7.2 definite integral — ગણિત ધોરણ 12 સાયન્સ — Question

Since $\sin \theta $ is positive in interval $(0,\pi )$

$\therefore I = \int_0^\pi {{{\sin }^3}\theta \,d\theta = \int_0^\pi {\sin \theta (1 - {{\cos }^2}\theta )\,\,d\theta } } $

$ = \int_0^\pi {\sin \theta \,d\theta + \int_0^\pi {( - \sin \theta )\,{{\cos }^2}\theta \,d\theta } } $

$ = [ - \cos \theta ]_0^\pi + \left( {\frac{{{{\cos }^3}\theta }}{3}} \right)_0^\pi = \frac{4}{3}$.