# Evaluate the indefinite integral integrate of (77x^6)((sec(x^7))^4)dx

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You should perform the following substitution such that:

`x^7 = t => 7x^6 dx = dt => x^6 dx = (dt)/7`

You need to change the variable such that:

`int 77x^6 sec^4(x^7) dx = int 77/7 sec^4 t dt`

You should use the following formula such that:

`int sec^k x dx = (sin x*sec^(k-1) x)/(k-1) + (k-1)/(k-2) int sec^(k-2) x dx`

Reasoning by analogy yields:

`int sec^4 t dt = (sin t*sec^3 t)/3 +2/3 int sec^2t dx`

`int sec^4 t dt = (sin t/cos t*sec^2 t)/3 + 2/3 int1/(cos^2 t)dx`

`int sec^4 t dt = (tan t*sec^2 t)/3 + 2/3 tan t + c`

You should factor out `(tan t)/3`  such that:

`int sec^4 t dt = (tan t)/3(sec^2 t + 2) + c`

Substituting back `x^7`  for t yields:

`int sec^4 t dt = (tan x^7)/3(sec^2x^7 + 2) + c`

Hence, evaluating the given integral,under the given conditions, yields `11int sec^4 t dt = 11(tan x^7)/3(sec^2 x^7 + 2) + c.`

jeew-m | College Teacher | (Level 1) Educator Emeritus

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`t = tan(x^7)`

`dt = sec^2(x^7)*7x^6dx`

`int (77x^6)((sec(x^7))^4)dx`

`= int11*7*x^6((sec(x^7))^2)((sec(x^7))^2)dx`

`= int11*(1+tan^2(x^7))*sec^2(x^7)*7x^6dx`

`= 11int (1+t^2)dt`

`= 11(t+t^3/3)+C` where C is a constant.

`= 11(tan(x^7)+(tan^3(x^7))/3)+C`

`int (77x^6)((sec(x^7))^4)dx = 11(tan(x^7)+(tan^3(x^7))/3)+C`

Sources:

quantatanu | Student, Undergraduate | (Level 1) Valedictorian

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Let us substitute,

tan(x^7) = 7 z

=> x^6 (sec(x^7))^2 dx = dz

=> 77 x^6 (sec(x^7))^4 dx =  77 (sec(x^7))^2 dz

= 77 [ 1 + (tan(x^7))^2] dz

= 77 [ 1 + (7z)^2 ] dz

= 77 [ 1 + 49 z^2 ] dz

Therefore,

Integral[ 77 x^6 (sec(x^7))^4 dx ]

= Integral [  77 [ 1 + 49 z^2 ] dz ]

= 77 [ z + 49* z^3 /3 ] +Constant

= 77 [tan(x^7) /7 + (49/3) (tan(x^7))^3 ] + Constant.

= 11 tan(x^7) + (3773/3) (tan(x^7))^3 +Constant

=  tan(x^7) [ 11 + (3773/3) (tan(x^7))^2 ] + Constant.