<s>[INST] $\int\limits_{-1/2}^{1/2}(\sin^{-1}(3x-4x^3)-\cos^{-1}(4x^3-3x))dx=$ $\int\limits_{-1/2}^{1/2}(\sin^{-1}(3x-4x^3)-\cos^{-1}(4x^3-3x))dx=$
$(A)0\hspace{1cm}(B)\frac{-\pi}{2}\hspace{1cm}(C)\frac{\pi}{2}\hspace{1cm}(D)\frac{7\pi}{2}$
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$$=3\int\limits_{-1/2}^{1/2}(\sin^{-1}(x)-\cos^{-1}(x))dx=3\int\limits_{-1/2}^{1/2}(\frac{\pi}{2}-2\cos^{-1}(x))dx=\frac{3\pi}{2}-6\int\limits_{-1/2}^{1/2}\cos^{-1}(x)dx$
$=\frac{3\pi}{2}+6\int\limits_{2\pi/3}^{\pi/3}t \sin t dt=\frac{-3\pi}{2}$
 [/INST] We use the facts that
$$\arccos(-x)=\pi-\arccos(x)$$
and
$$\arccos(x)+\arcsin(x)=\pi/2.$$
We have
\begin{align}
& \int_{-1/2}^{1/2}\left(\arcsin\left(3x-4x^3\right)-\arccos\left(4x^3-3x\right)\right)\mathrm{d}x \\
& \quad = \int_{-1/2}^{1/2}\left(\arcsin\left(3x-4x^3\right)-\pi+\arccos\left(3x-4x^3\right)\right)\mathrm{d}x \\
& \quad = \int_{-1/2}^{1/2}\left(-\pi+\pi/2\right)\mathrm{d}x=-\pi/2.
\end{align}
</s>