Find $$\int\frac{dx}{2\sin{x}+3\cos{x}+7}$$ My attempt: $$I= \int\frac{dx}{\sqrt{13}\sin{\left(x+\theta\right)}+{7}}$$ where $\sin{\theta}=\frac{3}{\sqrt{13}}$. Then, subsititute $$t = \tan{\frac{y}{2}}$$where $y=x +\theta$.
But, I think there is a simpler solution than above substitution. Would you help me?
Below is an alternative to the half-angle substitution
\begin{align} &\int\frac{1}{2\sin x+3\cos x+7}\ dx\\ =& \int\frac{1}{\sqrt{13}\sin y +7}\ dy =\int \frac{\sec^2 y \ (\sqrt{13}\sin y+7)}{\sec^2y\ (\sqrt{13}\sin y+7)^2} \ dy\\ =& \int \frac{\sqrt{13}\sec y \tan y + 7\sec^2 y}{13\tan^2 y + 49 \sec^2 y + 7\sqrt{13} \sec y \tan y} \ dy\\ =&\int \frac{d(\sqrt{13}\sec y+7\tan y)}{(\sqrt{13}\sec y+7\tan y)^2 +36} =\frac16 \tan^{-1}\frac {\sqrt{13}\sec y+7\tan y}6\\ = &\ \frac16 \tan^{-1}\frac {13\sec x +7(2\tan x+3)}{6(2-3\tan x)} \end{align}