<s>[INST] Integral Question $$
\int _{\large{ -\frac { \pi  }{ 2 } } }^{\large{ \frac { \pi  }{ 2 } } }{ \frac { \mathrm{d}x }{ \sin x-2\cos x+3 }  } 
$$
Wolframalpha suggests to evaluate indefinite integral by substituting $u = \tan(\frac { x }{ 2 })$. Is there a more basic way to evaluate this definite integral?
 [/INST] @ Hckr, we propose to calculate the full applying WA and then judge you whether to seek another method:
$u=\tan \frac{x}{2}=>\ dx = \frac{2}{u^2+1}\ du$
$$
\int _{\large{ -\frac { \pi  }{ 2 } } }^{\large{ \frac { \pi  }{ 2 } } }{ \frac { \mathrm{d}x }{ \sin x-2\cos x+3 }  }= \int _{-1}^{1}{ \frac {2 \mathrm{d}u }{ 5u^2+2u+1}  }=\frac{2}{5}\int _{-1}^{1}{ \frac { \mathrm{d}u }{ (u+\frac{1}{5})^2+(\frac{2}{5})^2}  }=$$
$$=\frac{2}{5}\cdot(\frac{2}{5})^{-1} \cdot \tan^{-1}\frac{5u+1}{2}|_{-1}^1=\tan^{-1}3-\tan^{-1}(-2)=\tan^{-1}3+\tan^{-1}2= \frac{3}{4}\pi 
$$
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