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#