How do you integrate #int sec^4(5x)#?
1 Answer
Explanation:
Start first with a simple substitution to simplify the argument of the secant functions: let
#intsec^4(5x)dx=1/5intsec^4(5x)(5dx)=1/5intsec^4(u)du#
When working with secant, it's important to keep the identities
In this case, we see that the integral
#=1/5intsec^2(u)sec^2(u)du=1/5int(1+tan^2(u))sec^2(u)du#
The point of doing this is that we have a function of tangent—
So, let
#=1/5int(1+v^2)dv#
We can now integrate term by term:
#=1/5(v+v^3/3)=v/5+v^3/15#
Returning to our original variable
#=tan(u)/5+tan^3(u)/15=tan(5x)/5+tan^3(5x)/15+C#