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

1 Answer
Apr 9, 2018

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

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 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