Evaluate the following integrals:

∫ sin7 x dx

∫ sin7 x dx =


=


= { since sin2x + cos2x = 1}


We know (a-b)3 = a3 - b3 - 3a2b + 3ab2


Here, a = 1 and b = cos2 x


Hence,


= ) {take sin xdx inside brackets)


= (separate the integrals)


Put cosx = t and -sinx dx = dt


=


=


=


Put back t = cos x


=


8