How do you integrate #tan^3(x) sec^5(x)dx#?

Question

24631 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-09T16:23:12

#I=1/7sec^7(x)-1/5sec^5(x)+C#

We want to solve

#I=inttan^3(x)sec^5(x)dx#

Rewrite the integrand in terms of sine and cosine

#I=intsin^3(x)/cos^8(x)dx#

#color(white)(I)=int(sin(x)(1-cos^2(x)))/cos^8(x)dx#

Now a substitution is clear #u=cos(x)=>du=-sin(x)dx#

#I=-int(1-u^2)/u^8du=intu^-6-u^-8du#

#color(white)(I)=-1/5u^-5+1/7u^-7+C#

Substitute back #u=cos(x)#

#I=-1/5(cos(x))^-5+1/7(cos(x))^-7+C#

#color(white)(I)=1/7sec^7(x)-1/5sec^5(x)+C#