MCQ
$\int_0^{2\pi } {\frac{{\sin 2\theta }}{{a - b\,\cos \theta }}\,d\theta = } $
  • A
    $1$
  • B
    $2$
  • C
    $\frac{\pi }{4}$
  • $0$

Answer

Correct option: D.
$0$
d
(d) $I = \int_0^{2\pi } {\frac{{\sin 2\theta }}{{a - b\cos \theta }}d\theta = \int_0^{2\pi } {\frac{{\sin (2\pi - 2\theta )}}{{a - b\cos (2\pi - \theta )}}d\theta } } $

==> I $ = - \int_0^{2\pi } {\frac{{\sin 2\theta }}{{a - b\cos \theta }}d\theta } $

$ \Rightarrow \,\,2I = 0 $

$\Rightarrow \,\,\int_0^{2\pi } {\frac{{\sin 2\theta }}{{a - b\cos \theta }}d\theta = 0} $..

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

What is the degree of the differential equation : $\text{y}=\text{x}\frac{\text{dy}}{\text{dx}}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^{-2}\ ?$
If $\text{x}=\text{at}^2,\text{y}=2\text{at}$ then $\frac{\text{d}^2\text{y}}{\text{dx}^2}=$
Let $M$ be a $3 \times 3$ matrix satisfying $M\left[\begin{array}{l}0 \\ 1 \\ 0\end{array}\right]=\left[\begin{array}{c}-1 \\ 2 \\ 3\end{array}\right], \quad M\left[\begin{array}{c}1 \\ -1 \\ 0\end{array}\right]=\left[\begin{array}{c}1 \\ 1 \\ -1\end{array}\right]$, and $M\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=\left[\begin{array}{c}0 \\ 0 \\ 12\end{array}\right]$ Then the sum of the diagonal entries of $M$ is
Find the general solution of: $\frac{\text{dy}}{\text{dx}}=\text{y}\sin\text{x:}$
Which of the following function  $(s)$ not defined at $x = 0$ has/have removable discontinuity at $x = 0$ ?
Find the values of $\sin \left(\frac{\pi}{3}-\sin ^{-1}\left(-\frac{1}{2}\right)\right)$ is equal to
Choose the correct answer from the given four options.
Let f : [0, 1] → [0, 1] be defined by $\text{f}(\text{x})=\begin{cases}\text{x, if x is rational}\\1-\text{x, if x is irrational}\end{cases}$ Then (fof)x is:
Let $[x]$ denote the greatest integer less than or equal to $x$. If $f(x)=\sin ^{-1} x, g(x)=\left[x^2\right]$ and $h(x)=2 x, \frac{1}{2} \leq x \leq \frac{1}{\sqrt{2}}$, then
The order and degree of the differential equation, $\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^{\frac{1}{4}}+\text{x}^{\frac{1}{5}}=0$ respectively are :
Domain of the function $f(x) =$ $\frac{1}{{\sqrt {\ln \,{{\cot }^{ - 1}}x} }}$ is