How do you find the definite integral for: #2(pi)x(cos^(-1)(x))dx# for the intervals #[0, 1]#?

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Answer 1 · 2018-06-09T11:24:34

#I=pi^2/4#

Here,

#I=2piint_0^1 xcos^-1x#

Subst. #cos^-1x=u=>x=cosu=>dx=-sinudu#

#x=0=>cosu=0=>u=pi/2and#

# x=1=>cosu=1=>u=0#

So,

#I=2piint_(pi/2)^0 cosu*u(-sinu)du#

#=2piint_0^(pi/2) usinucosudu.to[becauseint_a^bf(x)dx=-int_b^af(x)dx]#

#=piint_0^(pi/2)usin2udu...to[becausesin2theta=2sinthetacostheta]#

#"Using "color(red)"Integratio by Parts :"#

#color(blue)(intu*vdx=u*intvdx-int(u'intvdx)dx#

Take, #u=u andv=sin2u=>u'=1 and intvdu=-(cos2u)/2#

#I=pi{[u*(-cos2u)/2]_0^(pi/2)-int_0^(pi/2)(-cos2u)/2du}#

#=pi{[pi/2*(-cospi)/2-0]+1/2[(sin2u)/2]_0^(pi/2)}#

#=pi{pi/2(1/2)+1/4[sinpi-sin0]}#

#=pi{pi/4}#

#=pi^2/4#