Question
$\int_0^{2\pi } {\frac{{\sin 2\theta }}{{a - b\,\cos \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} $..
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$f(x)=\left\{\begin{array}{cc}\left(\frac{8}{7}\right)^{\frac{\tan 8 x}{\tan 7 x}}, & 0 < x < \frac{\pi}{2} \\ a-8, & x=\frac{\pi}{2} \\ (1+\mid \cot x)^{\frac{b}{a}|\tan x|}, & \frac{\pi}{2} < x < \pi\end{array}\right.$
Where $a, b \in Z$. If $f$ is continuous at $x=\frac{\pi}{2}$, then $\mathrm{a}^2+\mathrm{b}^2$ is equal to ..........