# Stuck on Indefinite Integral

$$\int \frac{1}{13\cos x+ 12}\,\mathrm{d}x$$

I appreciate any and all help. Thank you.

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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

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This was an excellent hint. Solving the problem was only secondary in comparison to learning a new technique for integration. Thank you very much! –  A is for Ambition Jun 15 '14 at 12:01

$\bf{My\; Solution::}$ 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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Was going to add this as a separate answer:) Good job. –  lab bhattacharjee Jun 15 '14 at 11:39
Soory lab bhattacharjee actually I did not seen yrs Hint. Thanks –  juantheron Jun 15 '14 at 11:43
Nothing to be sorry for. –  lab bhattacharjee Jun 15 '14 at 11:44


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