Question

If $u_n={\int }_0^1{x}^n{\tan }^{-1}xdx$ then find $(n+1)u_n+(n-1)u_{n-2}$

Answer 1

$u_n={\int }_0^1{x}^n{\tan }^{-1}xdx$

$u_n=ta{n}^{-1}x{\int }_0^1{x}^n-{\int }_0^1\frac{dta{n}^{-1}x}{dx}(\int x^{n}dx)dx$

$u_n=ta{n}^{-1}x{\left[\frac{x^{n+1}}{n+1}\right]}_0^1-{\int }_0^1\frac{dta{n}^{-1}x}{dx}\left(\frac{x^{n+1}}{n+1}\right)dx$

$(n+1)u_n=ta{n}^{-1}x\left(x^{n+1}\right)-{\int }_0^1\frac{dta{n}^{-1}x}{dx}\left(x^{n+1}\right)dx$

So,

$(n-1)u_{n-2}=ta{n}^{-1}x\left(x^{n-1}\right)-{\int }_0^1\frac{dta{n}^{-1}x}{dx}\left(x^{n-1}\right)dx$

$(n+1)u_n+(n-1)u_{n-2}=ta{n}^{-1}x\left(x^{n+1}\right){]}_0^1-{\int }_0^1\frac{dta{n}^{-1}x}{dx}\left(x^{n+1}\right)dx+ta{n}^{-1}x\left(x^{n-1}\right){]}_0^1-{}^{}{\int }_0^1\frac{dta{n}^{-1}x}{dx}\left(x^{n-1}\right)dx$

$\Rightarrow \frac{\pi }{4}\left(1^{n+1}\right)-{\int }_0^1\frac{dta{n}^{-1}x}{dx}\left[\right(x^{n+1})+(x^{n-1}\left)dx\right]+\frac{\pi }{4}\left(1^{n-1}\right)$

$\Rightarrow \frac{\pi }{2}-{\int }_0^1\frac{1}{1+x^2}{x}^{n-1}(1+x^2)dx$

$\Rightarrow \frac{\pi }{2}-{\int }_0^1{x}^{n-1}dx$

$\Rightarrow \frac{\pi }{2}-\frac{x^n}{n}{|}_0^1$

$\Rightarrow \frac{\pi }{2}-\frac{1}{n}$