Satyajit Samal
Last Activity: 10 Years ago
pi/2 to -pi/2 ∫ sqrt (cosx – cos 3x) dx
= – [-pi/2 to pi/2 ∫ sqrt (cosx – cos 3x) dx]
Since the function sqrt (cos x – cos 3x) is an even function, the difinite integral is equal to
= – 2[ 0 to pi/2 ∫ sqrt (cosx – cos 3x) dx]
= – 2[ 0 to pi/2 ∫ sqrt (2sin2x sinx) dx]
= – 2 [ 0 to pi/2 ∫ sqrt (4 sin2x cosx) dx]
= – 2 [ 0 to pi/2 ∫ 2 sinx sqrt(cosx) dx]
= – 4[ 0 to pi/2 ∫ sinx sqrt(cosx) dx]
Put cos x = t , – sin x dx = dt
= 4 [ 1 to 0 ∫ sqrt(t) dt
= – 4 [ 0 to 1 ∫ sqrt(t) dt
= – 4 * 2/3 [t3/2] ( 0 to 1)
= -8/3