Question

If the value of definite integral ${\int }_{\frac{-1}{2}}^{\frac{1}{2}}\left(si{n}^{-1}\right(3x-4x^3)-co{s}^{-1}(4x^3-3x\left)\right)dx$ is $\frac{-\pi }{k}$, then the value of 3K is

• A. 2
• B. 4
• C. 6
• D. 8

Answer 1

The correct option is A 2

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

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

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

$=3{[{\sin }^{-1}x-{\cos }^{-1}x+2\sqrt{1-x^2}]}_{\frac{-1}{2}}^{\frac{1}{2}}$

$=3\left(\frac{-\pi }{2}\right)$

$=\frac{-\pi }{\frac{2}{3}}$

$\therefore k=\frac{2}{3}$

$\Rightarrow 3k=2$