How do you integrate int sec^6(3x)∫sec6(3x)?
1 Answer
Explanation:
First let
I=intsec^6(3x)dx=1/3intsec^6(3x)3dx=1/3intsec^6(u)duI=∫sec6(3x)dx=13∫sec6(3x)3dx=13∫sec6(u)du
We will now make use of the identity
I=1/3intsec^4(u)sec^2(u)du=1/3int(sec^2(u))^2sec^2(u)duI=13∫sec4(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
I=int(1+v^2)^2dvI=∫(1+v2)2dv
Expand
I=int(1+2v^2+v^4)dv=v+2/3v^3+1/5v^5I=∫(1+2v2+v4)dv=v+23v3+15v5
From
I=tan(u)+2/3tan^3(u)+1/5tan^5(u)I=tan(u)+23tan3(u)+15tan5(u)
From
I=tan(3x)+2/3tan^3(3x)+1/5tan^5(3x)+CI=tan(3x)+23tan3(3x)+15tan5(3x)+C