નિશ્ચિત સંકલન int_{-1/2}^{1/2} (sin^{-1}(3x - | vedclass

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Question

(B) ધારો કે $I = \int_{-1/2}^{1/2} (\sin^{-1}(3x - 4x^3) - \cos^{-1}(4x^3 - 3x)) dx$.આપણે જાણીએ છીએ કે $x \in [-1/2, 1/2]$ માટે $\sin^{-1}(3x - 4x^3)

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) ધારો કે $I = \int_{-1/2}^{1/2} (\sin^{-1}(3x - 4x^3) - \cos^{-1}(4x^3 - 3x)) dx$.આપણે જાણીએ છીએ કે $x \in [-1/2, 1/2]$ માટે $\sin^{-1}(3x - 4x^3) = 3\sin^{-1}(x)$.વળી,$\cos^{-1}(4x^3 - 3x) = \cos^{-1}(-(3x - 4x^3)) = \pi - \cos^{-1}(3x - 4x^3)$.$\sin^{-1}(	heta) + \cos^{-1}(	heta) = \frac{\pi}{2}$ હોવાથી,$\cos^{-1}(3x - 4x^3) = \frac{\pi}{2} - \sin^{-1}(3x - 4x^3)$.આ કિંમત મૂકતા,$\cos^{-1}(4x^3 - 3x) = \pi - (\frac{\pi}{2} - \sin^{-1}(3x - 4x^3)) = \frac{\pi}{2} + \sin^{-1}(3x - 4x^3)$.હવે,સંકલ્ય પદ: $\sin^{-1}(3x - 4x^3) - (\frac{\pi}{2} + \sin^{-1}(3x - 4x^3)) = -\frac{\pi}{2}$ થાય છે.તેથી,$I = \int_{-1/2}^{1/2} -\frac{\pi}{2} dx = -\frac{\pi}{2} [x]_{-1/2}^{1/2} = -\frac{\pi}{2} (\frac{1}{2} - (-\frac{1}{2})) = -\frac{\pi}{2} (1) = -\frac{\pi}{2}$.

નિશ્ચિત સંકલન $\int_{-1/2}^{1/2} (\sin^{-1}(3x - 4x^3) - \cos^{-1}(4x^3 - 3x)) dx$ ની કિંમત શોધો.

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