Question

How to solve this problem??

`intln(sec^(-1)x)/(xsqrt(x^2-1))dx=?`

Answer 1

`I=sec^-1x*[ln(sec^-1x)-1]+c`

Explanation:

`I=int(ln(sec^-1x))/(xsqrt(x^2-1))=intln(sec^-1x)*1/(xsqrt(x^2-1))dx`
Let, `color(blue)(sec^-1x=t=>1/(xsqrt(x^2-1))dx=dt)`
`:.I=intln(t)*dt=int1*ln(t)dt`
Using, `color(red)( int(u*v)dt=u*intvdt-int(u^'*intvdt)dt)`
Take, `u=ln(t)andv=1=>u^'=(1/t)andintvdt=t`
`:.I=ln(t)*t-int(1/t*t)dt=t*ln(t)-int1dt`
`=>I=t*ln(t)-t+c`
`:.I=t*[ln(t)-1]+c`
`:.I=sec^-1x*[ln(sec^-1x)-1]+c`

Answer 2

`int ln(sec^(-1)x)/(xsqrt(x^2-1))dx = arcsec x (ln arcsec x- 1)+ C`

Explanation:

Restricting the integrand to the interval `x in (1,+oo)` substitute:

`x= sect`
`dx = sect tant dt`

so:

`int ln(sec^(-1)x)/(xsqrt(x^2-1))dx = int ln(sec^(-1)(sec t))/(sectsqrt(sec^2t-1)) sect tant dt`

`int ln(sec^(-1)x)/(xsqrt(x^2-1))dx = int lnt/(sectsqrt(sec^2t-1)) sect tant dt`

Now use the trigonometric identity:

`sec^2t -1 = tan^2t`

and consider that for `x in (1,+oo)` `t in (0,pi/2)` and `tan t` is positive:

`int ln(sec^(-1)x)/(xsqrt(x^2-1))dx = int lnt/(sectsqrt(tan^2t)) sect tant dt`

`int ln(sec^(-1)x)/(xsqrt(x^2-1))dx = int lnt/(sect tant) sect tant dt`

`int ln(sec^(-1)x)/(xsqrt(x^2-1))dx = int lntdt`

Integrating by parts:

`int ln(sec^(-1)x)/(xsqrt(x^2-1))dx = t lnt - int t dt/t`

`int ln(sec^(-1)x)/(xsqrt(x^2-1))dx = t (lnt - 1)+ C`

and undoing the substitution:

`int ln(sec^(-1)x)/(xsqrt(x^2-1))dx = arcsec x (ln arcsec x- 1)+ C`

Answer 3

The answer is `=arcsecx(ln(arcsecx)-1)+C`

Explanation:

Perform the substitution

`u=arcsecx`,

`secu=x`

`sec^2u=1+tan^2u`

`tan^2u=x^2-1`

`tanu=sqrt(x^2-1)`

`(du)/dxsecutanu=1`

`du=dx/(xsqrt(x^2-1))`

Therefore, the integral is

`int(ln(arcsecx)dx)/(xsqrt(x^2-1))=intlnudu`

Perform the integration by parts

`intp'q=pq-intpq'`

`p'=1`, `=>`, `p=u`

`q=lnu`, `=>`, `q'=1/u`

Therefore,

`int(ln(arcsecx)dx)/(xsqrt(x^2-1))=u lnu-int1/u*udu`

`=u lnu -u`

`=arcsecxln(arcsecx)-arcsecx+C`

`=arcsecx(ln(arcsecx)-1)+C`