Question

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

• A. $0$
• B. $-\frac{\pi }{2}$
• C. $\frac{7\pi }{2}$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is B $-\frac{\pi }{2}$

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

$\Rightarrow {\int }_{-1/2}^{1/2}-{\sin }^{-1}(4x^3-3x)-{\cos }^{-1}(4x^3-3x)$

$\Rightarrow -{\int }_{-1/2}^{1/2}{\sin }^{-1}(4x^3-3x)+{\cos }^{-1}(4x^3-3x)$

We know that ${\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}$

$\Rightarrow -{\int }_{-1/2}^{1/2}\frac{\pi }{2}$

$\Rightarrow -\frac{\pi }{2}\left(1\right)=-\frac{\pi }{2}$