Question

What is `int tan^2(2x) sec^4(2x) dx`?

Answer 1

`I=1/30(3tan^5(2x)+5tan^3(2x))+C`

Explanation:

We want to solve

`I=inttan^2(2x)sec^4(2x)dx`

Rewrite the integrand using the trig identity

`color(blue)(sec^2(a)=1+tan^2(a)`

Thus

`I=inttan^2(2x)(1+tan^2(2x))sec^2(2x)dx`

Make a substitution `u=tan(2x)=>du=2sec^2(2x)dx`

`I=1/2intu^2(1+u^2)du`

`color(white)(I)=1/2intu^4+u^2du`

`color(white)(I)=1/2(1/5u^5+1/3u^3)+C`

`color(white)(I)=1/30(3u^5+5u^3)+C`

Substitute back `u=tan(2x)`

`I=1/30(3tan^5(2x)+5tan^3(2x))+C`

Answer 2

`inttan^2(2x)sec^4(2x)dx=tan^3(2x)/6+tan^5(2x)/10+C`

Explanation:

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

= `inttan^2(2x)(1+tan^2(2x))^2dx`

if `u=tan(2x)` then `du=2sec^2(2x)dx=2(1+u^2)dx`

or `dx=(du)/(2(1+u^2))`

and `inttan^2(2x)(1+tan^2(2x))^2dx`

= `intu^2(1+u^2)^2(du)/(2(1+u^2))`

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

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

= `u^3/6+u^5/10+C`

= `tan^3(2x)/6+tan^5(2x)/10+C`