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

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.

Thanks

• Multiply and divide by cosx+1 Commented Apr 6, 2020 at 17:42
• @Butane not sure what you mean. Could you please explain further? Commented Apr 6, 2020 at 17:43
• Peter, it means: $\int \frac{1}{\cos x -1}\cdot \frac{\cos x+1}{\cos x +1} dx = \int \frac{\cos x +1}{\cos^2 x-1}\,dx$ Commented Apr 6, 2020 at 17:44
• Basically using trigonometry,multiply(1/(cosx)-1) with ((cosx)+1)/((cosx)+1) Commented Apr 6, 2020 at 17:45
• Butane: No problem, we all make typos :-) ...just glad to catch you so you could edit. Commented Apr 6, 2020 at 17:50

$$\cos (x)=1-2 \sin ^{2}(\frac{x}{2})$$ so your denominator is equal to $$-2 \sin ^{2}(\frac{x}{2})$$. Now integral is quite easy to calculate.
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$$\cot x+\csc x=\frac{1+\cos x}{\sin x}=\frac{2\cos^2\frac{x}{2}}{2\sin\frac{x}{2}\cos\frac{x}{2}}=\cot\frac{x}{2}.$$
$$\int \frac{dx}{\cos x-1} =-\int \frac{dx}{2\sin^2\frac x2} =-\frac12 \int \csc^2\frac x2 dx = \cot\frac x2+C$$
$$\int \frac{dx}{\cos x-1}=-\int \frac{(1+\cos x)dx}{(1+\cos x)(1-\cos x)}\\ =-\int \frac{(1+\cos x)dx}{\sin^{2} x}=\int\frac{-dx}{\sin^{2} x}+\int\frac{-\cos x}{\sin^{2} x}dx\\ =\int d(\cot x)+\int d(\frac{1}{\sin x})=\cot x+\frac{1}{\sin x}+c$$