# Evaluating the trigonometric integral &#x222B;<!-- ∫ --> cos &#x2061;<!-

Evaluating the trigonometric integral $\int \frac{\mathrm{cos}x}{1+\mathrm{cos}x}\phantom{\rule{thinmathspace}{0ex}}dx$
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gutinyalk
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
$\int \frac{\mathrm{cos}x}{1+\mathrm{cos}x}\phantom{\rule{thinmathspace}{0ex}}dx=\int \left(1-\frac{1}{1+\mathrm{cos}x}\right)dx=x-\int \frac{dx}{1+\mathrm{cos}x}.$
Now, the remaining integral can be handled by exploiting the Pythagorean identity:
$\frac{1}{1+\mathrm{cos}x}=\frac{1}{1+\mathrm{cos}x}\cdot \frac{1-\mathrm{cos}x}{1-\mathrm{cos}x}=\frac{1-\mathrm{cos}x}{1-{\mathrm{cos}}^{2}x}=\frac{1-\mathrm{cos}x}{{\mathrm{sin}}^{2}x}.$
$\int \frac{1-\mathrm{cos}x}{{\mathrm{sin}}^{2}x}dx=\int \frac{dx}{{\mathrm{sin}}^{2}x}-\int \frac{\mathrm{cos}x}{{\mathrm{sin}}^{2}x}\phantom{\rule{thinmathspace}{0ex}}dx.$
The first integral is elementary, and the second can be handled with a straightforward substitution.
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Dayanara Terry
Hint $\mathrm{cos}\left(x\right)=\frac{1-{\mathrm{tan}}^{2}\left(x/2\right)}{1+{\mathrm{tan}}^{2}\left(x/2\right)}$ so the integral reduces to $0.5\int \left(1-{\mathrm{tan}}^{2}\left(x/2\right)\right)dx$ where we can then use identity ${\mathrm{tan}}^{2}\left(x/2\right)={\mathrm{sec}}^{2}\left(x/2\right)-1$ to get the final integral which is $x-\mathrm{tan}\left(x/2\right)+C$