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
=