Question

If ${\int }_{-1}^1\frac{x}{\sqrt{1-x^2}}si{n}^{-1}\left(2x\sqrt{1-x^2}\right)dx=k(\sqrt{2}-1),$ then find the value of $k$.

• A. $2$
• B. $4$
• C. $\sqrt{2}-1$
• D. $\sqrt{2}+1$

Answer 1

The correct option is B $4$

${\int }_{-1}^1\frac{x}{\sqrt{1-x^2}}si{n}^{-1}\left(2x\sqrt{1-x^2}\right)dx=k(\sqrt{2}-1)$
Let $I={\int }_{-1}^1\frac{x}{\sqrt{1-x^2}}si{n}^{-1}\left(2x\sqrt{1-x^2}\right)dx$

Substitute $x=\sin \theta$
$dx=\cos \theta d\theta$
$\Rightarrow I={\int }_0^{\pi /2}\sin \theta {\sin }^{-1}\left(\sin 2\theta \right)d\theta$

$={\int }_0^{\pi /4}2\theta \sin \theta d\theta +{\int }_{\pi /4}^{\pi /2}(\pi -2\theta )\sin \theta d\theta$
$\Rightarrow I=4{\int }_0^{\pi /4}\theta \sin \theta d\theta +2\pi \int {\pi /4}^{\pi /2}\sin \theta d\theta -4\int {\pi /4}^{\pi /2}\theta \sin \theta d\theta$
$I=4(\sqrt{2}-1)$
$\Rightarrow k=4$