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
$\endgroup$ 63 Answers
$\begingroup$$\cos (x)=1-2 \sin ^{2}(\frac{x}{2})$ so your denominator is equal to $-2 \sin ^{2}(\frac{x}{2})$. Now integral is quite easy to calculate.
$\endgroup$ $\begingroup$Since $\cos^2x-1=-\sin^2x$, your integral is$$-\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}.$$
$\endgroup$ $\begingroup$Note,
$$\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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