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I would appreciate if somebody could help me with the following problem

Q: How to integrate this integral $$\int\frac{1}{\sin x+ 3\cos x}dx$$

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  • $\begingroup$ Perhaps you could use a more informative title? $\endgroup$ – davidlowryduda Dec 10 '13 at 1:44
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    $\begingroup$ Perhaps by using the tangent half-angle substitution ? $\endgroup$ – Lucian Dec 10 '13 at 1:54
  • $\begingroup$ @Lucian that's the answer to every question :) $\endgroup$ – Igor Rivin Dec 10 '13 at 2:09
  • $\begingroup$ @IgorRivin: Even to this one ? :-) $\endgroup$ – Lucian Dec 10 '13 at 2:42
  • $\begingroup$ @Lucian especially this one... $\endgroup$ – Igor Rivin Dec 10 '13 at 2:58
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\begin{align*} \int \frac 1 {\sin x +3\cos x}\ dx &= \int \frac 1 {2\sin \frac x 2\cos \frac x 2 +3\cos^2 \frac x 2 - 3 \sin^2 \frac x 2}\ dx \\ &= \int \frac 1 {\cos^2 \frac x 2} \cdot \frac 1 {3 + 2\tan \frac x 2 - 3 \tan^2 \frac x 2}\ dx \end{align*}

Switching $y=\tan \frac x 2 \implies\frac {dy}{dx} = \frac 1 2 \frac 1 {\cos^2 \frac x 2}$

\begin{align*} \int \frac 1 {\sin x +3\cos x}\ dx &= \int \frac 1 {3 + 2y - 3 y^2}2\ dy \\ &= 6\int \frac 1 {(3y-1-\sqrt{10})(3y-1+\sqrt{10})} \ dy \\ &= \frac 3 {\sqrt{10}}\int \frac 1 {3y-1-\sqrt{10}} - \frac 1 {3y-1+\sqrt{10}} \ dy \\ \end{align*}

I'll leave the rest.

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$$1\cdot \sin x +3 \cdot \cos x=\sqrt{10} \cdot \sin (x+a)$$ where $a=\tan ^{-1}(3)$

So, now can you just apply the known integral of $\csc(x+a) $?

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