Find the definite integral $\int\limits_{0}^{2\pi }{\dfrac{\sin 2\theta }{a-b\cos \theta }}d\theta =$
A. $1$
B. $2$
C. $\dfrac{\pi }{4}$
D. $0$

Hint: In this question, we are to find the given integral. To do this, the properties of the definite integral are applied. By the appropriate property, the integral is to be evaluated.



Formula Used:Definite integral:
Consider a function $f(x)$ is defined on $[a,b]$. If the integral of this function, $\int{f(x)dx=F(x)}$, then $F(b)-F(a)$ is called as the definite integral of the function $f(x)$ over $[a,b]$.
I.e.,$\int\limits_{a}^{b}{f(x)dx}=\left. F(x) \right|_{a}^{b}=F(b)-F(a)$
Here $a$ - lower limit and $b$ - upper limit.
Some of the properties of the definite integrals are:
1) Interchanging the limits: \[\int\limits_{a}^{b}{f(x)dx=-}\int\limits_{b}^{a}{f(x)dx}\]
2) If $a3) $\int\limits_{0}^{a}{f(x)dx}=\int\limits_{0}^{a}{f(a-x)dx}$
4) $\int\limits_{-a}^{a}{f(x)dx}=2\int\limits_{0}^{a}{f(x)dx}$ if $f(x)$ is an even function
$\int\limits_{-a}^{a}{f(x)dx}=0$ if $f(x)$ is an odd function
5) $\begin{align}
  & \int\limits_{0}^{2a}{f(x)dx}=2\int\limits_{0}^{a}{f(x)dx};\text{if }f(2a-x)=f(x) \\
 & \text{ }=0\text{ if }f(2a-x)=-f(x) \\
\end{align}$



Complete step by step solution:Given definite integral is
$\int\limits_{0}^{2\pi }{\dfrac{\sin 2\theta }{a-b\cos \theta }}d\theta $
Here the upper limit is $2\pi $ and the lower limit is $0$.
So, the property we can use is
$\begin{align}
  & \int\limits_{0}^{2a}{f(x)dx}=2\int\limits_{0}^{a}{f(x)dx};\text{if }f(2a-x)=f(x) \\
 & \text{ }=0\text{ if }f(2a-x)=-f(x) \\
\end{align}$
Consider $f(\theta )=\dfrac{\sin 2\theta }{a-b\cos \theta }$
Then,
$f(2a-x)=f(2\pi -\theta )=\dfrac{\sin (2\pi -2\theta )}{a-b\cos (2\pi -\theta )}$
But we know that,
$\begin{align}
  & \sin (2n\pi -\theta )=-\sin \theta \\
 & \cos (2n\pi -\theta )=\cos \theta \\
\end{align}$
So, we get
$\begin{align}
  & f(2\pi -\theta )=\dfrac{\sin (2\pi -2\theta )}{a-b\cos (2\pi -\theta )} \\
 & \text{ }=\dfrac{-\sin 2\theta }{a-b\cos \theta } \\
 & \text{ }=-f(\theta ) \\
\end{align}$
Therefore, the definite integral of the function is zero.
Thus, the given integral is
$\int\limits_{0}^{2\pi }{\dfrac{\sin 2\theta }{a-b\cos \theta }d\theta }=0$



Option ‘D’ is correct



Note: Here we need to check that the given function is an even or odd function. If the limits are in the form of $a=0;b=2a$, we need to check the given function for $f(2a-x)$. According to the result, the integration is evolved by using the properties of definite integrals.