Question

11, sin2x dx

Answer 1

The integral is given as,

y= ∫ −π 2 π 2 sin 2 x dx

We have to calculate the integral of y.

Consider g( x )= sin 2 x,

g( x )= sin 2 x g( −x )= sin 2 ( −x ) g( −x )= ( −sinx ) 2 g( −x )= sin 2 x

It can be observed that g( x )=g( −x ). So, sin 2 x is an even function.

Use the property of even function ∫ −b b g( x )dx =2 ∫ 0 b g( x )dx to solve the integral.

y=2 ∫ 0 π 2 sin 2 x dx =2 ∫ 0 π 2 ( 1−cos 2x 2 ) dx = ∫ 0 π 2 ( 1−cos2x ) dx

Simplify further,

y= [ x− sin2x 2 ] 0 π 2 = π 2 − sin2( π 2 ) 2 −0+ sin2( 0 ) 2 = π 2

Thus, the value of integral is π 2 .