Evaluate the definite integral ?sin2xdx

Question

Evaluate the definite from -π/4 to π/4 ∫sin2xdx

 

 

-π/4 to π/4= -pi/4 and pi/4

Answer 1

because the function is odd (symmetric with respect to the origin), the integral is zero whenever the integration is from -a to +a as is the case here. So there is no need to actually do the integration step.

Answer 2

Integral sin2x dx

=(-1/2)(cos 2x).        take limits from -pi/4. to pi/4

=(-1/2)[ cos (pi/2)-cos(-pi/2)]

=-1/2*0

=0

Answer 3

Evaluate ∫sin2xdx from -pi/4 to pi/4

 

Let u = 2x

Hence du = 2 dx  (dx=du/2)

 

∫sin2xdx = (1/2)∫sinudu = -(1/2)cos(2x) = -(1/2)[cos(-2pi/4)-cos(2pi/4)] = 0