Value of the definite integral,intlimits_{ - | vedclass

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(B) Let $f(x) = \sin^{-1}(3x - 4x^3) - \cos^{-1}(4x^3 - 3x)$.We know that $\cos^{-1}(y) = \pi - \sin^{-1}(y)$ for $y \in [-1, 1]$.Thus,$f(x) = \sin^{-

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$ - \frac{\pi }{2}$

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(B) Let $f(x) = \sin^{-1}(3x - 4x^3) - \cos^{-1}(4x^3 - 3x)$.We know that $\cos^{-1}(y) = \pi - \sin^{-1}(y)$ for $y \in [-1, 1]$.Thus,$f(x) = \sin^{-1}(3x - 4x^3) - [\pi - \sin^{-1}(4x^3 - 3x)]$.Since $4x^3 - 3x = -(3x - 4x^3)$,we have $\sin^{-1}(4x^3 - 3x) = \sin^{-1}(-(3x - 4x^3)) = -\sin^{-1}(3x - 4x^3)$.Substituting this into the expression for $f(x)$:$f(x) = \sin^{-1}(3x - 4x^3) - \pi + \sin^{-1}(4x^3 - 3x) = \sin^{-1}(3x - 4x^3) - \pi - \sin^{-1}(3x - 4x^3) = -\pi$.Wait,let us re-evaluate using the identity $\sin^{-1}(3x - 4x^3) = 3\sin^{-1}x$ for $x \in [-\frac{1}{2}, \frac{1}{2}]$.And $\cos^{-1}(4x^3 - 3x) = \cos^{-1}(-(3x - 4x^3)) = \pi - \cos^{-1}(3x - 4x^3)$.Using $\sin^{-1}u + \cos^{-1}u = \frac{\pi}{2}$,we have $\cos^{-1}(3x - 4x^3) = \frac{\pi}{2} - \sin^{-1}(3x - 4x^3)$.So,$\cos^{-1}(4x^3 - 3x) = \pi - (\frac{\pi}{2} - \sin^{-1}(3x - 4x^3)) = \frac{\pi}{2} + \sin^{-1}(3x - 4x^3)$.Then $f(x) = \sin^{-1}(3x - 4x^3) - (\frac{\pi}{2} + \sin^{-1}(3x - 4x^3)) = -\frac{\pi}{2}$.Therefore,$I = \int_{-1/2}^{1/2} -\frac{\pi}{2} dx = -\frac{\pi}{2} [x]_{-1/2}^{1/2} = -\frac{\pi}{2} (\frac{1}{2} - (-\frac{1}{2})) = -\frac{\pi}{2}$.

Value of the definite integral,$\int\limits_{ - \frac{1}{2}}^{\frac{1}{2}} {\,(\,\,{{\sin }^{ - 1}}(3x - 4{x^3})\,\, - \,\,{{\cos }^{ - 1}}(4{x^3} - 3x)\,\,)\,dx} \,\,$

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