Stuck on Indefinite Integral Please help me. I have been stuck on this for ages :(
$$\int \frac{1}{13\cos x+ 12}\,\mathrm{d}x$$
I appreciate any and all help. Thank you.
 A: $\bf{My\; Solution::}$ Given $$\displaystyle \int\frac{1}{13\cos x+12}dx = \int\frac{1}{13(1+\cos x)-1}dx$$
Now Using $$\displaystyle 1+\cos x = 2\cos^2 \frac{x}{2}\;,$$ we get
$$\displaystyle \int\frac{1}{26\cos^2 \frac{x}{2}-1}dx$$
Now Divide both $\bf{N_{r}}$ and $\bf{D_{r}}$ by $\displaystyle \cos^2 \frac{x}{2}$
$$\displaystyle \int\frac{\sec^2 \frac{x}{2}}{26-1-\tan^2 \frac{x}{2}}dx = \int\frac{\sec^2 \frac{x}{2}}{5^2-\tan^2\frac{x}{2}}dx$$
Now Let $\displaystyle \tan \frac{x}{2} = t\;,$ Then $\displaystyle \sec^2\frac{x}{2}dx = 2dt$
So Integral is $$\displaystyle 2\int\frac{1}{5^2-t^2}dt = \frac{1}{10}\ln \left|\frac{5+t}{5-t}\right|+\mathbb{C}$$
Where $\displaystyle t = \cos^2 \frac{x}{2}$
A: HINT:
For the integrals of the form $\displaystyle\frac1{a\sin x+b\cos x+c},$  where $a,b,c$ are arbitrary constants 
try setting $\displaystyle\tan\frac x2=t$ and use  Weierstrass substitution
A: $\newcommand{\+}{^{\dagger}}
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\begin{align}
&\color{#66f}{\large\int{1 \over 13\cos\pars{x} + 12}\,\dd x}
=\int{1 \over 26\cos^{2}\pars{x/2} - 1}\,\dd x
=\int{\sec^{2}\pars{x/2} \over 26 - \sec^{2}\pars{x/2}}\,\dd x
\\[3mm]&=\
\overbrace{\int{\sec^{2}\pars{x/2} \over 25 - \tan^{2}\pars{x/2}}\,\dd x}
^{\ds{\tan\pars{x/2} \equiv t}}\ =\
2\int{\dd t \over 25 - t^{2}}
={1 \over 5}\int\pars{{1 \over t + 5} - {1 \over t - 5}}\,\dd t
\\[3mm]&={1 \over 5}\,\ln\pars{\verts{t + 5 \over t - 5}}
=\color{#66f}{\large%
{1 \over 5}\,\ln\pars{\verts{\tan\pars{x/2} + 5 \over \tan\pars{x/2} - 5}}}
+ \mbox{a constant.}
\end{align}
