How do you integrate #int sec^4(5x)#?

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Answer 1 · 2017-01-14T20:04:47

#tan(5x)/5+tan^3(5x)/15+C#

Start first with a simple substitution to simplify the argument of the secant functions: let #u=5x# which implies that #du=5dx#. Then:

#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 #d/dxsecx=secxtanx#, #d/dxtanx=sec^2x#, and #1+tan^2x=sec^2x# in mind.

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—#1+tan^2(u)#—paired with the derivative of tangent—#sec^2(u)#.

So, let #v=tan(u)# so that #dv=sec^2(u)du#. This gives us:

#=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 #x# using #v=tan(u)# and #u=5x# then adding the constant of integration:

#=tan(u)/5+tan^3(u)/15=tan(5x)/5+tan^3(5x)/15+C#