Question

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

Answer 1

Answer:

0

Step-by-step explanation:

Concept:

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

Now,

$\int\limits^{\frac{\pi}{2}}_0{cos2x}\:dx\\\\=[\frac{sin2x}{2}]^{\frac{\pi}{2}}_0\\\\=\frac{1}{2}[sin2x]^{\frac{\pi}{2}}_0\\\\=\frac{1}{2}[sin2(\frac{\pi}{2})-sin2(0)]\\\\=\frac{1}{2}[sin(\pi)-sin(0)]\\\\=\frac{1}{2}[0-0]\\\\=0$