Question

Evaluate the definite integral: $∫^{\frac{\pi}{4}}_0 sin2xdx$

Answer 1

Let I=$∫^{\frac{\pi}{4}}_0 sin2xdx$

$∫sin2xdx=\frac{-cos2x}{2}=F(x)$

By second fundamental theorem of calculus,we obtain

$I=F(\frac{π}{4})-F(0)$

$=-\frac{1}{2}[cos2(\frac{π}{4})-cos0]$

$=-\frac{1}{2}[cos(\frac{π}{2})-cos0]$

$=-\frac{1}{2}[0-1]$

$=\frac{1}{2}$