How do you integrate #sec(5/2x) tan(5/2x) dx#?

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Answer 1 · 2015-04-21T19:52:59

Personnaly i don't use the #sec(x)# function

#u = 5/2x#
#du = 5/2#

#int1/cos(5/2x)*sin(5/2x)/cos(5/2x)dx = 2/5int1/cos(u)*sin(u)/cos(u)du#

#2/5intsin(u)/cos^2(u)du =-2/5int- sin(u)*cos^-2(u)du#

We have : #u'*u^n#

Remember the formula #intu'u^n = 1/(n+1)*u^(n+1)#

#-2/5int1/(-2+1)*cos^(-2+1)(u) = -2/5int-1/cos(u)du#

#=>2/5[1/cos(5/2x)]+C#