How can I compute definite integral of the following function?
$\int_{x(0)}^{0} \frac{dx}{k_2\,\sin x + k_1\frac{\cos x - 1}{\sin 2x}}$
$k_1$ and $k_2$ are positive constants.
At this point, I know that the function is odd. How should I approach the solution? Any help would be highly appreciated.
With some trigonometric identities we can rewrite the integral as
$$\int_{x(0)}^{0} \frac{dx}{k_2\,\sin x + k_1\frac{\cos x - 1}{\sin 2x}} = \int_{x(0)}^{0} \frac{\sin \left(2x\right) \,dx}{k_2\,\sin \left(x\right) \, \sin \left(2x\right) + k_1 \left(\cos \left(x\right) - 1\right)} = \int_{x(0)}^{0} \frac{2\sin \left(x\right)\cos \left(x\right) \,dx}{k_2\, 2\sin^2 \left(x\right)\cos \left(x\right) + k_1 \left(\cos \left(x\right) - 1\right)} = \int_{x(0)}^{0} \frac{2\sin \left(x\right)\cos \left(x\right) \,dx}{2 k_2\, \left(1 - \cos^2 \left(x\right)\right)\cos \left(x\right) + k_1 \cos \left(x\right) - k_1} \stackrel{\cos \left(x\right)\mapsto u}{=} 2 \int_{\cos\left(x(0)\right)}^{1} \frac{u \,dx}{2 k_2 u\, \left(u^2 - 1\right) - k_1 u + k_1} $$ Is now trivial evaluate the integral using partial fraction decomposition
