How do you integrate int sec^6(3x)sec6(3x)?

1 Answer
Jan 10, 2017

tan(3x)+2/3tan^3(3x)+1/5tan^5(3x)+Ctan(3x)+23tan3(3x)+15tan5(3x)+C

Explanation:

First let u=3xu=3x so that du=3dxdu=3dx. Then we have:

I=intsec^6(3x)dx=1/3intsec^6(3x)3dx=1/3intsec^6(u)duI=sec6(3x)dx=13sec6(3x)3dx=13sec6(u)du

We will now make use of the identity tan^2(theta)+1=sec^2(theta)tan2(θ)+1=sec2(θ).

I=1/3intsec^4(u)sec^2(u)du=1/3int(sec^2(u))^2sec^2(u)duI=13sec4(u)sec2(u)du=13(sec2(u))2sec2(u)du

I=1/3int(1+tan^2(u))^2sec^2(u)duI=13(1+tan2(u))2sec2(u)du

Now we should perform the substitution v=tan(u)v=tan(u). Note that dv=sec^2(u)dudv=sec2(u)du, which is why we left a sec^2(u)sec2(u) floating around in the integrand.

I=int(1+v^2)^2dvI=(1+v2)2dv

Expand (1+v^2)^2=(1+v^2)(1+v^2)(1+v2)2=(1+v2)(1+v2) and then integrate term by term:

I=int(1+2v^2+v^4)dv=v+2/3v^3+1/5v^5I=(1+2v2+v4)dv=v+23v3+15v5

From v=tan(u)v=tan(u):

I=tan(u)+2/3tan^3(u)+1/5tan^5(u)I=tan(u)+23tan3(u)+15tan5(u)

From u=3xu=3x:

I=tan(3x)+2/3tan^3(3x)+1/5tan^5(3x)+CI=tan(3x)+23tan3(3x)+15tan5(3x)+C