Value of the definite integral int_{-1/2}^{1/ | vedclass

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(B) Let $I = \int_{-1/2}^{1/2} (\sin^{-1}(3x - 4x^3) - \cos^{-1}(4x^3 - 3x)) dx$.We know that $\sin^{-1}(3x - 4x^3) = 3\sin^{-1}(x)$ for $x \in [-1/2,

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let $I = \int_{-1/2}^{1/2} (\sin^{-1}(3x - 4x^3) - \cos^{-1}(4x^3 - 3x)) dx$.We know that $\sin^{-1}(3x - 4x^3) = 3\sin^{-1}(x)$ for $x \in [-1/2, 1/2]$.Also,$\cos^{-1}(4x^3 - 3x) = \cos^{-1}(-(3x - 4x^3)) = \pi - \cos^{-1}(3x - 4x^3)$.Since $\sin^{-1}(	heta) + \cos^{-1}(	heta) = \frac{\pi}{2}$,we have $\cos^{-1}(3x - 4x^3) = \frac{\pi}{2} - \sin^{-1}(3x - 4x^3)$.Substituting this,$\cos^{-1}(4x^3 - 3x) = \pi - (\frac{\pi}{2} - \sin^{-1}(3x - 4x^3)) = \frac{\pi}{2} + \sin^{-1}(3x - 4x^3)$.Now,the integrand becomes: $\sin^{-1}(3x - 4x^3) - (\frac{\pi}{2} + \sin^{-1}(3x - 4x^3)) = -\frac{\pi}{2}$.Thus,$I = \int_{-1/2}^{1/2} -\frac{\pi}{2} dx = -\frac{\pi}{2} [x]_{-1/2}^{1/2} = -\frac{\pi}{2} (\frac{1}{2} - (-\frac{1}{2})) = -\frac{\pi}{2} (1) = -\frac{\pi}{2}$.

List-$I$	List-$II$
$I. \int_{-1}^1 x\|x\| dx$	$(a) \frac{\pi}{2}$
$II. \int_0^{\pi/2} \left(1 + \log \left(\frac{4+3\sin x}{4+3\cos x}\right)\right) dx$	$(b) \int_0^a 2f(x) dx$
$III. \int_0^a f(x) dx$	$(c) \int_0^a [f(x) + f(-x)] dx$
$IV. \int_{-a}^a f(x) dx$	$(d) 0$
	$(e) \int_0^a f(a-x) dx$

Value of the definite integral $\int_{-1/2}^{1/2} (\sin^{-1}(3x - 4x^3) - \cos^{-1}(4x^3 - 3x)) dx$

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