Question

${\int }_{\pi /6}^{\pi /3}\frac{{\cos }^{3}x}{{\sin }^{3}x+{\cos }^{3}x}d{x}_{=}$

• A. $\frac{\pi }{3}$
• B. $\frac{-\pi }{2}$
• C. $\frac{\pi }{6}$
• D. $\frac{\pi }{12}$

Answer 1

The correct option is D $\frac{\pi }{12}$

$I={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{{\cos }^{3}x}{{\sin }^{3}x+{\cos }^{3}x}dx$

${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f\left[\right(a-b)-x]dx$

$I={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{{\cos }^{3}x}{{\sin }^{3}x+{\cos }^{3}x}dx={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{{\sin }^{3}x}{{\sin }^{3}x+{\cos }^{3}x}dx=I$

$2I={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}1\cdot dx$

$2I=\frac{\pi }{6}$

$I=\frac{\pi }{12}$