How do you integrate $(\frac{tan x}{sec x}) d x$ $(tanx/secx) dx$ ?

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How do you integrate $(\frac{tan x}{sec x}) d x$ $(tanx/secx) dx$ ?

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Answer 1 · 2016-11-05T23:28:55

The integral is $y = - cos x + C$ $y = -cosx+ C$ .

Explanation:

Start by simplifying the function using the identities $tan θ = \frac{sin θ}{cos θ}$ $tantheta = sintheta/costheta$ and $sec θ = \frac{1}{cos θ}$ $sectheta = 1/costheta$ .

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

If you know your basic integrals, $\int (sin x) d x = - cos x + C$ $int(sinx)dx = -cosx + C$ .

Hopefully this helps!

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How do you integrate $(\frac{tan x}{sec x}) d x$ $(tanx/secx) dx$ ?

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