int sin^{-1}(3x - 4x^3) , dx = | vedclass

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(D) Let $x = \sin 	heta$,then $dx = \cos 	heta \, d	heta$. Substituting this into the integral,we get: $\int \sin^{-1}(3\sin 	heta - 4\sin^3 	het

Vedclass · Accepted Answer

$3[x\sin^{-1}x + \sqrt{1 - x^2}] + c$

Vedclass · Answer

(D) Let $x = \sin 	heta$,then $dx = \cos 	heta \, d	heta$. Substituting this into the integral,we get: $\int \sin^{-1}(3\sin 	heta - 4\sin^3 	heta) \cos 	heta \, d	heta$ Since $\sin(3	heta) = 3\sin 	heta - 4\sin^3 	heta$,the integral becomes: $\int \sin^{-1}(\sin 3	heta) \cos 	heta \, d	heta = \int 3	heta \cos 	heta \, d	heta$ Using integration by parts: $3 \int 	heta \cos 	heta \, d	heta = 3 [	heta \sin 	heta - \int \sin 	heta \, d	heta]$ $= 3 [	heta \sin 	heta + \cos 	heta] + c$ Substituting back $	heta = \sin^{-1}x$ and $\cos 	heta = \sqrt{1 - x^2}$: $= 3 [x \sin^{-1}x + \sqrt{1 - x^2}] + c$.

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

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