How do you evaluate the integral $\int x arcsec (x^{2})$ $int x"arcsec"(x^2)$ ?

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How do you evaluate the integral $\int x arcsec (x^{2})$ $int x"arcsec"(x^2)$ ?

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Answer 1 · 2018-05-20T13:09:20

$\int x \cdot a r c sec (x^{2}) \cdot d x$ $int x*arcsec(x^2)*dx$

= $\frac{1}{2} x^{2} \cdot a r c sec (x^{2}) - \frac{1}{2} ln (x^{2} + \sqrt{x^{4} - 1}) + C$ $1/2x^2*arcsec(x^2)-1/2ln(x^2+sqrt(x^4-1))+C$

Explanation:

$\int x \cdot a r c sec (x^{2}) \cdot d x$ $int x*arcsec(x^2)*dx$

= $\frac{1}{2} \int 2 x \cdot a r c sec (x^{2}) \cdot d x$ $1/2int 2x*arcsec(x^2)*dx$

After using $y = x^{2}$ $y=x^2$ and $2 x \cdot d x = d y$ $2x*dx=dy$ transforms, this integral became

$\frac{1}{2} \int a r c sec y \cdot d y$ $1/2int arcsecy*dy$

After using $z = a r c sec y$ $z=arcsecy$ , $y = sec z$ $y=secz$ and $d y = sec z \cdot tan z \cdot d z$ $dy=secz*tanz*dz$ transforms, it became

$\frac{1}{2} \int z \cdot sec z \cdot tan z \cdot d z$ $1/2int z*secz*tanz*dz$

= $\frac{1}{2} z \cdot sec z - \frac{1}{2} \int sec z \cdot d z$ $1/2z*secz-1/2int secz*dz$

= $\frac{1}{2} z \cdot sec z - \frac{1}{2} \int \frac{sec z \cdot (sec z + tan z) \cdot d z}{sec z + tan z}$ $1/2z*secz-1/2int (secz*(secz+tanz)*dz)/(secz+tanz)$

= $\frac{1}{2} z \cdot sec z - \frac{1}{2} ln (sec z + tan z) + C$ $1/2z*secz-1/2ln(secz+tanz)+C$

For $y = sec z$ $y=secz$ , $tan z$ $tanz$ must be equal to $\sqrt{y^{2} - 1}$ $sqrt(y^2-1)$ . Thus,

$\frac{1}{2} y \cdot a r c sec y - \frac{1}{2} ln (y + \sqrt{y^{2} - 1}) + C$ $1/2y*arcsecy-1/2ln(y+sqrt(y^2-1))+C$

= $\frac{1}{2} x^{2} \cdot a r c sec (x^{2}) - \frac{1}{2} ln (x^{2} + \sqrt{x^{4} - 1}) + C$ $1/2x^2*arcsec(x^2)-1/2ln(x^2+sqrt(x^4-1))+C$

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How do you evaluate the integral $\int x arcsec (x^{2})$ $int x"arcsec"(x^2)$ ?

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