Question

Find the antiderivative of ..?

`8cosxtanx`

Answer 1

`int8cosxtanxdx=-8cosx+"C"`

Explanation:

Given: `int8cosxtanxdx`

Take out the constant `8` and rewrite `tanx` as `sinx/cosx` to get the following integral:

`8intcosxsinx/cosxdx`

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

`8*intsinxdx`

The integral `intsinxdx` is a common integral whose antiderivative is `-cosx+"C"`

So

`8*intsinxdx=8*-cosx+"C"=-8cosx+"C"`

Answer 2

The answer is `-8cos(x)+C`.

Explanation:

Remember that the definition of `tan(theta)` is `sin(theta)/cos(theta)`.

`int8cos(x)*tan(x)*dx`

`8*intcancel(cos(x))*sin(x)/cancel(cos(x))*dx`

`8*intsin(x)`

`8*-cos(x)`

`-8cos(x)+C`