How do you find #int x*secx^2dx#?

1 Answer

#int (x* sec x^2) dx=1/2*ln (sec x^2+tan x^2)+C#

Explanation:

What we have here as given is

#int (x* sec x^2) dx#

take note: differential of #x^2# is #d(x^2)=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#

from the formula

#int sec u* du=ln (sec u+tan u)+C#

#color(red)(int (x* sec x^2) dx=1/2int ( sec x^2)*(2x) dx=#
#color(red)(1/2*ln (sec x^2+tan x^2)+C)#

Let us do a little checking by differentiation

from our answer

#1/2*ln (sec x^2+tan x^2)+C#

Let us perform differentiation

#d/dx(1/2*ln (sec x^2+tan x^2)+C)#

#d/dx(1/2*ln (sec x^2+tan x^2))+d/dx(C)#

#1/2*(1/(sec x^2+tan x^2))*d/dx(sec x^2+tan x^2)+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))+0#

#1/2*(1/(sec x^2+tan x^2))*(sec x^2*tan x^2*2x+sec^2 x^2*2x)#

factoring the common monomial factor #2x*sec x^2#

#1/2*((2x*sec x^2)/(sec x^2+tan x^2))*(sec x^2+tan x^2)#

#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.