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

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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,

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

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

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

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