Sam Longoria

2022-01-05

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

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Step 1
to evaluate: $\int \frac{dx}{\mathrm{cos}x-1}$
Solution:
We have formula:
$1-\mathrm{cos}2x=2{\mathrm{sin}}^{2}x$
Therefore,
$1-\mathrm{cos}x=2{\mathrm{sin}}^{2}\left(\frac{x}{2}\right)$
And also we have $\int {\mathrm{csc}}^{2}xdx=-\mathrm{cot}x$
Step 2
Evaluate the above integral:
$\int \frac{dx}{\mathrm{cos}x-1}=\int \frac{1}{-\left(1-\mathrm{cos}x\right)}dx$
$=\int \frac{1}{-{\mathrm{sin}}^{2}\left(\frac{x}{2}\right)}dx$
$=\int -{\mathrm{csc}}^{2}\left(\frac{x}{2}\right)dx$
$=\frac{\mathrm{cot}\left(\frac{x}{2}\right)}{\left(\frac{1}{2}\right)}+c$ (Using $\int -{\mathrm{csc}}^{2}x=\mathrm{cot}x$)
$=2\mathrm{cot}\left(\frac{x}{2}\right)+c$
Hence, required answer is $\int \frac{dx}{\mathrm{cos}x-1}=2\mathrm{cot}\left(\frac{x}{2}\right)+c$.

$\int \frac{1}{\mathrm{cos}x-1}dx$
Multiply both the numerator and the denominator by $\mathrm{cos}x+1$
$=\int \frac{1}{\mathrm{cos}x-1}\cdot \frac{\mathrm{cos}x+1}{\mathrm{cos}x+1}dx$
$=\int \frac{\mathrm{cos}x+1}{{\mathrm{cos}}^{2}x-1}dx$
$=\int \frac{\mathrm{cos}x+1}{-{\mathrm{sin}}^{2}x}dx$
$=\int -\frac{\mathrm{cos}x}{{\mathrm{sin}}^{2}x}-\frac{1}{{\mathrm{sin}}^{2}x}dx$
$=\int -\frac{\mathrm{cos}x}{\mathrm{sin}x}\cdot \frac{1}{\mathrm{sin}x}-\frac{1}{{\mathrm{sin}}^{2}x}dx$
$=\int -\mathrm{csc}x\mathrm{cot}x-{\mathrm{csc}}^{2}xdx$
The antiderivative of $\mathrm{csc}x\mathrm{cot}x$ and ${\mathrm{csc}}^{2}x$ are $-\mathrm{csc}x$ and $-\mathrm{cot}x$ respectively.
$=\mathrm{csc}x+\mathrm{cot}x+C$
Result:
$\mathrm{csc}x+\mathrm{cot}x+C$

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