# `int sec^4 (2x) dx` Find the indefinite integral

Given to solve

`int sec^4(2x)dx`

let `u= 2x`

=> `du = 2dx => dx = (1/2)du`

so,

`int sec^4(2x)dx`

= `int sec^4(u) (1/2) du`

=` (1/2) int sec^4(u) du`

let us sovle

`int sec^4(u) du`

as by the formulae

`int sec^n (x) dx `

= `(sec^(n-1) (x) (sinx))/(n-1) +((n-2)/(n-1))*(int...

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Given to solve

`int sec^4(2x)dx`

let `u= 2x`

=> `du = 2dx => dx = (1/2)du`

so,

`int sec^4(2x)dx`

= `int sec^4(u) (1/2) du`

=` (1/2) int sec^4(u) du`

let us sovle

`int sec^4(u) du`

as by the formulae

`int sec^n (x) dx `

= `(sec^(n-1) (x) (sinx))/(n-1) +((n-2)/(n-1))*(int sec^(n-2) (x) dx)`

so,

`int sec^4(u) du `

=`(sec^(4-1) (u) (sin u))/(4-1) +((4-2)/(4-1))*(int sec^(4-2) (u) du)`

=`(sec^(3) (u) (sin u))/(3) +(2/3)*(int sec^(2) (u) du)`

`= (sec^(3) (u) (sin u))/(3) +(2/3)*(tan u)`

so,

`int sec^4(2x)dx`

= `(1/2) int sec^4(u) du`

=`(1/2)[(sec^(3) (u) (sin u))/(3) +(2/3)*(tan u)]`

but `u= 2x`

so,

`(1/2)[(sec^(3) (u) (sin u))/(3) +(2/3)*(tan u)]`

= `(1/2)[(sec^(3) (2x) (sin (2x)))/(3) +(2/3)*(tan (2x))]`

so,

`int sec^4(2x)dx `

`=(1/2)[(sec^(3) (2x) (sin (2x)))/(3) +(2/3)*(tan (2x))] +c`

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