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
=