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I'm trying to solve the following integral:

$\int \frac{1}{\cos (x)-1}dx$

I can solve it using the Weierstrass substitution, but that isn't something we learned about in our calculus class, so there must be a simpler solution.

Could you please help me find a solution without the Weierstrass substitution?

Thanks

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    $\begingroup$ Multiply and divide by cosx+1 $\endgroup$ – asd.123 Apr 6 '20 at 17:42
  • $\begingroup$ @Butane not sure what you mean. Could you please explain further? $\endgroup$ – Peter Apr 6 '20 at 17:43
  • $\begingroup$ Peter, it means: $\int \frac{1}{\cos x -1}\cdot \frac{\cos x+1}{\cos x +1} dx = \int \frac{\cos x +1}{\cos^2 x-1}\,dx$ $\endgroup$ – amWhy Apr 6 '20 at 17:44
  • $\begingroup$ Basically using trigonometry,multiply(1/(cosx)-1) with ((cosx)+1)/((cosx)+1) $\endgroup$ – asd.123 Apr 6 '20 at 17:45
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    $\begingroup$ Butane: No problem, we all make typos :-) ...just glad to catch you so you could edit. $\endgroup$ – amWhy Apr 6 '20 at 17:50
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Since $\cos^2x-1=-\sin^2x$, your integral is$$-\int\frac{\cos x+1}{\sin^2x}dx=\cot x-\int\frac{\cos x dx}{\sin^2x}=\cot x+\csc x+C.$$Edit: @Quanto gives a somewhat more elegant antiderivative, $\cot\frac{x}{2}+C$. The two are equal because$$\cot x+\csc x=\frac{1+\cos x}{\sin x}=\frac{2\cos^2\frac{x}{2}}{2\sin\frac{x}{2}\cos\frac{x}{2}}=\cot\frac{x}{2}.$$

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$\cos (x)=1-2 \sin ^{2}(\frac{x}{2})$ so your denominator is equal to $-2 \sin ^{2}(\frac{x}{2})$. Now integral is quite easy to calculate.

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Note,

$$\int \frac{dx}{\cos x-1} =-\int \frac{dx}{2\sin^2\frac x2} =-\frac12 \int \csc^2\frac x2 dx = \cot\frac x2+C$$

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