Question

Question #9e6e3

Answer 1

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

Explanation:

We want to solve

`I=int1/cos^6(x)dx=intsec^2(x)(sec^2(x))^2dx`

Use the identity `sec^2(x)=1+tan^2(x)`

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

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

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

`=intu^4+2u^2+1du`

`=1/5u^5+2/3u^3+u+C`

Substitute back `u=tan(x)`

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