Question

πsín 2 x dr3.sin2 xcos2 x

Answer 1

The function is given as,

y =  ∫ 0 π 2 sin 3 2 x sin 3 2 x+ cos 3 2 x dx (1)

From the property of integration,

∫ 0 b f( x ) dx= ∫ 0 b f( b−x ) dx

y =  ∫ 0 π 2 sin 3 2 ( π 2 −x ) sin 3 2 ( π 2 −x )+ cos 3 2 ( π 2 −x ) dx y= ∫ 0 π 2 cos 3 2 x cos 3 2 x+ sin 3 2 x dx (2)

We have to add equations (1) and (2) to get the solution.

y+y= ∫ 0 π 2 sin 3 2 x sin 3 2 x+ cos 3 2 x dx + ∫ 0 π 2 cos 3 2 x cos 3 2 x+ sin 3 2 x dx 2y= ∫ 0 π 2 sin 3 2 x+ cos 3 2 x sin 3 2 x+ cos 3 2 x dx 2y= ∫ 0 π 2 1⋅dx 2y= π 2

Simplify further,

y= π 4

Thus, the value of the integral is π 4 .