`inttan^3(2t)sec^3(2t)dt`

Apply integral substitution: `u=2t`

`du=2dt`

`inttan^3(2t)sec^3(2t)dt=inttan^3(u)sec^3(u)(du)/2`

Take the constant out,

`=1/2inttan^3(u)sec^3(u)du`

Rewrite the integral as,

`=1/2intsec^3(u)tan^2(u)tan(u)du`

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

`=1/2intsec^3(u)(sec^2(u)-1)tan(u)du`

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

`dv=sec(u)tan(u)du`

`=1/2intv^2(v^2-1)dv`

`=1/2int(v^4-v^2)dv`

Apply the sum and power rule,

`=1/2(intv^4dv-intv^2dv)`

`=1/2{(v^(4+1)/(4+1))-(v^(2+1)/(2+1))}`

`=1/2(v^5/5-v^3/3)`

Substitute back `v=sec(u)` and `u=2t`, and add a constant C to the solution,

`=1/2((sec^5(2t))/5-(sec^3(2t))/3)+C`

## See eNotes Ad-Free

Start your **48-hour free trial** to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Already a member? Log in here.