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

2 Answers
Apr 8, 2018

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

Apr 8, 2018

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