Evaluate the following integrals:

∫ cos7 x dx

∫ cos7 x dx =


=


= { since sin2x + cos2x = 1}


We know (a-b)3 = a3b3- 3a2b + 3ab2


Here, a = 1 and b = sin2 x


Hence,


= ) {take cos xdx inside brackets)


= (separate the integrals)


Put sinx = t and cos xdx = dt


=


=


=


Put back t = sin x


=


6