Question

The value of the integral $\underset{0}{\overset{1}{\int }}xco{t}^{-1}(1-x^2+x^4)dx$ is equal to $\frac{\pi }{4}-\frac{k}{2}ln2$,(natural $logln$) then $k$is

Answer 1

Answer: 1

$\underset{0}{\overset{1}{\int }}xco{t}^{-1}(1-x^2+x^4)dx=\underset{0}{\overset{1}{\int }}xta{n}^{-1}(\frac{1}{1+x^4-x^2})$
$=\underset{0}{\overset{1}{\int }}xta{n}^{-1}(\frac{x^2-(x^2-1)}{1+x^2(x^2-1)})dx$
$=\frac{1}{2}\underset{0}{\overset{1}{\int }}1ta{n}^{-1}{t}^{2}dx-\frac{1}{2}\underset{-1}{\overset{0}{\int }}1ta{n}^{-1}kdx$
Put $x^2=t$
$\Rightarrow 2xdx=dt$ in the first integral
and $x^2-1=k$
$\Rightarrow 2xdx=dk$ in the second integral.
$=\frac{1}{2}\underset{0}{\overset{1}{\int }}1ta{n}^{-1}tdt-\frac{1}{2}\underset{0}{\overset{1}{\int }}tta{n}^{-1}kdk$
$=\frac{1}{2}\left(tta{n}^{-1}t{|}_0^1-\underset{0}{\overset{1}{\int }}\frac{t}{1+t^2}dt\right)$
$-\frac{1}{2}(kta{n}^{-1}k{|}_0^1-\underset{-1}{\overset{0}{\int }}\frac{k}{1+k^2}dk)$
$=\frac{1}{2}\left(\frac{\pi }{4}-\left(\frac{1}{2}ln\left(1+t^2\right){|}_0^1\right)\right)$
$-\frac{1}{2}\left(-\frac{\pi }{4}-\left(\frac{1}{2}ln\left(1+k^2\right){|}_{-1}^0\right)\right)$
$=\left(\frac{\pi }{8}-\frac{1}{4}ln2\right)-\left(\frac{-\pi }{8}-\frac{1}{4}10-ln2\right)=\frac{\pi }{4}-\frac{1}{2}ln2$
$\Rightarrow \frac{\pi }{4}-\frac{1}{2}In2=\frac{\pi }{4}-\frac{k}{2}In2$
$\Rightarrow k=1$