# Question #4323f

Jan 12, 2018

$\int t g \left(x\right) \mathrm{dx} = t G \left(x\right) + c , c \in \mathbb{R}$, where $G \left(x\right)$ is an antiderivative of $g \left(x\right)$. Whether a substitution is needed depends on $g$'s formula.

#### Explanation:

Assuming $t$ is a constant with respect to $x$, all you have to do is find an antiderivative of $g \left(x\right)$. If we call that antiderivative $G \left(x\right)$, then you have

$\int t g \left(x\right) \mathrm{dx} = t \int g \left(x\right) \mathrm{dx} = t G \left(x\right) + c .$

Jan 12, 2018

$\int \setminus \tan \left(x\right) \setminus \mathrm{dx} = - \ln | \cos \left(x\right) | + C$

#### Explanation:

I assume that you want to find the integral of the tangent function, or $\int \setminus \tan \left(x\right) \setminus \mathrm{dx}$.

First, using trigonometric identities, rewrite it to
$\int \setminus \sin \frac{x}{\cos} \left(x\right) \setminus \mathrm{dx}$

Now, we need to use u-substitution. For this, it would be best if we could find some expression in the integrand that can be substituted to a single variable and which derivative, when divided into the integrand, can simplify so that no $x$'s occur.

In this example, we will choose $u = \cos \left(x\right)$, because the integrand divided by its derivative $\frac{\mathrm{du}}{\mathrm{dx}} = - \sin \left(x\right)$ will simplify dramatically.

Then, we have
$\int \setminus \sin \frac{x}{\cos} \left(x\right) \frac{\mathrm{dx}}{\mathrm{du}} \setminus \mathrm{du} = - \int \setminus \sin \frac{x}{u} \frac{1}{\sin} \left(x\right) \setminus \mathrm{du} = - \int \setminus \frac{\mathrm{du}}{u}$.

This is simply
$- \ln | u | + C$.

Substitute back $u = \cos \left(x\right)$ to get the final answer of
$- \ln | \cos \left(x\right) | + C$.

Jan 12, 2018

$\int \text{tg"x dx = ln|secx|+"c}$

#### Explanation:

Firstly, we shall let $\text{tg} \equiv \tan$

So our integral is $\int \tan x \mathrm{dx} = \int \frac{\sec x \tan x \mathrm{dx}}{\sec} x$

Now let $u = \sec x$ and $\mathrm{du} = \sec x \tan x \mathrm{dx}$. We then transform the integral to get it as $\int \frac{1}{u} \mathrm{du} = \ln | u | = \ln | \sec x | + \text{c}$