Question

Evaluate the definite integrals: $\int^\frac{\pi}{4}_0 {sin\ 2x} \, dx$

Answer 1

Answer:

$\frac{1}{2}$

Step-by-step explanation:

Concept:

$\int{sinax}\:dx=\frac{-1}{a}cosax+c$

Now,

$\int\limits^{\frac{\pi}{4}}_0{sin2x}\:dx \\\\=[\frac{-cos2x}{2}]^{\frac{\pi}{4}}_0 \\\\=\frac{-1}{2}[cos2x]^{\frac{\pi}{4}}_0 \\\\=\frac{-1}{2}[cos2(\frac{\pi}{4})-cos2(0)] \\\\=\frac{-1}{2}[cos(\frac{\pi}{2})-cos(0)] \\\\=\frac{-1}{2}[0-1] \\\\=\frac{1}{2}$