Question
Evaluate the following integrals:$\int\sin^{-1}(3\text{x}-4\text{x}^3)\text{dx}$

Answer

Let $\text{I}=\int\sin^{-1}(3\text{x}-4\text{x}^3)\text{dx}$
Let $\text{x}=\sin\theta$
$\text{dx}=\cos\theta \text{d}\theta$
$=\int\sin^{-1}(3\sin\theta-4\sin^3\theta)\cos\theta\text{d}\theta$
$=\int\sin^{-1}(\sin3\theta)\cos\theta\text{d}\theta$
$=\int3\theta\cos\theta\text{d}\theta$
$=3[\theta\int\cos\theta\text{d}\theta-\int(1\int\cos\theta\text{d}\theta)\text{d}\theta]$
$=3[\theta\sin\theta-\int\sin\theta\text{d}\theta]$
$=3[\theta\sin\theta+\cos\theta]+\text{C}$
$\text{I}=3\Big[\text{x}\sin^{-1}\text{x}+\sqrt{1-\text{x}^2}\Big]+\text{C}$

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