int cos^{-1}(2x^2-1) , dx = | vedclass

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Question

(C) Let $I = \int \cos^{-1}(2x^2-1) \, dx$. Substitute $x = \cos 	heta$,then $dx = -\sin 	heta \, d	heta$. The integral becomes $I = \int \cos^{-1}

Vedclass · Accepted Answer

$2(x \cos^{-1} x - \sqrt{1-x^2}) + c$

Vedclass · Answer

(C) Let $I = \int \cos^{-1}(2x^2-1) \, dx$. Substitute $x = \cos 	heta$,then $dx = -\sin 	heta \, d	heta$. The integral becomes $I = \int \cos^{-1}(2 \cos^2 	heta - 1) \cdot (-\sin 	heta) \, d	heta$. Using the identity $\cos 2	heta = 2 \cos^2 	heta - 1$,we get $I = \int \cos^{-1}(\cos 2	heta) \cdot (-\sin 	heta) \, d	heta = \int 2	heta \cdot (-\sin 	heta) \, d	heta = -2 \int 	heta \sin 	heta \, d	heta$. Using integration by parts,$\int u \, dv = uv - \int v \, du$,where $u = 	heta$ and $dv = \sin 	heta \, d	heta$: $I = -2 [	heta(-\cos 	heta) - \int (-\cos 	heta) \, d	heta] = -2 [-	heta \cos 	heta + \sin 	heta] + c = 2	heta \cos 	heta - 2 \sin 	heta + c$. Since $x = \cos 	heta$,then $	heta = \cos^{-1} x$ and $\sin 	heta = \sqrt{1 - \cos^2 	heta} = \sqrt{1-x^2}$. Substituting these back,$I = 2x \cos^{-1} x - 2\sqrt{1-x^2} + c = 2(x \cos^{-1} x - \sqrt{1-x^2}) + c$.

$\int \cos^{-1}(2x^2-1) \, dx =$

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