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

Question

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

Answer 1

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$$\cot x+\csc x=\frac{1+\cos x}{\sin x}=\frac{2\cos^2\frac{x}{2}}{2\sin\frac{x}{2}\cos\frac{x}{2}}=\cot\frac{x}{2}.$$

Answer 2

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

Answer 3

Note,

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

Answer 4

$$\int \frac{dx}{\cos x-1}=-\int \frac{(1+\cos x)dx}{(1+\cos x)(1-\cos x)}\\ =-\int \frac{(1+\cos x)dx}{\sin^{2} x}=\int\frac{-dx}{\sin^{2} x}+\int\frac{-\cos x}{\sin^{2} x}dx\\ =\int d(\cot x)+\int d(\frac{1}{\sin x})=\cot x+\frac{1}{\sin x}+c$$