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

=