`int 4arccosx dx` Find the indefinite integral

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Given ,

`y= int 4 cos^(-1)x dx`

By Applying the integration by parts we get,

this solution

so,

let` u=cos^(-1)x=> u'= (cos^(-1)x )'`

as we know `(d/dx)cos^(-1)x =(-1)/(sqrt(1-x^2))`

and `v'=1 =>v =x`

now by Integration by parts ,

`int uv' dx= uv-int u'v dx`

so , now

`int (cos^(-1)x )dx`

...

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Given ,

`y= int 4 cos^(-1)x dx`


By Applying the integration by parts we get,

this solution

so,

let` u=cos^(-1)x=> u'= (cos^(-1)x )'`

as we know `(d/dx)cos^(-1)x =(-1)/(sqrt(1-x^2))`

and `v'=1 =>v =x`

now by Integration by parts ,

`int uv' dx= uv-int u'v dx`

so , now

`int (cos^(-1)x )dx`

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

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

let `1-x^2 = q`

=> `-2x dx= dq`

so ,

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

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

=` (-1/2)int (1/(sqrt(q))) dq = (-1/2)q^((-1/2)+1)/((-1/2)+1) =-sqrt(q) = - sqrt(1-x^2)`

 so ,now

`int (cos^(-1)x )dx`

= `(cos^(-1)x )(x)  -(sqrt(1-x^2))`

and now

`int 4(cos^(-1)x )dx`

=`4int (cos^(-1)x )dx`

=`4(x(cos^(-1)x )  -(sqrt(1-x^2)))+c`

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