What is the antiderivative of $tan x d x$ $tanx dx$ ?

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What is the antiderivative of $tan x d x$ $tanx dx$ ?

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Answer 1 · 2015-10-07T21:57:16

The antiderivative of $tan (x)$ $tan(x)$ is $- ln (cos (x))$ $-ln(cos(x))$

Explanation:

We know $tan (x) = \frac{sin (x)}{cos (x)}$ $tan(x) = sin(x)/cos(x)$ .

$\int tan (x) d x = \int \frac{sin (x)}{cos (x)} d x$ $inttan(x)dx = intsin(x)/cos(x)dx$

We can use substitution to simplify the integral.

$u = cos (x)$ $u = cos(x)$ and $d u = - sin (x) d x$ $du = -sin(x)dx$

$\int tan (x) d x = \int \frac{sin (x)}{cos (x)} d x = - \int \frac{1}{u} d u = - ln (u) + C$ $inttan(x)dx = intsin(x)/cos(x)dx = -int1/udu = -ln(u) + C$

To get the answer, just plug $u$ $u$ back in to get:

$- ln (cos (x))$ $-ln(cos(x))$

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What is the antiderivative of $tan x d x$ $tanx dx$ ?

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