Question

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

Answer 1

`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`