# Find the antiderivative of ..?

## $8 \cos x \tan x$

Jan 26, 2018

$\int 8 \cos x \tan x \mathrm{dx} = - 8 \cos x + \text{C}$

#### Explanation:

Given: $\int 8 \cos x \tan x \mathrm{dx}$

Take out the constant $8$ and rewrite $\tan x$ as $\sin \frac{x}{\cos} x$ to get the following integral:

$8 \int \cos x \sin \frac{x}{\cos} x \mathrm{dx}$

The cosines cancel and we are left with this simple integral:

$8 \cdot \int \sin x \mathrm{dx}$

The integral $\int \sin x \mathrm{dx}$ is a common integral whose antiderivative is $- \cos x + \text{C}$

So

$8 \cdot \int \sin x \mathrm{dx} = 8 \cdot - \cos x + \text{C"=-8cosx+"C}$

Jan 26, 2018

The answer is $- 8 \cos \left(x\right) + C$.

#### Explanation:

Remember that the definition of $\tan \left(\theta\right)$ is $\sin \frac{\theta}{\cos} \left(\theta\right)$.

$\int 8 \cos \left(x\right) \cdot \tan \left(x\right) \cdot \mathrm{dx}$

$8 \cdot \int \cancel{\cos \left(x\right)} \cdot \sin \frac{x}{\cancel{\cos \left(x\right)}} \cdot \mathrm{dx}$

$8 \cdot \int \sin \left(x\right)$

$8 \cdot - \cos \left(x\right)$

$- 8 \cos \left(x\right) + C$