How do you find the integral of #arccos(x)x#?

1 Answer
maganbhai P.
Aug 9, 2018

#I=x^2/2arc cosx+1/4arc sinx-x/4sqrt(1-x^2)+C#

Explanation:

Here ,

#I=intarc cosx*xdx#

Using Integration by parts:

#I=arc cosx intxdx-int(d/(dx)(arc cosx)intxdx)dx#

#I=arc cosx(x^2/2)-int(-1)/sqrt(1-x^2)xxx^2/2dx#

#I=x^2/2arc cosx+1/2intx^2/sqrt(1-x^2)dx#

#color(red)(I=x^2/2arc cosx+1/2I_1............to(A)#

where, # I_1=intx^2/sqrt(1-x^2)dx#

Subst. #color(blue)(x=sinu=>dx=cosudu#

So,

#I_1=intsin^2u/sqrt(1-sin^2u)cosudu#

#I_1=intsin^2u/cosucosudu#

#I_1=intsin^2udu#

#I_1=int(1-cos2u)/2du#

#I_1=1/2[u-(sin2u)/2]+c#

#I_1=1/2[u-sinucosu]+c#

#I_1=1/2[u-sinusqrt(1-sin^2u)]+c#

Subst. back #color(blue)(sinu=x and u=arcsinx#

#I_1=1/2[arc sinx-xsqrt(1-x^2)]+c#

Subst. value of #I_1# into #color(red)((A)# we get

#I=x^2/2arc cosx+1/2{1/2[arc sinx-xsqrt(1-x^2)]}+C#

#I=x^2/2arc cosx+1/4arc sinx-x/4sqrt(1-x^2)+C#

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