Find out the answer to "What is the antiderivative of tan⁡(x) ?" - Plainmath

nemired9

Answered question

2021-12-18

What is the antiderivative of

\tan (x)

?

Answer & Explanation

Papilys3q

Beginner2021-12-19Added 34 answers

Recall:

\int \frac{g^{'} (x)}{g (x)} d x = \ln | g (x) | + C

(You can verify this by substitution

u = g (x)

.)
Now, let us look at the posted antiderivative.
By the trig identity

\tan x = \frac{\sin x}{\cos x}

\int \tan x d x = \int \frac{\sin x}{\cos x} d x

by rewriting it a bit further to fit the form above,

= - \int \frac{- \sin x}{\cos x} d x

by the formula above,

= - \ln | \cos x | + C

or by

r \ln x = {\ln x}^{r}

= \ln {| \cos x |}^{- 1} + C = \ln | \sec x | + C

I hope that this was helpful.

Foreckije

Beginner2021-12-20Added 32 answers

By

\tan x = \frac{\sin x}{\cos x}

\int \tan x d x = \int \frac{\sin x}{\cos x} d x

Let

u = \cos x \Rightarrow \frac{d u}{d x} = - \sin x \Rightarrow d x = \frac{d u}{- \sin x}

By substitution,

= \int \frac{\sin x}{u} \cdot \frac{d u}{- \sin x}

By cancelling

\sin x

= - \int \frac{1}{u} d u

By finding an antiderivative,

= - \ln | u | + C

By plugging

\cos x

back in for u

= - \ln | \cos x | + C

By the log property

r \ln x = {\ln x}^{r}

= \ln {| \cos x |}^{- 1} + C

By

{(\cos x)}^{- 1} = \frac{1}{\cos x} = \sec x

= \ln | \sec x | + C

RizerMix

Expert2021-12-29Added 656 answers

We want to find $\int \tan x d x$
$\int \tan x d x = \int \frac{\tan x \sec x}{\sec x} d x$
Now let $u = \sec x$ and $d u = \sec x \tan x d x$ . Then
$\int \frac{\tan x \sec x}{\sec x} d x = \int \frac{1}{u} d u$
This is a standard integral which evaluates to
$\ln | u | + C = \ln | \sec x | + C$

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