Question

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

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

Answer 1

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

Let $I={\int }_{\pi /6}^{\pi /3}\frac{{\sin }^{3}x}{{\sin }^{3}x+{\cos }^{3}x}dx$.....(1)

Now using ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$

$I={\int }_{\pi /6}^{\pi /3}\frac{{\sin }^3(\frac{\pi }{2}-x)}{{\sin }^3(\frac{\pi }{2}-x)+{\cos }^3(\frac{\pi }{2}-x)}dx={\int }_{\pi /6}^{\pi /3}\frac{{\cos }^{3}x}{{\cos }^{3}x+{\sin }^{3}x}dx$......(2)

Now add (1) and (2)

$2I={\int }_{\pi /6}^{\pi /6}dx=[x{]}_{\pi /6}^{\pi /3}=\frac{\pi }{3}-\frac{\pi }{6}=\frac{\pi }{6}$

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