Question

Solve:$I=2{\int }_0^{1/2}{\sin }^{-1}(3x-4x^3)dx$

Answer 1

$I=2{\int }_0^{1/2}{\sin }^{-1}(3x-{4x}^2)dx$

Let $x=\sin \theta$

$\Rightarrow dx=\cos \theta d\theta$

when $x=0,\theta =0$

$x=1/2,\theta =\pi /6$

$=2{\int }_0^{\pi /6}{\sin }^{-1}(3\sin \theta -4{\sin }^3\theta )\cos \theta d\theta$

$=2{\int }_0^{\pi /6}{\sin }^{-1}\left(\sin \left(3\theta \right)\right)\cos \theta d\theta$

$=2{\int }_0^{\pi /6}\left(3\theta \right)\left(\cos \theta \right)d\theta$

$=6{\int }_0^{\pi /6}\theta \cos \theta d\theta =6\{{\left[\theta \sin \theta \right]}_0^{\pi /6}-{\int }_0^{\pi /6}\sin \theta d\theta \}$

$=6\{\frac{\pi }{12}+{\left[\cos \theta \right]}_0^{\pi /6}\}=6\{\frac{\pi }{12}+(\frac{\sqrt{3}}{2}-1)\}$

$=\frac{\pi }{2}-6+3\sqrt{3}$