# `int tan^5(2x)sec^4(2x) dx` Find the indefinite integral

`inttan^5(2x)sec^4(2x)dx`

Let's apply integral substitution:`u=2x`

`(du)=2dx`

`inttan^5(2x)sec^4(2x)dx=inttan^5(u)sec^4(u)(du)/2`

Take the constant out and rewrite the integral as,

`=1/2intsec^2(u)sec^2(u)tan^5(u)du`

Now use the trigonometric identity :`sec^2(x)=1+tan^2(x)`

`=1/2int(1+tan^2(u))sec^2(u)tan^5(u)du`

Again apply the integral substitution:`v=tan(u)`

`dv=sec^2(u)du`

`=1/2int(1+v^2)v^5dv`

`=1/2int(v^5+v^7)dv`

apply the sum rule and power rule,

`=1/2(intv^5dv+intv^7dv)`

`=1/2{(v^(5+1)/(5+1))+(v^(7+1)/(7+1))}`

`=1/2(v^6/6+v^8/8)`

substitute back `v=tan(u)` and  `u=2x`

`=1/2((tan^6(2x))/6+(tan^8(2x))/8)`

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

`inttan^5(2x)sec^4(2x)dx`

Let's apply integral substitution:`u=2x`

`(du)=2dx`

`inttan^5(2x)sec^4(2x)dx=inttan^5(u)sec^4(u)(du)/2`

Take the constant out and rewrite the integral as,

`=1/2intsec^2(u)sec^2(u)tan^5(u)du`

Now use the trigonometric identity :`sec^2(x)=1+tan^2(x)`

`=1/2int(1+tan^2(u))sec^2(u)tan^5(u)du`

Again apply the integral substitution:`v=tan(u)`

`dv=sec^2(u)du`

`=1/2int(1+v^2)v^5dv`

`=1/2int(v^5+v^7)dv`

apply the sum rule and power rule,

`=1/2(intv^5dv+intv^7dv)`

`=1/2{(v^(5+1)/(5+1))+(v^(7+1)/(7+1))}`

`=1/2(v^6/6+v^8/8)`

substitute back `v=tan(u)` and  `u=2x`

`=1/2((tan^6(2x))/6+(tan^8(2x))/8)`

Add a constant C to the solution,

`=1/2(1/6tan^6(2x)+1/8tan^8(2x))+C`

Approved by eNotes Editorial Team