Question

Integral trigonometri. Dimohon dengan cara ya.

Answer 1

∫ (π...0) x² . sin x dx

def .....intr
x²........sin x
2x......-cos x
2 .......- sin x
0 .......cos x

= |- x²  cos x + 2x  sin x + 2 cos x |(π...0)

= -(π² cos(π) - 0 cos 0 ) + 2(π-0)(sinπ - sin 0) + 2 {cos π - cos 0}
= -(-π² ) + 2(π)(0-0) + 2(-1 - 1)
= π² + 0 - 4
= π² - 4