Question

Evaluate $I={\int }_{\pi /6}^{\pi /3}\sin xdx$

• A. $\frac{1-\sqrt{3}}{2}$
• B. $\frac{\sqrt{3}+1}{2}$
• C. $\frac{\sqrt{3}-1}{2\sqrt{3}}$
• D. None of these
  

Answer 1

The correct option is A $\frac{1-\sqrt{3}}{2}$

Given : $I={\int }_{\pi /6}^{\pi /3}\sin xdx$

Integeration of $\sin xdx$ is $-cosxdx+c$

$I=-cosxdx$

Substuting the upper and lower limit values we get,

$I=-cos\frac{\pi }{3}+cos\frac{\pi }{6}$

$I=\frac{-\sqrt{3}}{2}+\frac{1}{2}$

$I=\frac{1-\sqrt{3}}{2}$