How do you integrate tan^3(x) sec^5(x)dx?
1 Answer
Apr 9, 2018
Explanation:
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
I=-int(1-u^2)/u^8du=intu^-6-u^-8du
color(white)(I)=-1/5u^-5+1/7u^-7+C
Substitute back
I=-1/5(cos(x))^-5+1/7(cos(x))^-7+C
color(white)(I)=1/7sec^7(x)-1/5sec^5(x)+C