<s>[INST] 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.
 [/INST] $\bf{My\; Solution::}$ 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 $\displaystyle t = \cos^2 \frac{x}{2}$
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