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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Expert Answer

porschomcl
Answered 2022-01-06 Author has 1051 answers
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 \(\displaystyle\int{\frac{{{\left.{d}{x}\right.}}}{{{\cos{{x}}}-{1}}}}={2}{\cot{{\left({\frac{{{x}}}{{{2}}}}\right)}}}+{c}\).
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Nadine Salcido
Answered 2022-01-07 Author has 1385 answers
\(\displaystyle\int{\frac{{{1}}}{{{\cos{{x}}}-{1}}}}{\left.{d}{x}\right.}\)
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:
\(\displaystyle{\csc{{x}}}+{\cot{{x}}}+{C}\)
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karton
Answered 2022-01-11 Author has 8659 answers

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

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