Question

The value of the integral $\int\limits_0^1 x\,cot^{-1} ( 1 - x^2 + x^4)dx$ is equal to $\frac{\pi}{4} - \frac{k}{2} ln 2$,(natural $log \,ln$) then $k $is

Answer 1

$\int\limits_0^1 x\,cot^{-1} ( 1 - x^2 + x^4)dx = \int\limits_0^1 x\,tan^{-1} (\frac{1}{1 + x^4 - x^2})$
$ = \int\limits_0^1 x\,tan^{-1} ( \frac{x^2 - (x^2 - 1)}{1 + x^2 ( x^2 - 1)})dx$
$=\frac{1}{2} \int\limits_0^1 1\,tan^{-1} t^2 dx - \frac{1}{2} \int\limits_{-1}^0 1\,tan^{-1} k \,dx$
Put $x^2 = t $
$\Rightarrow 2xdx = dt$ in the first integral
and $x^2 - 1 = k$
$\Rightarrow 2xdx = dk$ in the second integral.
$ = \frac{1}{2} \int\limits_0^1 1 \,tan^{-1} t dt-\frac{1}{2} \int\limits_0^1 t\,tan^{-1} k dk$
$ = \frac{1}{2}\left(t \,tan^{-1} t|_{0}^{1} - \int\limits_{0}^{1} \frac{t}{1 + t^{2}} dt\right)$
$ - \frac{1}{2}(k\,tan^{-1}\,k|_{0}^{1} - \int\limits_{-1}^{0} \frac{k}{ 1 + k^2} dk)$
$ = \frac{1}{2}\left( \frac{\pi}{4} - \left(\frac{1}{2} ln\left( 1 + t^{2}\right)|_{0}^{1}\right)\right)$
$-\frac{1}{2}\left(-\frac{\pi}{4} -\left(\frac{1}{2} ln\left( 1 +k^{2}\right)|_{-1}^{0}\right)\right)$
$= \left(\frac{\pi}{8} -\frac{1}{4} ln \,2\right)-\left(\frac{-\pi}{8} - \frac{1}{4} 10 - ln \,2\right) = \frac{\pi}{4}-\frac{1}{2} ln\,2$
$\Rightarrow \frac{\pi}{4} - \frac{1}{2} In\,2 = \frac{\pi}{4} - \frac{k}{2} In\,2$
$\Rightarrow k = 1$