Question

limit zero to pi by 2 Sin 2x integration​

Answer 1

Solution :

$\displaystyle \mathrm{Now,\:\int_{0}^{\frac{\pi}{2}}sin2x\:dx}$

$\displaystyle \mathrm{=-\frac{1}{2}[cos2x]_{0}^{\frac{\pi}{2}}}$

$\displaystyle \mathrm{=-\frac{1}{2}(cos\pi-cos0)}$

$\displaystyle \mathrm{=-\frac{1}{2}(-1-1)}$

$\displaystyle \mathrm{=-\frac{1}{2}(-2)}$

= 1