mohon bantuannya ya teman2

jawab
∫ 2 sin² x  cos x dx | 0...π/2)
= 2/3 (sin x)³ ](π/2 ...0)
= 2/3 [(sin π/2)³ - (sin 0)³]
= 2/3 ( 1³ - 0³)
= 2/3

∫ sin² t cos t dx | 0..π/2)
= 1/3 (sin t)³ ] (π/2 ...0)
= 1/3 { (sin π/2)³ - (sin 0)³}
= 1/3 ( 1 -0)
= 1/3