Question

Solve $\int {\sin }^{-1}(3x-4x^3)dx$

Answer 1

$\int si{n}^{-1}(3x-4x^3)dx$

put $x=sin\theta$

$\therefore dx=cos\theta .d\theta$

$=\int si{n}^{-1}(3sin\theta -4si{n}^3\theta ).cos\theta .d\theta$

$=\int si{n}^{-1}\left(sin\right(3\theta \left)\right).cos\theta .d\theta$ $[\because 3sinx-4si{n}^{3}x=sin3x]$

$=\int 3\theta .cos\theta .d\theta$

Integration by parts

$=3[\theta \int cos\theta .d\theta -\int (\frac{d\theta }{d\theta }\int cos\theta .d\theta \left)d\theta \right]$

$=3[\theta .sin\theta -\int sin\theta .d\theta ]$

$=3[\theta .sin\theta +cos\theta ]+c$