Evaluating the trigonometric integral $\int \frac{\mathrm{cos}x}{1+\mathrm{cos}x}\phantom{\rule{thinmathspace}{0ex}}dx$

Addison Trujillo
2022-07-10
Answered

Evaluating the trigonometric integral $\int \frac{\mathrm{cos}x}{1+\mathrm{cos}x}\phantom{\rule{thinmathspace}{0ex}}dx$

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gutinyalk

Answered 2022-07-11
Author has **11** answers

Hint Here's one method: Rewrite the integrand as

$\frac{\mathrm{cos}x}{1+\mathrm{cos}x}=1-\frac{1}{1+\mathrm{cos}x},$

so that the integral becomes

$\int \frac{\mathrm{cos}x}{1+\mathrm{cos}x}\phantom{\rule{thinmathspace}{0ex}}dx=\int (1-\frac{1}{1+\mathrm{cos}x})dx=x-\int \frac{dx}{1+\mathrm{cos}x}.$

Now, the remaining integral can be handled by exploiting the Pythagorean identity:

$\frac{1}{1+\mathrm{cos}x}=\frac{1}{1+\mathrm{cos}x}\cdot \frac{1-\mathrm{cos}x}{1-\mathrm{cos}x}=\frac{1-\mathrm{cos}x}{1-{\mathrm{cos}}^{2}x}=\frac{1-\mathrm{cos}x}{{\mathrm{sin}}^{2}x}.$

Additional hint Now,

$\int \frac{1-\mathrm{cos}x}{{\mathrm{sin}}^{2}x}dx=\int \frac{dx}{{\mathrm{sin}}^{2}x}-\int \frac{\mathrm{cos}x}{{\mathrm{sin}}^{2}x}\phantom{\rule{thinmathspace}{0ex}}dx.$

The first integral is elementary, and the second can be handled with a straightforward substitution.

$\frac{\mathrm{cos}x}{1+\mathrm{cos}x}=1-\frac{1}{1+\mathrm{cos}x},$

so that the integral becomes

$\int \frac{\mathrm{cos}x}{1+\mathrm{cos}x}\phantom{\rule{thinmathspace}{0ex}}dx=\int (1-\frac{1}{1+\mathrm{cos}x})dx=x-\int \frac{dx}{1+\mathrm{cos}x}.$

Now, the remaining integral can be handled by exploiting the Pythagorean identity:

$\frac{1}{1+\mathrm{cos}x}=\frac{1}{1+\mathrm{cos}x}\cdot \frac{1-\mathrm{cos}x}{1-\mathrm{cos}x}=\frac{1-\mathrm{cos}x}{1-{\mathrm{cos}}^{2}x}=\frac{1-\mathrm{cos}x}{{\mathrm{sin}}^{2}x}.$

Additional hint Now,

$\int \frac{1-\mathrm{cos}x}{{\mathrm{sin}}^{2}x}dx=\int \frac{dx}{{\mathrm{sin}}^{2}x}-\int \frac{\mathrm{cos}x}{{\mathrm{sin}}^{2}x}\phantom{\rule{thinmathspace}{0ex}}dx.$

The first integral is elementary, and the second can be handled with a straightforward substitution.

Dayanara Terry

Answered 2022-07-12
Author has **4** answers

Hint $\mathrm{cos}(x)=\frac{1-{\mathrm{tan}}^{2}(x/2)}{1+{\mathrm{tan}}^{2}(x/2)}$ so the integral reduces to $0.5\int (1-{\mathrm{tan}}^{2}(x/2))dx$ where we can then use identity ${\mathrm{tan}}^{2}(x/2)={\mathrm{sec}}^{2}(x/2)-1$ to get the final integral which is $x-\mathrm{tan}(x/2)+C$

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