$\int\limits_{-1/2}^{1/2}(\sin^{-1}(3x-4x^3)-\cos^{-1}(4x^3-3x))dx=$ $\int\limits_{-1/2}^{1/2}(\sin^{-1}(3x-4x^3)-\cos^{-1}(4x^3-3x))dx=$
$(A)0\hspace{1cm}(B)\frac{-\pi}{2}\hspace{1cm}(C)\frac{\pi}{2}\hspace{1cm}(D)\frac{7\pi}{2}$
I tried and got the answer but my answer is not matching the options given.Is my method not correct?
$\int\limits_{-1/2}^{1/2}(\sin^{-1}(3x-4x^3)-\cos^{-1}(4x^3-3x))dx=\int\limits_{-1/2}^{1/2}(3\sin^{-1}(x)-3\cos^{-1}(x))dx$$=3\int\limits_{-1/2}^{1/2}(\sin^{-1}(x)-\cos^{-1}(x))dx=3\int\limits_{-1/2}^{1/2}(\frac{\pi}{2}-2\cos^{-1}(x))dx=\frac{3\pi}{2}-6\int\limits_{-1/2}^{1/2}\cos^{-1}(x)dx$
$=\frac{3\pi}{2}+6\int\limits_{2\pi/3}^{\pi/3}t \sin t dt=\frac{-3\pi}{2}$

Hint:
$$\arcsin{a} - \arccos{(-a)} = \arcsin{a}+\arcsin{(-a)} - \frac{\pi}{2} = - \frac{\pi}{2}$$
