# Evaluate the indefinite integral integrate of (7x^6)((sec(x^7))^4)dx All the other previous answers to this question have been incorrect

*print*Print*list*Cite

### 1 Answer

You need to use the following substitution such that:

`x^7 = t => 7x^6 dx = dt`

Changing the variable yields:

`int 7x^6 (sec x^7)^4 dx = int (sec t)^4 dt`

You need to substitute `1/ cos t` for sec t such that:

`int (sec t)^4 dt = int (1/cos t)^4 dt`

You should use the following trigonometric substitution such that:

`1 + tan^2 t = 1/(cos^2 t)`

`int (1/cos t)^4 dt = int (1 + tan^2 t)*1/(cos^2 t) dt`

You need to use the following substitution such that:

`tan t = u => 1/(cos^2 t) dt = du`

Changing the variable yields:

`int (1 + u^2)*du`

Using the property of linearity of integral yields:

`int (1 + u^2)*du = int du + int u^2 du`

`int (1 + u^2)*du = u + u^3/3 + c`

Substituting back `tan t` for `u` yields:

`int (1/cos t)^4 dt = tan t + (tan t)^3/3 + c`

Substituting back `x^7` for `t` yields:

`int 7x^6 (sec x^7)^4 dx = tanx^7 + (tan x^7)^3/3 + c`

**Hence, evaluating the given integral using substitution yields `int 7x^6 (sec x^7)^4 dx = tan x^7 + (tan x^7)^3/3 + c.` **