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

1 Answer
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