Question

`IfI_n=int_0^1x^n(tan^(-1)x)dx ,t h e np rov et h a t` `(n+1)I_n+(n-1)I_(n-2)=-1/n+pi/2`

Answer 1

Correct Answer - NA
`I_(n)=int_(0)^(1) x^(n)(tan^(-1)x)dx=int_(0)^(1)x^(n-1)(xtan^(-1)x)dx`
`=[x^(n-1)((x^(2))/2"tan"^(-1)x-(x^(2))/2+(tan^(-1)x)/2)]_(0)^(1)`
`=(n-1)int_(0)^(1)x^(n-2)((x^(2))/2"tan"^(-1)x-x/2+(tan^(-1)x)/2)dx`
`=(pi)/4-1/2-((n-1))/2 I_(n)+((n-1))/2int_(0)^(1)x^(n-1)dx-1/2(n-1)I_(n-2)`
or `((n+1))/2I_(n)=(pi)/4-1/2+1/2 1/(2n)-1/2(n-1)I_(n-2)`
or `(n+1)I_(n)+(n-1)I_(n-2)=-1/n+(pi)/2`