Question

How do you integrate `int 4arccosx` by parts?

Answer 1

`int4arccosxdx=4xarccosx-4(1-x^2)^(1/2)+C`

Explanation:

IBP formula

`I=intu(dv)/(dx)dx=uv-intv(du)/(dx)dx`

we have

`I=int4arccosxdx`

`I=int4cos^(-1)xdx=4int(1xxcos^(-1)x)dx`#

the success of using IBP is the correct identification of the `u" "`&`" "(dv)/(dx)" "`

in this case

`u=cos^(-1)x=>(du)/(dx)=-1/sqrt(1-x^2`

`(dv)/(dx)=1=>v=x`

`I=intu(dv)/(dx)dx=uv-intv(du)/(dx)dx`

becomes

`I=4[xcos^(-1)x-int(-x/sqrt(1-x^2)dx]`

now the second integral can be done by inspection

`int(-x/sqrt(1-x^2))dx=int (-x(1-x^2)^(-1/2))dx`

the outside is a multiple of the bracket differentiated

so we can guess the integral as the bracket to the power +1

ie`" "(1-x^2)^(1/2)`

try this out

`d/(dx)((1-x^2)^(1/2))`

`=1/2xx(-2x)(1-x^2)^(-1/2)`

`=-x(1-x^2)^(-1/2)`

which is the integral we want

so the final integral is;

`int4arccosxdx=4xarccosx-4(1-x^2)^(1/2)+C`