# Evaluate the integral. \int \frac{dx}{\cos x-1}

Sam Longoria 2022-01-05 Answered
Evaluate the integral.
$$\displaystyle\int{\frac{{{\left.{d}{x}\right.}}}{{{\cos{{x}}}-{1}}}}$$

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porschomcl
Step 1
to evaluate: $$\displaystyle\int{\frac{{{\left.{d}{x}\right.}}}{{{\cos{{x}}}-{1}}}}$$
Solution:
We have formula:
$$\displaystyle{1}-{\cos{{2}}}{x}={2}{{\sin}^{{{2}}}{x}}$$
Therefore,
$$\displaystyle{1}-{\cos{{x}}}={2}{{\sin}^{{{2}}}{\left({\frac{{{x}}}{{{2}}}}\right)}}$$
And also we have $$\displaystyle\int{{\csc}^{{{2}}}{x}}{\left.{d}{x}\right.}=-{\cot{{x}}}$$
Step 2
Evaluate the above integral:
$$\displaystyle\int{\frac{{{\left.{d}{x}\right.}}}{{{\cos{{x}}}-{1}}}}=\int{\frac{{{1}}}{{-{\left({1}-{\cos{{x}}}\right)}}}}{\left.{d}{x}\right.}$$
$$\displaystyle=\int{\frac{{{1}}}{{-{{\sin}^{{{2}}}{\left({\frac{{{x}}}{{{2}}}}\right)}}}}}{\left.{d}{x}\right.}$$
$$\displaystyle=\int-{{\csc}^{{{2}}}{\left({\frac{{{x}}}{{{2}}}}\right)}}{\left.{d}{x}\right.}$$
$$\displaystyle={\frac{{{\cot{{\left({\frac{{{x}}}{{{2}}}}\right)}}}}}{{{\left({\frac{{{1}}}{{{2}}}}\right)}}}}+{c}$$ (Using $$\displaystyle\int-{{\csc}^{{{2}}}{x}}={\cot{{x}}}$$)
$$\displaystyle={2}{\cot{{\left({\frac{{{x}}}{{{2}}}}\right)}}}+{c}$$
Hence, required answer is $$\displaystyle\int{\frac{{{\left.{d}{x}\right.}}}{{{\cos{{x}}}-{1}}}}={2}{\cot{{\left({\frac{{{x}}}{{{2}}}}\right)}}}+{c}$$.
###### Not exactly what you’re looking for?
$$\displaystyle\int{\frac{{{1}}}{{{\cos{{x}}}-{1}}}}{\left.{d}{x}\right.}$$
Multiply both the numerator and the denominator by $$\displaystyle{\cos{{x}}}+{1}$$
$$\displaystyle=\int{\frac{{{1}}}{{{\cos{{x}}}-{1}}}}\cdot{\frac{{{\cos{{x}}}+{1}}}{{{\cos{{x}}}+{1}}}}{\left.{d}{x}\right.}$$
$$\displaystyle=\int{\frac{{{\cos{{x}}}+{1}}}{{{{\cos}^{{{2}}}{x}}-{1}}}}{\left.{d}{x}\right.}$$
$$\displaystyle=\int{\frac{{{\cos{{x}}}+{1}}}{{-{{\sin}^{{{2}}}{x}}}}}{\left.{d}{x}\right.}$$
$$\displaystyle=\int-{\frac{{{\cos{{x}}}}}{{{{\sin}^{{{2}}}{x}}}}}-{\frac{{{1}}}{{{{\sin}^{{{2}}}{x}}}}}{\left.{d}{x}\right.}$$
$$\displaystyle=\int-{\frac{{{\cos{{x}}}}}{{{\sin{{x}}}}}}\cdot{\frac{{{1}}}{{{\sin{{x}}}}}}-{\frac{{{1}}}{{{{\sin}^{{{2}}}{x}}}}}{\left.{d}{x}\right.}$$
$$\displaystyle=\int-{\csc{{x}}}{\cot{{x}}}-{{\csc}^{{{2}}}{x}}{\left.{d}{x}\right.}$$
The antiderivative of $$\displaystyle{\csc{{x}}}{\cot{{x}}}$$ and $$\displaystyle{{\csc}^{{{2}}}{x}}$$ are $$\displaystyle-{\csc{{x}}}$$ and $$\displaystyle-{\cot{{x}}}$$ respectively.
$$\displaystyle={\csc{{x}}}+{\cot{{x}}}+{C}$$
Result:
$$\displaystyle{\csc{{x}}}+{\cot{{x}}}+{C}$$
karton

$$\text{Step 1} \\I=\int \frac{dx}{\cos x-1} \\\text{Multiply and divide by}\ (\cos x+1) \\I=\int \frac{dx}{\cos x-1}*\frac{\cos x+1}{\cos x+1} \\=\int \frac{\cos x+1}{\cos^{2}x-1}dx \\\text{We know that}\ \sin^{2}x+\cos^{2}x=1\Rightarrow \cos^{2}x-1=-\sin^{2}x \\I=-\int \frac{\cos x+1}{\sin^{2}x}dx \\=-\int (\frac{\cos x}{\sin^{2}x}+\frac{1}{\sin^{2}x})dx \\=-\int (\frac{\cos x}{\sin x*\sin x}+\frac{1}{\sin^{2}x})dx \\=-\int (\cot x\csc x+\csc^{2}x)dx \\=\int -\cot x\csc x dx-\int \csc^{2}xdx \\=\csc x+\cot x+c \\\text{Result:} \\\int \frac{dx}{\cos x-1}=\csc x+\cot x+c$$

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