How to solve this problem??

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

3 Answers
Mar 13, 2018

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

Mar 13, 2018

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

Mar 13, 2018

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