Question

Value of the definite integral ${\int }_{-1/2}^{1/2}\left(si{n}^{-1}\right(3x-4x^3)-co{s}^{-1}(4x^3-3x\left)\right)dx$

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

Answer 1

Answer: (B)

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

$I={\int }_{\frac{-1}{2}}^{\frac{1}{2}}\left({\sin }^{-1}\right(3x-4x^3)-{\cos }^{-1}(4x^3-3x\left)\right)dx$

using definite integral properties:

$I={\int }_{\frac{-1}{2}}^{\frac{1}{2}}(-{\sin }^{-1}(3x-4x^3)-{\cos }^{-1}(3x-4x^3\left)\right)dx$

${\cos }^{-1}(-x)=\pi -{\cos }^{-1}\left(x\right)$

${\sin }^{-1}(-x)=-{\sin }^{-1}\left(x\right)$

$I={\int }_{\frac{-1}{2}}^{\frac{1}{2}}(-{\sin }^{-1}(3x-4x^3)+(-\pi )+{\cos }^{-1}(4x^3-3x\left)\right)$

$2I={\int }_{\frac{-1}{2}}^{\frac{-1}{2}}-\pi$

$2I=-\pi [\frac{1}{2}+\frac{1}{2}]$

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