How do you evaluate the integral #int xsec(4x^2+7)#?

Question

1469 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-15T13:46:20

# ln|sec(4x^2+7)+tan(4x^2+7)|+C.#

If we subst. #4x^2+7=y," then, "8xdx=dy.#

Hence,

#intxsec(4x^2+7)dx=1/8int(sec(4x^2+7))8xdx,#

#=1/8intsecydy,#

#=ln|secy+tany|,#

#=ln|sec(4x^2+7)+tan(4x^2+7)|+C.#

Answer 2 · 2017-08-15T13:51:11

The answer is #=1/8ln(tan(4x^2+7)+sec(4x^2+7))+C#

We need

#intsecxdx=ln(tanx+secx)+C#

We perform this integral by substitution

Let

#u=4x^2+7#, #=>#, #du=(8x)dx#

Therefore,

#intxsec(4x^2+7)dx=1/8intsecudu#

#=1/8ln(tanu + secu)#

#=1/8ln(tan(4x^2+7)+sec(4x^2+7))+C#