Question

What is the derivative of sec^3(2x)-3sec(2x)?

Answer 1

`1/4sec(2x)tan(2x)-5/(4)ln|tan(2x)+sec(2x)|+c`

Explanation:

`intsec^3(2x)-3sec(2x)dx`

Apply sum rule and take the constant out,

`color(red)(1/2intsec^3(2x)dx)-color(blue)(3/2intsec(2x)dx`

Apply integral reduction (refer to picture 1),

`color(red)(1/2((sec^(3-2)(2x)tan(2x))/(3-1)+(3-2)/(3-1)intsec^(3-2)(2x)))-color(blue)(3/2intsec(2x)dx`
`=color(red)(1/2((sec(2x)tan(2x))/(2)+(1)/(2)intsec(2x)))-color(blue)(3/2intsec(2x)dx`

Integrate `sec` (refer to picture 2),

`color(red)(1/2((sec(2x)tan(2x))/(2)+(1)/(2)ln|tan(2x)+sec(2x)|)-color(blue)(3/2ln|tan(2x)+sec(2x)|`

Simplify,

`color(red)(1/4sec(2x)tan(2x)+(1)/(4)ln|tan(2x)+sec(2x)|)-color(blue)(3/2ln|tan(2x)+sec(2x)|`
`=1/4sec(2x)tan(2x)-5/(4)ln|tan(2x)+sec(2x)|+c`

Reference:

Picture 1: Integral reduction

Picture 2: Integral of `secx`

Hope you have a nice day!

Answer 2

`dy/dx=6sec^3(2x)tan(2x)-6sec(2x)tan(2x)`

Explanation:

.

`y=sec^3(2x)-3sec(2x)`

Let `2x=u, :. 2dx=du`

`(du)/dx=2`

Let `sec(2x)=z, :. secu=z`

`secutanu(du)=dz`

`(dz)/(du)=secutanu=sec(2x)tan(2x)`

`y=z^3-3z`

`dy/dz=3z^2-3=3sec^2(2x)-3`

The Chain Rule says:

`dy/dx=(dy/(dz))((dz)/(du))((du)/dx)`

`dy/dx=(3sec^2(2x)-3)(sec(2x)tan(2x))(2)`

`dy/dx=6sec^3(2x)tan(2x)-6sec(2x)tan(2x)`