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

=