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Last updated date: 05th Dec 2023
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MVSAT Dec 2023

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

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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, \[\sin \left( 180-\theta \right)=\sin \theta \]
                   \[\sin \left( \pi -x \right)=\sin x\]
  & 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} \\
\[I=\int\limits_{0}^{\pi }{\dfrac{\pi \sin x}{1+\sin x}dx}-\int\limits_{0}^{\pi }{\dfrac{x\sin x}{1+\sin x}dx}\]
\[I=\int\limits_{0}^{\pi }{\dfrac{\pi \sin x}{1+\sin x}dx}-I\]
  & \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} \\
Multiply numerator and denominator with \[\left( 1-\sin x \right)\].
  & 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} \\
We know, \[\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}\].
  & {{\sin }^{2}}x+{{\cos }^{2}}x=1 \\
 & \therefore {{\cos }^{2}}x=1-{{\sin }^{2}}x \\
 & \tan x=\dfrac{\sin x}{\cos x} \\
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, \[{{\tan }^{2}}x={{\sec }^{2}}x-1\].
  & =\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] \\
\[\sec \pi =-1\] \[\tan \pi =0\]
\[\sec 0=+1\] \[\tan 0=0\]
  & =\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} \\
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.