Question

Using properties of integral evaluate: $\int\limits_{0}^{\pi }{\dfrac{x}{1+\sin x}dx}$.

Answer 1

Hint: The given definite integral requires two different theorems to narrow it down to its simplest form. We apply $\int\limits_{0}^{a}{f\left( x \right)dx}=\int\limits_{0}^{a}{f\left( a-x \right)dx}$ and $\int\limits_{0}^{2a}{f\left( x \right)dx}=2\int\limits_{0}^{a}{f\left( x \right)dx}$ to remove the x component from the numerator part. Then we use trigo integral form to find the solution of the problem.

Complete step-by-step solution
We have been given a definite integral. We assume $I=\int\limits_{0}^{\pi }{\dfrac{x}{1+\sin x}dx}$.
We have the theorem of definite integral $\int\limits_{0}^{a}{f\left( x \right)dx}=\int\limits_{0}^{a}{f\left( a-x \right)dx}$.
For our given integral $a=\pi ,f\left( x \right)=\dfrac{x}{1+\sin x}$.
We try to find the value of $f\left( a-x \right)=f\left( \pi -x \right)=\dfrac{\pi -x}{1+\sin \left( \pi -x \right)}=\dfrac{\pi -x}{1+\sin x}$.
So, $I=\int\limits_{0}^{\pi }{\dfrac{x}{1+\sin x}dx}=\int\limits_{0}^{\pi }{\dfrac{\pi -x}{1+\sin x}dx}$.
We add these two integrals to remove the x in the numerator of the integral.
We have the theorem $\int\limits_{a}^{b}{f\left( x \right)dx}+\int\limits_{a}^{b}{g\left( x \right)dx}=\int\limits_{a}^{b}{\left[ f\left( x \right)+g\left( x \right) \right]dx}$.
So, $2I=\int\limits_{0}^{\pi }{\dfrac{x}{1+\sin x}dx}+\int\limits_{0}^{\pi }{\dfrac{\pi -x}{1+\sin x}dx}=\int\limits_{0}^{\pi }{\dfrac{x+\pi -x}{1+\sin x}dx}=\int\limits_{0}^{\pi }{\dfrac{\pi }{1+\sin x}dx}$.
The integral becomes
\[\begin{align}
  & 2I=\int\limits_{0}^{\pi }{\dfrac{\pi }{1+\sin x}dx} \\
 & \Rightarrow I=\dfrac{\pi }{2}\int\limits_{0}^{\pi }{\dfrac{dx}{1+\sin x}} \\
\end{align}\]
We again apply the theorem $\int\limits_{0}^{2a}{f\left( x \right)dx}=2\int\limits_{0}^{a}{f\left( x \right)dx}$ if $f\left( 2a-x \right)=f\left( x \right)$.
In our integral if we assume \[{{I}_{1}}=\int\limits_{0}^{\pi }{\dfrac{dx}{1+\sin x}}\], we have $\sin \left( \pi -x \right)=\sin x$.
So, \[{{I}_{1}}=\int\limits_{0}^{\pi }{\dfrac{dx}{1+\sin x}}=2\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{dx}{1+\sin x}}\] which means \[I=\dfrac{\pi }{2}{{I}_{1}}=\pi \int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{dx}{1+\sin x}}\].
We multiply with $1-\sin x$ to both numerator and denominator and get
\[I=\pi \int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{dx}{1+\sin x}}=\pi \int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\left( 1-\sin x \right)}{\left( 1+\sin x \right)\left( 1-\sin x \right)}}dx=\pi \int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\left( 1-\sin x \right)}{{{\cos }^{2}}x}}dx\].
Now we use the trigonometric identities and find the integral as
\[I=\pi \int\limits_{0}^{\dfrac{\pi }{2}}{\left( {{\sec }^{2}}x-\tan x\sec x \right)}dx\]. We now use the integral theorem of trigo ratios.
\[\begin{align}
  & I=\pi \int\limits_{0}^{\dfrac{\pi }{2}}{\left( {{\sec }^{2}}x-\tan x\sec x \right)}dx \\
 & \Rightarrow \pi \left[ \tan x-\sec x \right]_{0}^{\dfrac{\pi }{2}} \\
 & \Rightarrow \pi \left[ \left( \tan \dfrac{\pi }{2}-\tan 0 \right)-\left( \sec \dfrac{\pi }{2}-\sec 0 \right) \right] \\
 & \Rightarrow \pi \left[ \left( \tan \dfrac{\pi }{2}-\tan 0 \right)-\left( \sec \dfrac{\pi }{2}-\sec 0 \right) \right] \\
 & \Rightarrow \pi \\
\end{align}\]
So, the integral value of $\int\limits_{0}^{\pi }{\dfrac{x}{1+\sin x}dx}$ is $\pi $.

Note: In the last part of the integral we had terms like \[\tan \dfrac{\pi }{2}\] and \[\sec \dfrac{\pi }{2}\] which have no exact value as they tends to infinity as they approach $\dfrac{\pi }{2}$. But the value of them gets similar as $x\to \dfrac{\pi }{2}$. That’s why we can eliminate them.