\dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x}dx} \\ & =\dfrac{\pi }{2}\int\limits_{0}^{\pi }{\dfrac{\tan x}{\cos x}dx}-\dfrac{\pi }{2}\int\limits_{0}^{\pi }{{{\tan }^{2}}xdx} \\ & =\dfrac{\pi }{2}\int\limits_{0}^{\pi }{\tan x.\sec x.dx}-\dfrac{\pi }{2}\int\limits_{0}^{\pi }{{{\tan }^{2}}x.dx} \\ \end{align}
We know, $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$.
\begin{align} & {{\sin }^{2}}x+{{\cos }^{2}}x=1 \\ & \therefore {{\cos }^{2}}x=1-{{\sin }^{2}}x \\ & \tan x=\dfrac{\sin x}{\cos x} \\ \end{align}
Which are basic, trigonometric formulae.
$\because \dfrac{1}{\cos x}=\sec x$
We know $\int{\tan x.\sec x=\sec x}$and $\int{{{\sec }^{2}}x=\tan x}$.
Similarly, ${{\tan }^{2}}x={{\sec }^{2}}x-1$.
\begin{align} & =\dfrac{\pi }{2}\int\limits_{0}^{\pi }{\tan x.\sec x.dx}-\dfrac{\pi }{2}\int\limits_{0}^{\pi }{\left( {{\sec }^{2}}-1 \right)dx} \\ & =\dfrac{\pi }{2}\left[ \sec x \right]_{0}^{\pi }-\dfrac{\pi }{2}\left[ \int\limits_{0}^{\pi }{{{\sec }^{2}}x}dx-\int\limits_{0}^{\pi }{1.dx} \right] \\ & =\dfrac{\pi }{2}\left[ \sec x \right]_{0}^{\pi }-\dfrac{\pi }{2}\left[ \left[ \tan x \right]_{0}^{\pi }-\left[ x \right]_{0}^{\pi } \right] \\ & =\dfrac{\pi }{2}\left[ \left[ \sec x \right]_{0}^{\pi }-\left[ \tan x \right]_{0}^{\pi }+\left[ x \right]_{0}^{\pi } \right] \\ & =\dfrac{\pi }{2}\left[ \left( \sec \pi -\sec 0 \right)-\left( \tan \pi -\tan 0 \right)+\left( \pi -0 \right) \right] \\ \end{align}
$\sec \pi =-1$ $\tan \pi =0$
$\sec 0=+1$ $\tan 0=0$
\begin{align} & =\dfrac{\pi }{2}\left[ \left[ -1-1 \right]+\pi \right] \\ & =\dfrac{\pi }{2}\left[ -2+\pi \right] \\ & =\dfrac{\pi \left( \pi -2 \right)}{2} \\ & \therefore I=\dfrac{\pi \left( \pi -2 \right)}{2} \\ \end{align}
Hence, by evaluating the integral, we get $\dfrac{\pi \left( \pi -2 \right)}{2}$.
Note:- Be careful while simplifying the integral. Open brackets, don’t mix up the sign. Remember the basic identities and trigonometric formulae. You should learn the integral values of $\tan x.\sec x,{{\sec }^{2}}x$ etc, which we have used in solving the integral. Finally substitute $\left( \pi ,0 \right)$and simplify the expression.