How do you integrate $\frac{tan x}{sec x + cos x}$ $tanx / (secx + cosx)$ ?

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How do you integrate $\frac{tan x}{sec x + cos x}$ $tanx / (secx + cosx)$ ?

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Answer 1 · 2016-08-01T05:07:17

$- arctan (cos x) + C$ $-arctan(cosx)+C$

Explanation:

We should first try to simplify the integrand.

$\frac{tan x}{sec x + cos x} = \frac{\frac{sin x}{cos x}}{\frac{1}{cos x} + cos x} = \frac{sin x}{cos x (\frac{1}{cos x} + cos x)} = \frac{sin x}{1 + {cos}^{2} x}$ $tanx/(secx+cosx)=(sinx/cosx)/(1/cosx+cosx)=sinx/(cosx(1/cosx+cosx))=sinx/(1+cos^2x)$

Thus:

$\int \frac{tan x}{sec x + cos x} d x = \int \frac{sin x}{1 + {cos}^{2} x} d x$ $inttanx/(secx+cosx)dx=intsinx/(1+cos^2x)dx$

We can use substitution here. Let $u = cos x$ $u=cosx$ . This implies that $d u = - sin x d x$ $du=-sinxdx$ .

$= - \int \frac{- sin x}{1 + {cos}^{2} x} d x = - \int \frac{1}{1 + u^{2}} d u$ $=-int(-sinx)/(1+cos^2x)dx=-int1/(1+u^2)du$

This is the arctangent integral!

$= - arctan (u) + C = - arctan (cos x) + C$ $=-arctan(u)+C=-arctan(cosx)+C$

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