7.2 definite integral — ગણિત ધોરણ 12 સાયન્સ — Question

$I = \int_{ - \pi }^\pi {({{\cos }^2}ax + {{\sin }^2}bx - 2\cos \,ax\,\,\sin bx)\,\,dx} $

$I = \int_{ - \pi }^\pi {({{\cos }^2}ax + {{\sin }^2}bx)\,\,dx} - \int_{ - \pi }^\pi {2\cos ax\sin bx\,\,dx} $

$I = 2\int_0^\pi {({{\cos }^2}ax + {{\sin }^2}bx)\,\,dx} - 0$

$I = 2\int_0^\pi {\left( {\frac{{1 + \cos 2ax}}{2} + \frac{{1 - \cos 2bx}}{2}} \right)\,dx} $

$I = \int_0^{\pi \,} {\left( {2 + \cos 2ax - \cos 2bx} \right)\,dx} = 2\pi .$