What we have here as given is
int (x* sec x^2) dx∫(x⋅secx2)dx
take note: differential of x^2x2 is d(x^2)=2x*dxd(x2)=2x⋅dx
We do a numerical preparation here by placing a 2 inside the integral and 1/2 outside the integral.
int (x* sec x^2) dx=1/2int ( sec x^2)*(2x) dx∫(x⋅secx2)dx=12∫(secx2)⋅(2x)dx
from the formula
int sec u* du=ln (sec u+tan u)+C∫secu⋅du=ln(secu+tanu)+C
color(red)(int (x* sec x^2) dx=1/2int ( sec x^2)*(2x) dx=∫(x⋅secx2)dx=12∫(secx2)⋅(2x)dx=
color(red)(1/2*ln (sec x^2+tan x^2)+C)12⋅ln(secx2+tanx2)+C
Let us do a little checking by differentiation
from our answer
1/2*ln (sec x^2+tan x^2)+C12⋅ln(secx2+tanx2)+C
Let us perform differentiation
d/dx(1/2*ln (sec x^2+tan x^2)+C)ddx(12⋅ln(secx2+tanx2)+C)
d/dx(1/2*ln (sec x^2+tan x^2))+d/dx(C)ddx(12⋅ln(secx2+tanx2))+ddx(C)
1/2*(1/(sec x^2+tan x^2))*d/dx(sec x^2+tan x^2)+012⋅(1secx2+tanx2)⋅ddx(secx2+tanx2)+0
1/2*(1/(sec x^2+tan x^2))*(sec x^2*tan x^2*d/dx(x^2)+sec^2 x^2*d/dx(x^2))+012⋅(1secx2+tanx2)⋅(secx2⋅tanx2⋅ddx(x2)+sec2x2⋅ddx(x2))+0
1/2*(1/(sec x^2+tan x^2))*(sec x^2*tan x^2*2x+sec^2 x^2*2x)12⋅(1secx2+tanx2)⋅(secx2⋅tanx2⋅2x+sec2x2⋅2x)
factoring the common monomial factor 2x*sec x^22x⋅secx2
1/2*((2x*sec x^2)/(sec x^2+tan x^2))*(sec x^2+tan x^2)12⋅(2x⋅secx2secx2+tanx2)⋅(secx2+tanx2)
1/cancel2*((cancel2x*sec x^2)/cancel(sec x^2+tan x^2))*cancel(sec x^2+tan x^2)
x * sec x^2" "which is the given integrand
God bless....I hope the explanation is useful.
