Question #9e6e3

Question

1293 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-20T09:37:50

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

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 $u=tan(x)$

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