$\int\limits_{-1/2}^{1/2}(\sin^{-1}(3x-4x^3)-\cos^{-1}(4x^3-3x))dx=$

$\int\limits_{-1/2}^{1/2}(\sin^{-1}(3x-4x^3)-\cos^{-1}(4x^3-3x))dx=$

$(A)0\hspace{1cm}(B)\frac{-\pi}{2}\hspace{1cm}(C)\frac{\pi}{2}\hspace{1cm}(D)\frac{7\pi}{2}$

I tried and got the answer but my answer is not matching the options given.Is my method not correct?

$\int\limits_{-1/2}^{1/2}(\sin^{-1}(3x-4x^3)-\cos^{-1}(4x^3-3x))dx=\int\limits_{-1/2}^{1/2}(3\sin^{-1}(x)-3\cos^{-1}(x))dx$
$=3\int\limits_{-1/2}^{1/2}(\sin^{-1}(x)-\cos^{-1}(x))dx=3\int\limits_{-1/2}^{1/2}(\frac{\pi}{2}-2\cos^{-1}(x))dx=\frac{3\pi}{2}-6\int\limits_{-1/2}^{1/2}\cos^{-1}(x)dx$
$=\frac{3\pi}{2}+6\int\limits_{2\pi/3}^{\pi/3}t \sin t dt=\frac{-3\pi}{2}$

• $\arcsin(a+b)\neq\arcsin a+\arcsin b$. Aug 9, 2015 at 14:08
• @Aretino,where have i used that?
– diya
Aug 9, 2015 at 14:10
• The substitution $3x-4x^{3} \to x$ is incorrect Aug 9, 2015 at 14:15
• @diya Sorry, I had misread your work. However your change of variables in the integral is incorrect. Aug 9, 2015 at 14:41
• I think $\arccos(4x^3-3x)$ is not $3\arccos(x)$ in this interval. For the angle $\arccos(x)$ is not far from $90$ degrees, so multiplying by $3$ we get something much bigger than $\arccos(4x^3-3x)$. Aug 9, 2015 at 14:42

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