How do you integrate int xarcsecx∫xarcsecx by parts from [2,4][2,4]?

1 Answer
Narad T.
Dec 9, 2016

The answer is =7.39=7.39

Explanation:

If y=arcsecxy=arcsecx

secy=xsecy=x

1/cosy=x1cosy=x

cosy=1/x=x^(-1)cosy=1x=x−1

siny=sqrt(x^2-1)/xsiny=√x2−1x

(-siny)dy/dx=-1/x^2(−siny)dydx=−1x2

dy/dx=1/(x^2siny)dydx=1x2siny

=1/(x^2*sqrt(x^2-1)/x)=1/(xsqrt(x^2-1))=1x2⋅√x2−1x=1x√x2−1

Let's do the integration by parts

intuv'=uv-intu'v

u=arcsecx, u'=1/(xsqrt(x^2-1))

v'=x, v=x^2/2

intxarcsecxdx=x^2/2arcsecx-int(x^2dx)/(2xsqrt(x^2-1))

int(x^2dx)/(2xsqrt(x^2-1))=1/2int(xdx)/sqrt(x^2-1)

Let u=x^2-1, du=2xdx, xdx=(du)/2

1/2int(xdx)/sqrt(x^2-1)=1/4int(du)/(sqrtu)

=1/4*sqrtu/(1/2)

=1/2sqrtu=1/2*sqrt(x^2-1)

So,

intxarcsecxdx=x^2/2arcsecx-1/2*sqrt(x^2-1)

int_2^4xarcsecxdx= [x^2/2arcsecx]_2^4 - [1/2*sqrt(x^2-1)]_2^4

=16/2arcsec4-4/2arcsec2-(1/2*sqrt15-1/2sqrt3)

=8*1.32-2*1.05-1/2(sqrt15-sqrt3)

=10.56-2.1-1.07

=7.39

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